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Chapter 5: Probability

5.1 What is probability anyway? Probability is a branch of mathematics which intrudes on everyday conversation perhaps more than any other (except for just plain arithmetic and counting). A TV weather report might announce that “there is a 70% chance of rain today.” When you hear that statement you automatically infer that there must be a 30% chance that it will not rain, don’t you? When you do that you’re intuitively using one of the basic properties of probability. Other common expressions that implicitly have to do with probability are statements like, “N.C. State is a 2 to 1 favorite to beat Wake Forest today,” or “I think there’s about a 50-50 chance that I’ll pass English this semester.” Often when we talk about probability in everyday conversation we speak in terms of “percentages” or “odds” or “chances.” When speaking mathematically we usually treat probabilities as fractions (or decimal numbers) rather than percentages, and we usually use the word “probability.” The three statements in the paragraph above are equivalent to these: (1) The probability of rain today is .7. (2) The probability that N.C. State will beat Wake Forest today is 2/3. (3) I think the probability that I will pass English this semester is about 1/2. Many of the examples which are most useful in developing our understanding of probabilities involve simple experiments. With each experiment there are certain outcomes for the experiment. For instance, if a coin is tossed then the possible outcomes are HEADS and TAILS. If a class holds an election to elect a class president, then the possible outcomes are the individual members of the class, any of whom could be elected president. When talking about probabilities, it’s always a good idea to have a firm understanding of just what the possibilities are. In fact, this idea is so important that there is a special name for the set of all possibilities. It’s called the sample space.

Definition: Sample space The sample space for an experiment is the set of all possible outcomes.

Example 5.1. A coin is tossed twice. In this case the sample space could be considered to be the set S = {HH,HT,TH,TT}. The expression TH, for example, would

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represent the outcome in which a “tails” occurs on the first toss and a “heads” occurs on the second toss. Example 5.2. A card is drawn from a standard deck of 52 cards. Here the possible outcomes are the 52 cards in the deck, so the sample space is the set S = {2♣, 2♦,2♥,2♠,3♣,...,K♠,Α♣,A♦,A♥,A♠}. Don’t get the idea that you have to be an artist, however, in order to represent the elements of the sample space. The important concept is that the sample space consists of the 52 cards. Whatever notation you wish to use to represent the 52 cards is your business. Example 5.3. Two people are elected from a class of 35 to represent the class on a school committee. What is the sample space for this experiment? This is a “combinations” question since it is simply a matter of recording which 2 of the 35 are elected. The number of “possibilities” is the number of ways of choosing 2 students from the 35 in the class, and so the number of possibilities is C(35,2). In this example the sample space is obviously too big for us to want to list its elements, or even to have names for all the elements. Nevertheless, it is important to have an understanding for what the sample space is. If we want to verbally describe the sample space we can just say that it is made up of “all the possible pairs of students that could be chosen from the 35 in the class.” From Chapter 4 we know that the number of elements in the sample space is C(35,2) = 595. In ordinary conversation we sometimes say things like, “Do you think that event will occur?” The words “event” and “occur” used in this sentence are used in probability as mathematical terms, but the way in which they are used is consistent with the way they are used in everyday conversation.

Definition: Event An event is a subset of the sample space. To say that an event occurs means that the observed outcome is an element of this subset.

Example 5.4. An ordinary 6-sided die is rolled. The sample space then can be viewed as just the set of numbers S = {1,2,3,4,5,6}. What do we mean when we say something like “an odd number occurs”? First of all it means we’re talking about the event A = {1,3,5} = set of all odd numbers in S. And what we’re saying when we say “an odd number occurs” can be rephrased by saying that “the observed outcome (when the experiment is carried out) belongs to the set A which is the set of odd numbers in the sample space.” You notice that this last way of

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saying it is very long and awkward, and furthermore nobody really talks that way. That’s why we like to be able to say just that “an odd number occurs.” But when we talk this way you need to understand conceptually that we mean that the observed outcome belongs to the set A above. If you are ever confused about what a particular event is, you might try writing out the set of elements that make up the event. Notice also that the event A above is described by listing its elements and also described verbally. In most examples it will be difficult or impossible to list the elements of the sample space, or of events, because there are too many. In these cases we will have to use verbal descriptions of the sample space and of the relevant events in order to convey the necessary information. Example 5.5. Two dice are rolled, one red and one green. (See Example 4.23) The event E = {(6,3),(5,4),(4,5),(3,6)} can easily be described as “the event consisting of all outcomes for which the sum of the numbers appearing on the dice is equal to nine.” Or more briefly we might just say “the event that the sum is nine.” Example 5.6. Let’s suppose that a class of 35 students has 19 males and 16 females. If two students are elected to represent the class on a committee, then there are C(35,2) different ways of choosing which two students will be on the committee. This means we need to think of a sample space of C(35,2) possible outcomes for the election. Now what does it mean to talk about the “event that two women are elected”? This event consists of all possible outcomes in which both people elected come from the 16 women in the class. There are C(16,2) ways that two women can be chosen. So when we talk about “the event that two women are elected,” we are talking about an event consisting of C(16,2) elements that is a subset of the sample space of C(35,2) elements. Sometimes a sample space will need to be based on some kind of existing data. The next example illustrates this. Example 5.7. A television manufacturer has kept data on all televisions it has sold during the past decade to determine how many complete years of service the TV gave before it had to be brought in for repairs. Here is their data: Years 0 1 2 3 4 5 or more

% of televisions 12% 26% 33% 14% 9% 6%

Notice that the very form in which the data is tabulated forces us to view the sample space in a particular way. For the data as presented, the natural sample space is

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S = {0, 1, 2, 3, 4, 5 or more}. The sample space shows six possibilities because that’s how the information was grouped by the company. If they had kept monthly rather than yearly data, then a much more detailed sample space with many more possible outcomes would be possible. Example 5.8. Here is an example to show you that sometimes you have to make a choice in deciding what to consider your sample space to be. In a game like monopoly, you roll a pair of dice but what you’re interested in is not the individual numbers that appear on the two dice but the sum of the two numbers that appear. Should you think of the sample space as all 36 ordered pairs that can be formed from the numbers, 1,...,6, or should you think of the sample space as all possible sums that could appear, in which case we would just think of S as being S = {2,3,4,5,6,7,8,9,10,11,12}? You should understand that the question here is really just the question “how much information should I record when I roll the dice?” The two sample spaces are not in conflict, they just record different amounts of information. For example the outcome “4” in the latter choice for sample space corresponds to all the outcomes (1,3), (2,2) and (3,1) if the numbers on the individual dice are recorded. The best rule is to err on the side of caution and record all the information that you think you might need. It’s better to record too much information than too little. For example, if you chose the sample space that just shows the sum, you would be unable to talk about the event “at least one die shows a five” because your sample space doesn’t catalog that kind of information. In the sample space of 36 ordered pairs, the event “at least one die shows a five” consists of 11 of the 36 pairs. (Count them to convince yourself that this is correct.) Example 5.9. Sue and Ann are playing a tennis match. The winner will be the first player to win two sets. What could we use for the sample space in this situation? Solution: This is the kind of problem for which we used a tree diagram in Chapter 4. A tree diagram for the tennis match looks something like the following one: Sue

Ann Sue

Ann

Sue

Ann

Sue Ann

Sue

Ann

The six possible outcomes for the tennis match correspond to the six “leaves” of the

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tree diagram. So if we like we can think of the sample space as consisting of the six outcomes shown in this tree. Example 5.10. A company is going to build three stores, and there are 10 locations available. What is the sample space corresponding to the possibilities for the sites they select? Solution: Since they will be choosing 3 of the 10 sites, there are C(10,3) different possibilities for the sites that will be selected. This is obviously a sample space too large for us to list all the elements, but from Chapter 4 we know that there are C(10,3) = 120 possible choices, i.e. 120 “possible outcomes” in the sample space.

Problems 1.

The grades that may be obtained in a class are A, B, C, D, and F. (a) If a student takes this class, describe the sample space that would represent the possible grades the student might get in the class. (b) Describe the event “the student makes a passing grade.” (c) Describe the event “the student does not make an A in the class.”

2.

A game show contestant who loses has his choice of 2 out of 5 consolation prizes: a microwave oven, a food processor, a tape recorder, a camera, and a blender. (a) List the 10 elements that comprise the sample space of all possible choices the contestant might make. (b) List the elements of the event “the contestant chooses the camera.” (c) List the elements of the event “the contestant chooses the camera but not the blender.” (d) List the elements of the event “the contestant chooses either the food processor or the tape recorder.” (Remember that “or” allows for the possibility of both.)

3.

A TV station news manager is trying to decide whether to send a reporter to cover a local chemical spill. If she does decide to send someone, it will be Ed, Ann, or Sue. What might a sample space for this situation look like?

4.

A club is going to choose its president and vice-president from among these people: Ralph, Mark, Ruth, and Abraham. Give a sample space for this process.

5.

A coin is to be tossed 3 times. State the sample space. How many different elements are there for this sample space? How many different events are there?

6.

A pair of dice is rolled, one red and one green. Using the usual sample space of 36 possible outcomes, list the elements of each of these events: (a) the event that the sum of numbers on the two dice is 8. (b) the event that the product of the numbers on the two dice is 12.

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(c) the event that the two dice show the same number. (d) the event that the number on the red die is greater than the number on the green one. 7.

Carl and Hubert are going to play a tennis match. To win a match in men’s tennis a player must win 3 sets. Draw a tree diagram to represent all the possibilities for the match (see Example 5.9). How many different possible outcomes does your tree show for the tennis match?

8.

Three days after Halloween, Andy’s bag of candy contains 2 Milky Ways, 1 Snickers, and 1 Almond Joy. He pulls out candy bars one at a time without replacement until he has the Snickers. (a) Draw a tree diagram representing this situation. (b) Write out the elements of the sample space. (c) List the elements of the event E = set of all outcomes in which no more than 1 Milky Way is selected.

9.

A company is monitoring use of its parking lot. They have collected data on the number of cars that enter the lot during a one-hour time interval by observing the parking lot during 50 different one-hour time intervals. The data they have accumulated is shown below. n = number of cars observed frequency 0 ≤ n < 100 5 100 ≤ n < 200 7 200 ≤ n < 300 14 300 ≤ n < 400 12 400 ≤ n < 500 7 n ≥ 500 5 (a) Give an appropriate sample space for this data. (b) Suppose the company wants to “compress” the data into 3 cases: 0 ≤ n < 200, or 200 ≤ n < 400, or n ≥ 400. What sample space would be chosen then?

5.2 Properties of Probability The properties of probabilities are based on a few simple ideas: (1) Each element of the sample space is assigned a probability which is a non-negative number. (2) The sum of all these probabilities must be equal to one. (3) If we want to know the probability of an event, we compute it by simply adding up the probabilities of the outcomes that make up the event. Example 5.11. An ordinary 6-sided die is rolled, so that the sample space is S = {1,2,3,4,5,6}.

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Generally dice are made so that the various numbers have about the same probability of occurring. If all the probabilities are going to be equal, and if they are going to add up to one, then each outcome must be assigned probability 1/6. (A die that has this property is called a fair die.) The event E = {2,4,6} = set of even outcomes is an event in this sample space. The probability of the event E is denoted by P(E) and its probability is computed by adding the probabilities of the outcomes that make up E: P(E) = P(2) + P(4) + P(6) = 1/6 + 1/6 + 1/6 = 1/2. Example 5.12: Suppose on a winter day there are three possibilities for the state of the weather: rain, snow, and fair. We can think then of a sample space consisting of these three possibilities. S = {rain, snow, fair} Let’s suppose that there is a 30% chance of rain, a 20% chance of snow, and a 50% chance that the weather will be fair. This means that we will assign probabilities to the possible outcomes in S as follows: P(rain) = .3, P(snow) = .2, and P(fair) = .5. In this context the event E = event that precipitation occurs is simply the event E = {rain, snow}. So the probability of E is P(E) = P(rain) + P(snow) = .3 + .2 = .5. Sometimes students wonder things like, “Where do probabilities come from?” That’s a good question, and the answer is not always the same. Let’s think about one of the simplest of all situations: tossing a coin and observing whether it comes up heads or tails. So our sample space is clearly something like S = {H,T}. But where do we get the probabilities? Most people’s instinctive reaction is to say P(H) = 1/2 and P(T) = 1/2. One reason might be the symmetry of the coin. Assuming that the coin isn’t bent and that the two faces of the coin are basically flat, it seems reasonable to think that the coin shouldn’t have any more preference for coming up tails than for coming up heads, and vice versa. For that reason we would assign probability 1/2 to each outcome, since this is the only way that the probabilities can be assigned so that heads and tails are equally probable. Similarly, if we are rolling an ordinary six-sided die, we might well be inclined to assign probability 1/6 to each outcome because of the symmetry of the die. But there’s another intuitive idea that’s important too. It is that the probability of an event represents the relative frequency with which an event occurs. For example, in tossing an unbiased coin where the probabilities of heads and tails are 1/2, the fraction of tosses that result in heads should be close to 1/2 if we toss the coin a large number of times. Similarly, if we say that the probability of getting a “5” when we roll a die is 1/6, we mean that in a large number of rolls the relative frequency with which a “5” should

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occur is about 1/6 of the time. And if we say that the probability that Sue beats Ann in tennis is 2/3, we mean that in the long run Sue should win about 2/3 of the time when they play. So you can think of probability as a theoretical relative frequency. This of course doesn’t mean that if you toss a coin 10 times you will get exactly 5 heads. Nor does it mean that in 100 tosses you will get 50 heads. In fact it doesn’t even say that the nickel that might turn up in your pocket tomorrow really will have exactly a .5 probability of coming up heads. When we talk about a “fair” coin with exactly the same probabilities for heads and tails, we are talking about a mathematical model for a coin. Whether the coin in your pocket actually has that property or not isn’t even a question for mathematics. By tossing the coin 100 times and seeing if the number of heads is near 50, for example, you might gather some evidence as to whether the coin is “fair” or not. Testing such as this, and the significance of such tests, is studied in the branch of mathematics known as statistics. But there is no test that will prove once and for all that the coin in your pocket is a perfectly fair coin. This should not be a great concern for you though. How would you judge the value of a mathematical model for an airplane wing? The question isn’t whether the model is theoretically perfect, it’s whether the airplane will fly. In the same way we get lots of useful information about the way the real world behaves by examining simple mathematical models for things like coins, dice, or tennis matches without having to know that our models offer an absolutely perfect description of the coin in your pocket, the dice you bought at K-Mart, or the tennis players who live down the street. Here are some properties that will be useful in all our discussions of probability. These properties are all logical consequences of the fact that the probability of an event is simply the sum of the probabilities of the individual elements that make up the event.

Basic Properties of Probabilities 1. For any event E, P(E) ≥ 0. probability zero, i.e. P(Ø) = 0.

The empty set Ø is assigned

2. The probability of the entire sample space S is equal to 1,i.e. P(S) = 1. 3. If A and B are disjoint events then P(A ∪ B) = P(A) + P(B). (Remember that A and B are disjoint if A ∩ B = Ø. Another way of saying A and B are disjoint is to say that they are mutually exclusive.) 4. For any events A and B, P(A ∪ B) = P(A) + P(B) – P(A ∩ B). 5. For any event A, P(Ac) = 1 – P(A).

Property (1) is a consequence of the fact that we compute the probability of an

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event by adding up the probabilities of the individual elements that make up the event. Since these probabilities are non-negative numbers, their sum is non-negative. Assigning probability zero to the empty set is logically consistent with this. Property (2) is just a restatement of the fact that the probabilities of all outcomes have to add up to one. Property (3) is also a consequence of the fact that the probability of an event is the sum of the probabilities of the outcomes that make up the event. The reason is that A ∪ B includes all the elements of A and of B. So all their probabilities get included when we compute P(A ∪ B). Furthermore, since A ∩ B = Ø, none of the probabilities gets included more than once when we add P(A) + P(B). Property (4) includes an extra term on the right side of the equation that is not included in the equation of Property (3). The extra term P(A ∩ B) is needed when A and B are not disjoint. The following simple example illustrates why this extra term is needed in this case. Example 5.13. When a fair die is rolled, the sample space is S = {1,2,3,4,5,6}. Think about the two events A = {1,2,3} and B = {3,4}. In this case A ∪ B = {1,2,3,4}, and A ∩ B = {3}. The die is fair, so each outcome has probability 1/6. This means that 4 3 2 1 P(A ∪ B) = 6 = P(A) + P(B) – P(A ∩ B) = 6 + 6 – 6 . A glance at this equation shows that the reason that it is necessary to subtract off the term appearing as P(A ∩ B) on the right side is because otherwise the probability of the outcome “3” in A ∩ B would be included twice on the right side of the equation and only once on the left side. By including the term “– P(A ∩ B)” on the right side of the equation in Property (4), the proper “adjusting” is performed to compensate for the elements in A ∩ B. Notice that Property (5) is really just a special instance of Properties (2) and (3). This is because A ∪ Ac = S. Furthermore, since A and Ac are disjoint, P(A) + P(Ac) = P(A ∪ Ac) = P(S) = 1, so P(Ac) = 1 – P(A). Example 5.14. John thinks that there is a 90% chance that he will pass English, and only a 20% chance that he will make worse than a C. What is the probability that he will make a D? Solution: Let’s try to describe all the events of interest in terms of the two events F1 = event that John passes English, and F2 = event that John makes worse than a C. What are the things we know to start with? First we know that P(F1) = .9 and P(F2) = .2.

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Next convince yourself that F1 ∪ F2 includes all possible grades. The reason is that F1 = {A , B , C , D} and F2 = {D , F}. Since F1 ∪ F2 is the whole sample space, P(F1 ∪ F2) = 1. Now notice that F1 ∩ F2 = {D}. If we go back to Property (4) and make use of P(F1 ∪ F2) = P(F1) + P(F2) – P(F1 ∩ F2), this tells us 1 = .9 + .2 – P(F1 ∩ F2). So P(F1 ∩ F2) = .1 Example 5.15. A pair of dice is rolled. What is the probability that the sum of the numbers appearing on the two dice is different from 7? Solution. Of the 36 equally probable outcomes, the event that the sum is equal to 7 is the event E = {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}. So P(E) = 6/36 = 1/6. Since the event that the sum is different from 7 is simply Ec, we can obtain its probability by a simple subtraction from 1 [using Property (5)]: 1 5 P(Ec) = 1 – P(E) = 1 – 6 = 6 . There are many situations in which all possible outcomes should be treated as equally probable. Often we deliberately devise situations so that outcomes will be equally probable. For example, the whole purpose in shuffling a deck of cards before playing a card game is to mix up the deck and destroy any “pattern” that might be present in the arrangement of cards in the deck. It is natural to assume that a card dealt from a well-shuffled deck would have equal probability of being any of the 52 cards in the deck. Similarly, when an opinion poll is conducted an attempt is made to “randomize” the people questioned. For instance to determine the opinion of students in a school you might question 30 students chosen “at random.” This means that you would like one group of 30 students to be just as likely to be questioned as any other group of thirty students. If there happened to be 400 students in the school, there are C(400,30) different possible ways to choose 30 students. For “random sampling,” this means that each group of thirty should be just as likely to be chosen as any other group of 30, so since there are C(400,30) possibilities, each would be assigned a probability 1 C(400,30) . The terminology that is used to describe a situation in which all outcomes are assumed equally probable is to describe the sample space as a uniform sample space.

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Definition: Uniform sample space A uniform sample space is one in which all outcomes are assigned the same probability. So if there are n elements in the sample space, this means that each element is assigned probability 1/n.

As a reminder, however, that real data often doesn’t lead to a uniform sample space, let’s return to a situation that illustrates this fact. Example 5.16. The distribution of grades in a class is as follows Grade A B C D F

Number of students 4 9 17 6 3

If a student is randomly picked from the class what is the probability that the student is a C student? What is the probability that the student has a grade higher than C? Solution: Since there are 39 students in the class and 4 have A’s, the probability assigned to the outcome “A” should be 4/39. The reason is that if a student is randomly chosen from the class, there are 4 chances in 39 that the student is an A student, and this converts to a probability of 4/39. In the same way P(C student) = 17/39. And P(grade higher than C) = P(A student) + P(B student) = 4/39 + 9/39 = 13/39. The assignment of probabilities to the various outcomes such as A, B, C, D, and F in the above example is frequently referred to as specification of a probability distribution. The point we have just been observing is that depending on the situation the correct probability distribution may be quite different from a uniform probability distribution in which all outcomes are assigned the same probability. Depending upon the sample space that is used, a uniform probability distribution may or may not be correct. For instance, if two coins are tossed we can use the sample space given by S = {HH, HT, TH, TT} and assign probability 1/4 to each outcome. On the other hand, if it is only the number of heads we are interested in, we might be tempted to use the sample space S = {two heads, one head, zero heads}. If we want this latter viewpoint to be consistent with the former, we would have to assign probability 1/2 to the outcome “one head” and probability 1/4 to each of the other two outcomes. So in this case we would not be dealing with a uniform sample space. Often for ease of computation it is easiest to choose a uniform sample. This will be illustrated in the next

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section. Probabilities and Venn diagrams Sometimes Venn diagrams are useful in determining probabilities. Suppose for example that A and B are events with P(A ∩ B) = .2, P(A ∩ Bc) = .3, and P(B) = .4. If you need to find the probabilities of some other events that are described in terms of A and B, a good procedure is to place the probabilities into a Venn diagram in much the same manner we used with counting problems in Chapter 4. You can immediately place the information P(A ∩ B) = .2 and P(A ∩ Bc) = .3 in the Venn Diagram. Then knowing that P(B) = .4 you can fill in the probability of all the regions and end up with a picture looking like this one.

A

B .3

.2

.2 .3

Now if you need to know the probability of an event such as A ∪ Bc, for example, you can get it by finding the appropriate region in the Venn diagram. Thus P(A ∪ Bc) = .8.

Problems 1.

Suppose A and B are mutually exclusive events with P(A) = 1/3 and P(B) = 1/6. (a) Find P(A ∪ B) (b) Find P(A ∩ B)

2.

50% of the students in a school weigh more than 140 pounds, but 70% weigh less than 170. If a student is randomly selected, what is the probability the student will weigh between 140 and 170?

3.

Suppose P(E) = .4, P(F) = .3, and P(E ∩ F) = .2. Find (a) P(E ∩ Fc) (c) P(Ec ∪ Fc) (b) P(E ∪ F) (d) P(Ec ∪ F)

4.

A pair of dice is rolled, one red and one green. Suppose E = event consisting of all outcomes in which the red die shows the number 4 F = event consisting of all outcomes in which the sum of the numbers on the red and green dice is equal to 8 (a) Find P(E) (b) Find P(E ∩ F) (c) Find P(E ∪ F)

5.

John wants to go to a movie Friday and a concert Saturday.

He figures the

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Chapter 5: Probability probability that his parents let him go to the movie is .5, and the probability that they let him go to the concert is .7. However, he suspects that the probability that they let him go to both is only .3. What is the probability that he gets to go nowhere at all?

6.

P(A ∪ B) = .8, P(Ac ∩ B) = .1, and P(A ∩ Bc) = .2. (a) Find P(B) (b) Find P(A ∩ B)

(c) Find P(Ac ∩ Bc)

7.

The probability that Ann will go to the beach this summer is .6 and the probability that she will go to the mountains is .5. The probability she does both is .25. What is the probability that (a) she goes to the beach and not the mountains? (b) she doesn’t go to either place?

8.

Suppose S = {s1, s2, s3, s4, s5} is a sample space, and suppose the probability distribution for S is given by P(s1) = 1/3, P(s2) = 1/6, P(s3) = 1/4, P(s4) = 1/12, and P(s5) = 1/6. If A = {s1, s2, s3} and B = {s1, s3, s5}, find (a) P(A) (b) P(A ∩ B) (c) P(Ac ∪ Bc)

9.

Answer TRUE or FALSE. (a) P(A ∩ Bc) = P(A) – P(A ∩ B). (b) If A ! B then P(A ∩ B) = P(A). (c) P(A ∪ B) ≥ P(A). (d) If A and B are mutually exclusive then P(A ∩ Bc) = P(A).

10. An experiment is conducted in which the possible outcomes are the counting numbers from 1 to 10. Assign probability 1/10 to each possible outcome. In this sample space let A = {1,2,3}, B = {5,7,8}, C = {2,3,7,8}, and D = {1,2,3,4,7,8,9}. (a) Does P(A ∩ Cc) = P(A) – P(A ∩ C)? [See 9(a)] (b) Does P(A ∩ D) = P(A)? [see 9(b)] (c) Is P(A ∪ C) ≥ P(A)? [see 9(c)] (d) Does P(A ∩ Bc) = P(A)? [see 9(d)] 11. To say that the odds in favor of an event are m to n means that the probability of the event is m/(m+n). For instance if the odds are 3 to 4 it means that the probability of the event is 3/7. As another illustration, if a die is rolled the odds of getting a “4” are 1 to 5, since there is 1 way to get a “4” and 5 ways not to get a “4”. The probability of getting a “4” is 1/6. Give the odds of each of the following: (a) drawing a heart from a deck of cards. (b) getting a “heads” when a coin is tossed. (c) not getting a “4” when a die is rolled. 12. To say that the odds against an event are m to n means that the probability is m/(m+n) that the event will not occur. Find the odds against each of the following

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things happening: (a) getting a “4” when a die is rolled. (b) drawing a heart when a card is drawn from a deck. (c) Syracuse beating Stanford if Stanford is a 2 to 1 favorite. [Hint: to say that Stanford is a 2 to 1 favorite means that the odds of Stanford beating Syracuse are 2 to 1.] (d) Duke beating N.C. State if the probability is 2/3 that N.C. State will win.

5.3 Uniform Probability Distributions We have already seen numerous examples in which the probabilities of the various outcomes are equal. We describe the situation in which all outcomes have the same probability as a uniform probability distribution. Not only is this situation very common, there are some special ideas that apply to uniform probability distributions that don’t apply to other situations. Example 5.17. A president and vice-president are to be elected in a club which has 14 members including 6 men and 8 women. Assuming that all members are equally likely to be elected, what is the probability that both officers will be women? Solution: The first observation here is that the sample space is made up of all possible choices of 2 people from 14. There are C(14,2) different ways of specifying which two people are selected as officers. (For the moment we will ignore any distinction between the two offices, i.e. we will consider only which two people are selected and not worry about which one is president and which is vice-president.) Since C(14,2) = 91, we are visualizing a sample space of 91 possible outcomes. Furthermore, we are going to treat all of these 91 outcomes as equally likely. This means that we assign probability 1/91 to each one. Now the event we are interested in is the event that both officers are women: E = event that two women are elected Since there are 8 women, this can happen in C(8,2) = 28 different ways, and so E consists of 28 of the 91 possible outcomes. But remember that each outcome has probability 1/91. So the probability of E is just the sum of the probabilities of these 28 outcomes, or 1 4 P(E) = 28 × 91 = 13 Alternate Solution: Suppose we decide that we’d like to keep track of who is elected president and who is elected vice-president. If we consider the offices one at a time (with president being considered first), there are 14 choices for president, and then after the president is selected that leaves 13 possibilities for whoever is elected vicepresident. If you think about it you should realize that this is simply a matter of selecting 2 people from 14 keeping track of the order of selection. So the number of ways of doing

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this is P(14,2) = 14 × 13 = 182. Conclusion: We now have a sample space of 182 equally likely possible outcomes, which means we assign probability 1/182 to each. Still the event we are interested in is E = event that two women are elected Now, however, since we are considering the offices separately this means that there would be 8 choices for which woman is elected president and then 7 choices for which woman is elected vice-president. This gives a total of P(8,2) = 8 × 7 = 56 ways in which two women could be elected. Since each of these 56 possibilities now has probability 1/182, 1 56 4 P(E) = 56 × 182 = 182 = 13 Notice that the answers for these two solutions agree. What is this telling us? It points out that in this problem (as will be the case in many others) we have some freedom of choice about what kind of information to keep track of and what information to ignore. Here it simply doesn’t matter whether we keep track of who fills what office. This is equivalent to saying that we can either take into consideration or ignore the order in which the two people are selected. In other words, we have a choice as to whether we want to consider the sample space as “combinations” or “permutations.” This “freedom of choice” may be bothersome to you at first, because some students would rather be told exactly what they have to do. There’s not a simple rule to follow here, but a general guideline might be to keep track of order only when necessary. Following that guideline here means choosing the first solution over the second. But both are correct. What is important about both solutions, however, is to realize that the solution really reduces to two counting problems. In both instances number of elements of E n(E) P(E) = number of elements of the sample space = n(S) .

Uniform probability distributions number of elements of E n(E) P(E) = number of elements of the sample space = n(S) .

Keep in mind however, that this is true only when all elements of the sample space are equally probable, i.e. when we have a uniform probability distribution. Example 5.18. Each student at a school is assigned an ID number which is a 5-digit number with the first digit not equal to zero. What is the probability that John’s ID will consist of 5 different digits?

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Solution: How many possible ID numbers are there? The first digit must be one of the digits 1,2,3,4,5,6,7,8,9, and all the others may be any of these digits or a 0. So there are 9 possibilities for the first digit and 10 for each of the other four digits. The multiplication principle says then that there are 9 × 104 = 90,000 possible ID numbers. We will treat these possibilities as equally likely, i.e. we will assume a uniform probability distribution. If E denotes the event E = event that John’s ID has no repeated digits, then there are 9 possibilities for the first digit in John’s ID. Then there are 9 possibilities for the second digit (any of the other 8 non-zero digits plus 0), 8 possibilities for the third (any of the 8 not used as the first two digits), 7 for the fourth, and 6 for the fifth. The multiplication principle therefore gives n(E) = 9 × 9 × 8 × 7 × 6 = 27,216. n(E) 27,216 So P(E) = n(S) =90,000

≈ .302.

In other words, there is about a 30% chance that John’s ID will consist of 5 different digits. Example 5.19. Two cards are dealt from a well-shuffled standard deck of 52 cards. What is the probability that both cards are clubs? Solution: There are C(52,2) = 1326 ways to deal two cards from the deck. There are C(13,2) = 78 ways to deal two clubs from the deck. Therefore the probability of getting two clubs is C(13,2) 78 1 C(52,2) = 1326 = 17 . It would be a worthwhile exercise for you to redo this example keeping track of the order in which the cards are dealt. In other words, treat the possible outcomes as permutations rather than combinations. Example 5.20. Charlie has 4 pairs of white socks, 2 pairs of black socks, and 3 pairs of blue socks. They are all mixed up in a drawer in his dresser. One morning he gets out of bed in the dark and reaches into the drawer and pulls out two socks. What is the probability that the socks will be of the same color? Solution: Altogether Charlie has 9 pairs of socks, for a total of 18 socks. So there are C(18,2) = 153 possible ways to choose two socks from this jumble of socks. How many matching pairs are there, i.e. how many pairs of the same color? Since he has 8 white socks, that makes C(8,2) = 28 ways to pick a pair of white socks. Similarly there are C(4,2) = 6 ways to pick a pair of black socks and C(6,2) = 15 ways to pick a pair of blue socks. Altogether this gives 28 + 6 + 15 = 49 ways to pick a pair of matching socks. This is 49 ways out of the total of 153 possible ways, giving a probability

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P(socks match) =

C(8,2) + C(4,2) + C(6,2) 49 = 153 . C(18,2)

Make sure you understand why the combinations terms in the numerator of this fraction are added rather than multiplied. Example 5.21. Eric and Edith have applied for jobs at Sadlacks. There are 6 job openings available and 25 people have applied. (a) What is the probability that Edith gets a job? (b) What is the probability that both of them get jobs? (Assume that the 6 people to get jobs are to be randomly chosen from among the 25 applicants.) Solution: (a) Since 6 of the 25 applicants will get jobs, there are C(25,6) = 177100 ways that the applicants can be selected. The other counting problem that we have to solve is this: In how many of these cases will Edith be one of the applicants chosen? If Edith is chosen, this means that there are 5 remaining jobs to be distributed among the other 24 applicants. This can be done in C(24,5) = 42504 ways. So the probability that Edith gets a job is C(24,5) 42504 6 P(Edith gets a job) = C(25,6) = 177100 = 25 = .24. A moment’s thought should convince you that this answer is intuitively correct. Since 6 out of 25 applicants (or 24% of the applicants) get jobs, Edith’s chances are 6 out of 25 = .24. (b) Now what about the probability that they both get jobs? If both get jobs, this leaves only 4 other jobs to be distributed among the other 23 applicants. And this can be done in C(23,4) = 8855 ways. Therefore the probability that they both get jobs is C(23,4) 6∞5 P(both Eric and Edith get jobs) = C(25,6) = 25 ∞ 24 = .05. Example 5.22. Joyce is a junior in a club with 18 juniors and 14 seniors, and Pat is a senior in the club. (a) If 10 students are to be randomly chosen from the club to go on a trip, what is the probability that Joyce and Pat will both be chosen? (b) If it is decided that 5 juniors and 5 seniors are to go, what is the probability that Joyce and Pat will both be chosen? Solution: The first question is exactly like the previous example. C(30,8) 10 ∞ 9 45 P(Joyce and Pat are both chosen) = C(32,10) = 32 ∞ 31 = 496 ≈ .0907. In the second question, we need to consider the juniors and seniors separately. There are C(18,5) ways to choose the juniors that go on the trip and C(14,5) ways to choose the seniors. The multiplication principle therefore gives a total of C(18,5) × C(14,5) ways to choose all the trip participants. If Joyce is to be one of the juniors, this leaves C(17,4) ways to choose the other 4 juniors. And if Pat is to be one of the seniors

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chosen, then there are C(13,4) ways to choose the other seniors. So the number of ways to fill out the list in such a way that Pat and Joyce are both included is C(17,4) × C(13,4). This means that the probability that Joyce and Pat get to go is C(17,4) ∞ C(13,4) 5∞5 C(18,5) ∞ C(14,5) = 18 ∞ 14 ≈ .099. Example 5.23. Pud is playing 5-card draw (a poker game). When he picks up the hand that is dealt to him, what is the probability that it will be a “full house”? Solution: In this game each player is dealt 5 cards from the standard 52-card deck. A “full house” is a hand of 5 cards in which 3 of the cards are of one rank and the remaining 2 are of some other rank. For example, 3 sevens and 2 fives would be a full house. Since the 5 cards are dealt from a standard 52-card deck, there are C(52,5) different hands possible. Now how many ways are there to get a full house? This question has already been answered in Example 4.30. We know we must have 3 cards from one of the 13 ranks (like “sevens,” for example). And then the pair must be from one of the remaining 12 ranks (such as “fives”). So we can analyze the process like this: C(13,1) = 13 ways to choose the rank that the 3-of-a-kind come from C(12,1) = 12 ways to choose the rank that the pair comes from C(4,3) = 4 ways to choose the 3-of-a-kind from the designated rank C(4,2) = 6 ways to choose the pair from the designated rank So from the multiplication principle this gives 13 × 12 × 4 × 6 = 3744 ways to get a full house. The probability of a full house is therefore C(13,1) ∞ C(12,1)∞ C(4,3) ∞ C(4,2) 3744 = 2598960 ≈ .00144. C(52,5)

Problems 1.

If 3 people are randomly lined up in a row, what is the probability that the shortest is first and the tallest is last in the line?

2.

A license plate is made up of 3 letters followed by three digits, for example ZTP284. If they are made and given out randomly, what is the probability that your license plate will be ABC-123?

3.

In a group of 17 girls and 15 boys, what is the probability that the smartest member of the group will be a girl?

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4.

Mary wrote postcards to 4 friends in California and 6 friends in South Carolina. 4 of the 10 cards fell out of her purse and were lost. (a) What is the probability that all of the cards that were lost were cards to South Carolina? (b) What is the probability that exactly 3 of the lost cards were addressed to California?

5.

A company has 7 stores in Raleigh and 4 in Durham. If 3 of the stores are randomly selected as a site for a special display, what is the probability that all three selected will be in Raleigh?

6.

Ed, Carol, Ann, and Pinky are going to pair up randomly to play tennis. What is the probability that Ann and Pinky will be partners?

7.

If five friends (with different names) randomly seat themselves in a row of five theater seats, what is the probability that they will be seated alphabetically by first name?

8.

Sue bought 3 raffle tickets. 100 tickets were sold. (a) What is the probability she holds the winning ticket? (b) If there are 2 prizes and 2 winning tickets, what is the probability she wins at least one of the prizes?

9.

John has 4 red greeting cards, 2 blue ones, and 3 green ones. If he randomly picks 2 of the cards, what is the probability that they will be the same color?

10. A school teacher is going to assign four children to safety patrol duty on four different streets. The children are Eddie, Bob, Mike, and Alice. The streets are Elm, Park, Van Dyke, and Oakwood. (a) What is the probability that Mike will be assigned duty at Elm Street? (b) What is the probability that Mike gets Elm Street and Alice gets Oakwood? (c) What is the probability that Mike gets Elm, Alice gets Oakwood, and Bob gets Van Dyke? (d) What is the probability that Mike gets Elm Street, Alice gets Oakwood, Bob is assigned Van Dyke, and Eddie gets Park Street? 11. A survey of 100 college students revealed the following: 40 read TIME 15 read TIME and NEWSWEEK 30 read NEWSWEEK 12 read TIME and SOJOURNERS 25 read SOJOURNERS 10 read NEWSWEEK and SOJOURNERS 4 read all three (a) What is the probability that a randomly chosen student will read exactly one magazine? (b) What is the probability that a randomly chosen student will read exactly two magazines?

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(c) What is the probability that a randomly chosen student will read Sojourners and exactly one other magazine? 12. While talking about birthdays, Ann, Bob, and Terry discovered that they were all born on a Tuesday. (a) Assuming that the seven days of the week are equally likely as birthdays, what is the probability that 3 people would be born on the same day of the week? (b) What is the probability that 3 people would all be born on different days of the week? 13. If the letters of the word LITTLE are written in a random order, what is the probability that they will appear in alphabetical order? 14. Six married couples are attending a party where two door prizes are awarded to two different people. (a) What is the probability that a married couple wins both prizes? (b) What is the probability that one prize goes to a man and the other to a woman? 15. Teledex Corporation is going to hire 5 programmers from a pool of 10 female and 8 male applicants. Sam and Martha are two of the applicants. If the 5 to get jobs are randomly chosen, (a) what is the probability Sam and Martha both get jobs? (b) what is the probability they both get jobs if Teledex has decided to hire 3 women and 2 men? 16. Ten people are dividing up to play basketball. Two of the ten are brothers. (a) What is the probability the brothers will be on the same team if the players divide up arbitrarily? (b) What is the probability the brothers and their friend Max will all be on the same team? 17. In a lottery a player can win a prize if any of the digits in a 4-digit number chosen by the player matches the corresponding digit in the winning 4-digit number. Thus a prize can be won if all 4 digits match, if 3 corresponding digits match, if 2 match, or if only 1 matches. Assume that any or all of the digits may be zero. What is the probability that (a) all 4 digits match? (b) exactly 3 digits match? (c) exactly 2 digits match? (d) at least 1 digit matches? 18. The letters of the word MAMMAL are scrambled and written in a row. (a) What is the probability that the 3 M’s come first? (b) What is the probability that none of the M’s appear side by side? 19. Ann, Bill, Carla, Dave, and Ernie go to a basketball game. Assuming they seat

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Chapter 5: Probability themselves randomly, (a) What is the probability the men sit together and the women sit together? (b) What is the probability the men and the women sit alternately (no two people of the same sex sitting beside each other)? (c) What is the probability that Bill, Carla, and Ernie sit in 3 adjacent seats?

20. A class has 23 students. What is the probability that there are two people in the class who have the same birthday? [Assume that the 365 days of the year are equally likely for each person’s birthday. This means that we’ll assume that none of the students was born on Feb. 29 in a leap year.]

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5.4 Conditional Probability Sometimes additional information becomes available that forces us to modify the probabilities of events that we might be interested in. Example 5.24. Suppose you roll a pair of dice (one red and one green). The probability that the sum of the two numbers appearing is equal to 9 is 4/36 = 1/9 because 4 of the 36 possible outcomes show a sum of 9. Here are the outcomes where the sum is 9: {(6,3),(5,4),(4,5),(3,6)} Now what if we happen to see that the red die is showing the number 5, but we still haven’t seen the green die? How do we “process” this information? Since the red die has a 5 showing, we know that the outcome is one of these: {(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)} Among these outcomes, there is exactly one in which the sum is equal to 9. So the chances of getting a sum of 9 based on the new information are 1 chance in 6. We say that the conditional probability that the sum is 9 given that the red die shows the number 5 is equal to 1/6. Example 5.25. A class has the following distribution of grades: Grade A B C D F

Number of students 5 14 7 9 8

If a student is randomly picked from this class, what is the probability that the student is an A student if it is known that the student is passing the course? Solution: The probability that the student is an A student (given no information as to whether the student is passing or not) is 5/43. As soon as we know that the student is passing, the conditional probability that the student is an A student becomes 5/35, since 5 of the 35 passing students have an A average.

Definition: Conditional probability If A and B are events with P(A) ≠ 0, then the conditional probability of B given A is defined by P(B|A) =

P(B ↔ A) P(A)

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Let’s go back and check this definition on Examples 5.24 and 5.25. In Example 5.24 let’s take A to be the event “red die shows the number 5” and B to be the event “sum = 9.” Then P(A) = 1/6, and since A ∩ B consists of the single outcome (5,4), P(A ∩ B) = 1/36. Therefore P(A ↔ B) 1/36 1 P(B|A) = = 1/6 = 6 . P(A) In Example 5.25 let’s take E1 = “A students” and E2 = “students who are passing.” Then P(E2) = 35/43 since 35 of the 43 students are passing. What is E1 ∩ E2 in this case? It is simply the A students since all A students are passing. So P(E1 ∩ E2) = P(E1) = 5/43. Therefore, P(E1 ↔ E2) 5/43 1 P(E1|E2) = = 35/43 = 7 . P(E2 ) One important aspect of the definition of conditional probability above is that it enables you to express a conditional probability as a fraction where both the numerator and denominator are ordinary (unconditional) probabilities. Example 5.26. Barbara, Carol, Alice, Peggy, and Sabrina are competing for two roles in a play. We will assume that the two to get roles will be randomly chosen from the five girls. What is the conditional probability that Peggy gets a role if we know that Carol does not get a role? Solution: Let’s name the events that are involved here: A = event that Carol does not get a role B = event that Peggy does get a role This problem involves a small number of possible outcomes, and it is helpful to list them all. The ten possibilities for which 2 girls get the roles are as follows (all 10 of these possibilities are assumed equally likely, so each has probability 1/10): #1 #2 #3 #4 #5 #6 #7 #8 #9 #10

Barbara & Carol Barbara & Alice Barbara & Peggy Barbara & Sabrina Carol & Alice Carol & Peggy Carol & Sabrina Alice & Peggy Alice & Sabrina Peggy & Sabrina

The event A consists of outcomes #2,#3,#4,#8,#9, and #10, so P(A) = 6/10. And the event A ∩ B consists of outcomes #3, #8, and #10, so P(A ∩ B) = 3/10. Therefore

Section 5.4: Conditional Probability

P(B|A) =

185

P(B ↔ A) 3/10 1 = 6/10 = 2 . P(A)

It is not absolutely necessary to list all the 10 outcomes as done here in order to do this problem. What is necessary is to figure out P(B ∩ A) and P(A) in some way, and these can both be determined using the basic counting methods we have practiced. For example C(4,2) 6 P(A) = C(5,2) = 10 because if Carol does not get a role then the two roles must be given out among the other 4 girls. And C(3,1) 3 P(B ∩ A) = C(5,2) = 10 because saying that Carol does not get a role and that Peggy does get a role means that there is one other role that will go to one of the three other girls. Alternate solution: Let’s look at another approach to the same problem. If we know that Carol does not get a role, then we can just forget about her altogether and reformulate the problem. The question then becomes: If the two roles are assigned among the four other girls (Barbara, Alice, Peggy, and Sabrina), what is the probability that Peggy gets a role? There are C(4,2) = 6 ways to pick 2 of the 4 girls for the roles. Of these 6 possibilities, Peggy gets a role in 3 of the 6 cases. (The 3 cases in which Peggy gets a role are the case in which Peggy and Barbara are chosen, the case in which Peggy and Alice are chosen, and the case in which Peggy and Sabrina are chosen.) So from these considerations we see that 3 1 P(Peggy gets a role) = 6 = 2 . This answer is of course consistent with the answer obtained in the first solution. In this latter solution we used the given information (that Carol does not get a role) in constructing the sample space. So our sample space had C(4,2) = 6 possibilities rather than C(5,2) = 10 as in the first solution. Since our new reduced sample space has already been adjusted for the new information (that Carol does not get a role), the conditional probability question in our original sample space becomes an ordinary (not conditional) probability question in the new reduced sample space. Example 5.27. If two cards are dealt from a standard 52-card deck, what is the probability that both are hearts if we know that at least one of them is a heart? Solution: This problem gives us a chance to review a couple of useful ideas. The sample space consists of the C(52,2) = 1326 possible ways to choose 2 cards from 52. The two events involved are A = event that 2 hearts are dealt B = event that at least one heart is dealt

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Chapter 5: Probability C(13,2) 1 Determining P(A) is easy: P(A) = C(52,2) = 17 .

It is easiest to get P(B) by first finding P(Bc) and then using P(B) = 1 – P(Bc). Bc is just the set of outcomes for which it is not true that at least one heart is dealt, meaning that both cards dealt come from the other three suits (with a total of 39 cards in the three suits). The probability of this happening is C(39,2) 19 19 15 P(Bc) = C(52,2) = 34 . So P(B) = 1 – 34 = 34 . Now what is A ∩ B? In this problem A ∩ B is simply A. (Because any outcome in which two hearts are dealt is certainly an outcome in which at least one heart is dealt.) Therefore P(A ∩ B) = P(A) = 1/17. So P(A|B) =

P(A ↔ B) 1/17 2 = 15/34 = 15 . P(B)

One thing that is a bit curious is the fact that conditional probabilities are often useful in computing unconditional probabilities. The way in which this comes about is that we can turn around the definition of conditional probability and look at it in the form P(A ∩ B) = P(A) P(B|A). In some situations the two probabilities on the right are easier to obtain than the one on the left. Example 5.28. If 2 cards are dealt from a standard 52-card deck, what is the probability that both are hearts? Solution: We have done this problem previously using some of the basic counting techniques. This time we are going to look at it from the standpoint of conditional probability. It’s useful for us to think of the cards in terms of a first card and a second card. It really doesn’t matter whether they were actually dealt one at a time or not. We can think of them that way if we like. It really just amounts to labeling one of the cards with the label “first” and the other with the label “second.” We also need to name the events that we are going to be interested in: A = event that the first card is a heart B = event that the second card is a heart A ∩ B = event that both cards are hearts The idea is to compute P(A ∩ B) by using the fact that P(A ∩ B) = P(A) P(B|A). The term P(A) on the right is easy, because when we draw the first card it is coming from a deck of 52 cards including 13 hearts. Therefore P(A) = 13/52. Now what about P(B|A)? This is the conditional probability that the second card is a heart based on the information that the first card is a heart. Knowing that the first card is a heart means that when the second is drawn it is coming from a deck of 51 cards in which there are 12 hearts

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remaining. Therefore P(B|A) = 12/51. And so P(A ∩ B) is given by 13 12 1 P(A ∩ B) = P(A) P(B|A) = 52 × 51 = 17 . This of course agrees with the answer we got earlier for this problem by using the basic counting techniques. This idea can be extended to more than two events. For example, with three events the equation becomes P(A ∩ B ∩ C) = P(A) P(B|A) P(C | A ∩ B). The last term on the right here means the conditional probability of C given that both A and B occur. If three cards are dealt, we could use this idea to determine the probability that all three are hearts. The computation would be 13 12 11 P(A ∩ B ∩ C) = P(A) P(B|A) P(C | A ∩ B) = 52 × 51 × 50 ≈ .0129. The third factor 11/50 on the right is the conditional probability that the third card is a heart given that the first two cards are hearts. The explanation is that if we know that the first two are hearts, then there are 11 hearts left in the 50 remaining cards when the third card is to be drawn. In the next section we will see how to apply conditional probabilities to a wide variety of problems by placing them appropriately in tree diagrams.

Problems 1.

If P(A) = .6, P(B) = .4, and P(A ∩ B) = .3, (a) find P(B | A) (b) find P(A | B)

2.

If P(A) = .7, P(B) = .6 and P(A ∩ B) = .4, (a) find P(A | Bc) (b) find P(Bc | A)

(c) find P(Bc | Ac)

3.

If a card is drawn from a deck of 52 cards, what is the probability the card is a heart if you know it is not a club?

4.

If two dice are rolled (one red and one green), (a) what is the probability of getting a pair of sixes? (b) what is the conditional probability of getting a pair of sixes if you know that the red die shows a six?

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Chapter 5: Probability (c) what is the conditional probability of getting a pair of sixes if you know that the green die shows a six? (d) what is the conditional probability of getting a pair of sixes if you know that at least one of the two dice shows a six?

5.

A committee has 3 members from California, 4 from Florida, 2 from Colorado, and 2 from North Carolina. (a) If the committee president is picked randomly, what is the probability the president will be from California? (b) What is the conditional probability the president will be from California if the 2 members from Colorado remove themselves from the selection process?

6.

If a coin is tossed 3 times, (a) what is the probability that all three tosses come up heads given that at least two of the tosses come up heads? (b) what is the probability that all three tosses come up heads given that the first two tosses come up heads? [Hint: List the sample space for this experiment as well as the events related to each of these questions.]

7.

A political party has decided to hold political rallies in two of the following states: Missouri, Montana, Mississippi, Michigan, Alabama, Connecticut, and Pennsylvania. Assuming that the states have equal probabilities of being chosen, (a) what is the probability that both of the states chosen begin with the letter “M”? (b) what is the probability that both start with “M” given that at least one does? (c) what is the probability that both start with “M” given that one of the states is Michigan?

8.

Two men (Jim and Bob) and three women (Ann, Beth, and Carol) are on a committee. Two of the five are to be chosen to serve as officers. If the officers are chosen randomly, (a) what is the probability that both officers will be women? (b) what is the probability at least one officer will be a woman? (c) what is the probability that both officers will be women given that at least one is a woman? (d) what is the probability both officers are women if you know that Beth is an officer?

9.

Evaluate the following probabilities. (a) P(A | Ac) (b) P(A | A ∩ B) (c) P(A | A)

(d) P(A ∪ B | A)

10. Six married couples are attending a party where two door prizes are awarded to two different people. (a) If John’s wife Mary wins a prize, what is the probability that John will also win a prize? (b) If Mary does not win a prize, then what is the probability that John will?

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11. John, Sue, Ted, and Penny are seating themselves randomly in a row of four theater seats. (a) What is the probability that John will be sitting beside Sue? [Hint: Just list the possible ways that John can be sitting beside Sue.] (b) What is the conditional probability that John is sitting beside Sue if you know that Penny is going to sit beside John? 12. A baby has 5 blocks in a bucket. One block is red, one is yellow, one is green, one is blue, and one is black. The baby pulls out a block, looks at it, and puts it back in the bucket. If he does this 4 times before he gets bored and crawls away, what is the probability (a) the 4 blocks pulled out are red, yellow, blue, and green (in that specific order)? (b) the 4 blocks selected are all the same color? (c) the blocks pulled out are red, yellow, blue, and green (in that order) if all the blocks the baby chooses are different colors from each other? (d) the blocks the baby chooses are of exactly two different colors? 13. Mary has 6 Hershey Kisses wrapped in foil. 2 are in gold foil, 2 in red foil, and 2 in green foil. She also has 6 chocolate covered mints wrapped in foil. 2 of these are in gold, 2 in red, and 2 in green. If she chooses 2 candies at random and without replacement, what is the probability that (a) both are mints? (b) both are mints given that at least one is a mint? (c) both are mints given that at least one candy chosen is wrapped in red foil? 14. Carl and Ann are members of a 10 member club. Two members are to be selected to represent the club at a convention. If the selections are randomly made, (a) what is the probability Carl is selected? (b) what is the probability Carl is selected if we know that Ann is not selected? (c) what is the probability Carl is selected if we know that Ann is selected? 15. Ellen has asked for job interviews with IBM, Apple, Exxon, K-Mart, and Kodak. She has no control over the order in which the interviews are scheduled. (a) What is the probability the first interview will be with Apple? (b) What is the probability the first 2 interviews will be with the 2 computer companies? (c) If the Apple interview is first, what then is the conditional probability that the IBM interview is last? (d) What is the probability that the K-Mart interview will come sometime after the Kodak interview? Let A denote the event “IBM interviews Ellen first” and let B denote the event “Exxon interviews Ellen last. (e) Are A and B mutually exclusive?

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5.5 Using Conditional Probabilities in Tree Diagrams In this section we will learn a useful way of analyzing many situations by combining conditional probabilities with tree diagrams. Example 5.29. Once again let’s consider dealing two cards from a 52-card deck. If we are interested in recording whether each card is a heart, a tree diagram to represent the possible outcomes might look like this: c H H

H

c H c H

H

In this tree H represents hearts and Hc represents the complement of the hearts, i.e. the cards that belong to some other suit. The first level of branches represents “the first card dealt” and the second level represents “the second card.” When we combine the idea of conditional probabilities with this tree we get c H

H 12/51

39/51 13/51

H 13/52

c H

H c H

38/51

39/52

The numbers inserted on the branches are probabilities. At the lowest level the number 13/52 and 39/52 represent the probability that the first card is or is not a heart. At all higher levels in the tree the probabilities are conditional probabilities. For example, the 12/51 at the second level is the conditional probability that the second card is a heart given that the first one is a heart. Similarly the 13/51 at the second level is the conditional probability that the second card is a heart given that the first card is not a heart. The reason that this is so useful is that now we can obtain the probability of each of the four outcomes shown in the tree diagram simply by multiplying the numbers from the root of the tree up to each leaf. For example, 13/52 × 12/51 = 1/17, 13/52 × 39/51 = 13/68, etc. 1/17

13/68

13/68

19/34

H

c H

H

c H

12/51

39/51 13/51

H 13/52

c H

38/51

39/52

Now the tree diagram shows not only the four possible outcomes of the experiment, but also the probability of each. The four possible outcomes correspond to the four circles, and the probability of each is shown above.

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The theoretical basis for doing this “tree” computation is exactly what was observed in the last section. For example, when we multiply 13/52 × 12/51 to get the probability that both cards are hearts, we are multiplying the probability that the first is a heart times the conditional probability that the second is a heart given that the first is a heart. And the other four multiplications all have a similar explanation. Example 5.30. Michelle and Kim are playing a tennis match. The winner will be the first person to win two sets. Michelle is the better player, and in fact whenever they play a set the probability is 2/3 that Michelle will win the set. What is the probability that Michelle will win the match? If she wins the first set, what then is the probability that she will win the match? What is the probability that she wins the first set given that she wins the match? What is the probability that the match will last for three sets? Solution: The tree diagram for the tennis match, with all probabilities filled in, looks like this. 4/27

2/27

4/27

2/27

M

K

M

K

2/3

12/27 3/27

1/3

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1/3

2/3

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1/3

2/3

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1/3

Once again we have multiplied from the root of the tree up to each of the leaves to determine the probability of each outcome. For example the 2/27 at the upper right corner of the tree is computed as 1/3 × 2/3 × 1/3 = 2/27. This is an especially simple tree in the sense that the probabilities are always 1/3 and 2/3 no matter where we are in the match. What is the probability that Michelle will win the match? Simply look at the three outcomes that show Michelle winning the match and add up their probabilities. 4 12 4 20 P(Michelle wins match) = 27 + 27 + 27 = 27 . 4 12 16 Notice that P(Michelle wins first set and wins match) = 27 + 27 = 27 . So P(Michelle wins match | Michelle wins first set) =

P(Michelle wins match and first set) 16/27 8 = 2/3 = 9 . P(Michelle wins first set)

Similarly, P(Michelle wins first set | Michelle wins match)

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=

P(Michelle wins match and first set) 16/27 4 = 20/27 = 5 . P(Michelle wins match)

What is the probability that the match goes for three sets? Four of the six outcomes show the match going for three sets. Add their probabilities: 4 2 4 2 4 P(match lasts 3 sets) = 27 + 27 + 27 + 27 = 9 . An alternate way of computing this is, of course, to find the probability the match does not last 3 sets and subtract from 1: 12 3 5 5 4 P(match does not last 3 sets) = 27 + 27 = 9 , so P(match lasts 3 sets) = 1 – 9 = 9 . Example 5.31. Three good light bulbs and two defective ones have become mixed in a box. In order to find which are which, bulbs are going to be withdrawn from the box one at a time and tested until both bad ones or all three good ones have been found. What is the probability that more than three light bulbs will have to be tested? Solution: The important thing to remember in drawing this tree diagram is that we stop the testing process whenever we have found all the good bulbs or whenever we have found all the bad bulbs. G

B

G

B

G

B

1/2

1/2

1/2

1/2

1/2

1/2

G

B

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2/3

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1/3

B 2/3 G

B 2/4 G

2/4

B

G 2/3 G

B 1/4

1/3

3/4 B

3/5

2/5

An inspection of the tree shows that there are 10 possible outcomes (10 circles), and in this example it happens to turn out that all the outcomes have probability 1/10. (You should verify this by multiplying up the branches from the root to each leaf.) The reason that the probabilities are as shown in the tree should be apparent if you think about each situation. For example, look at the probability 1/4 that appears in the second level in the tree. At this point we know that the first bulb was bad, so the conditional probability that the next bulb will be bad is only 1/4 since there is only one bad bulb remaining among the 4 untested bulbs at that point. The probability that more than three bulbs must be tested is the sum of the probabilities of the six outcomes that show four bulbs being tested, and since each outcome has probability 1/10 this means that

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P(more than 3 bulbs must be tested) = 6/10. Example 5.32. A coin is to be tossed until either two heads have occurred or until three consecutive tails have occurred. What is the probability that exactly 3 tosses are required? What is the probability that more heads than tails are obtained? Solution: Once again, the most important step is to draw the tree correctly. The tree should look something like the one in the following picture. The probabilities are all 1/2 in this tree, which means that all the outcomes at the same level (all circles at the same level) will have the same probability. For example, there is only one outcome at level two in the tree (the lower left circle), and the probability of that outcome is 1/4. There are 3 outcomes that show 3 tosses being required, and they all have probability 1/8. So that provides the answer to the first question. The probability that exactly 3 tosses are required is 3/8. H T

H .5

.5 T

H

H

.5

.5 T

H

.5

.5 T

H

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H .5

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H .5

.5 .5

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H .5

H .5

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T .5

T

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What is the probability that more heads than tails are obtained? This question requires (1) that you first identify all the outcomes in the tree where more heads than tails occur, and (2) that you add up the probabilities of these outcomes. As step (1), you should identify the outcomes in which more heads than tails occurs as those that are shaded in the following figure:

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H T

H .5

.5 T

H

H

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.5 H .5 T

H

.5

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T

H

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H .5

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1 1 1 1 As step (2), sum their probabilities: P(more heads than tails) = 4 + 8 + 8 = 2

Problems 1.

There are 2 cookie jars in the kitchen. Jar #1 contains 3 ginger snaps and 6 oatmeal cookies. Jar #2 contains 3 ginger snaps and 3 oatmeal cookies. A child chooses a jar at random and pulls a cookie from the jar to eat. (a) Draw a tree diagram that represents the possible outcomes. (Include all the probabilities.) (b) What is the probability the child gets an oatmeal cookie? (c) If the child does get an oatmeal cookie, what is the probability it came from Jar #1?

2.

Aaron has 2 drawers containing t-shirts. The top drawer has 1 white t-shirt and 1 blue t-shirt. The bottom drawer has 2 white t-shirts and 2 red t-shirts. He selects a drawer at random and pulls out t-shirts one at a time (without replacement) until he has a white t-shirt. (a) Draw a tree diagram for this process (including probabilities in the tree). (b) What is the probability he pulls out exactly 2 t-shirts? (c) What is the probability he is taking t-shirts out of the top drawer if you know that he takes out exactly 2 t-shirts?

3.

Suppose in a certain county 55% of the registered voters are female and 45% are male. Among the females, 60% are Democrats and 40% are Republicans. Among the males, 50% are Democrats and 50% are Republicans. (a) If a voter is randomly picked from the registration books, what is the probability

Section 5.5: Using Conditional Probabilities in Tree Diagrams

195

that the voter is a Republican? (b) If a randomly selected voter is found to be a Republican, then what is the conditional probability that the voter is female? (c) If a randomly selected voter is found to be a Democrat, then what is the conditional probability that the voter is male? 4.

Tanya wants to pass a qualifying exam, and she is allowed 3 attempts at passing the exam if necessary. Each time she takes the exam there is a 40% chance she will pass. (a) Draw a tree diagram for this process. [Remember, if she passes the exam she doesn’t need to take it again.] (b) What is the probability she will pass the exam? (c) If she fails the first attempt, what then is the conditional probability she will pass?

5.

5% of the population of a region is thought to have diabetes. A standard diagnostic test will correctly identify 94% of the people who have the disease. However, the test also incorrectly diagnoses 8% of those who do not have the disease as having the disease. A person is tested and his test result comes back “positive,” indicating that he has the disease. What is the probability that he actually does have the disease?

6.

The students in a math course consist of 20% freshmen, 40% sophomores, 30% juniors, and 10% seniors. It turns out that 10% of the freshmen make an A, 40% of the sophomores make an A, 20% of the juniors do, and 20% of the seniors do. If the teacher randomly picks an “A” student in the class to represent the class in a math contest, what is the probability that a sophomore will be picked?

7.

Each person carries a pair of genes which determines eye color. If someone carries (b,b), he has blue eyes. If he carries (B,B) or the mixed pair (B,b), he has brown eyes. The percentages of (B,B), (B,b), and (b,b) in the general population are 16%, 48%, and 36% respectively. If Tom has brown eyes, what is the probability that he carries the (B,B) pair of genes?

8.

Identical twins come from the same egg and must therefore be of the same sex. Fraternal twins, on the other hand, come from different eggs, so the sex of one twin is independent from the sex of the other (just as in the case of separate births). In general, 2/3 of all twin births are fraternal. Find the probability that a set of twins (a) has the same sex. [Assume males and females are equally likely.] (b) are fraternal, given that they are of the same sex.

9.

40% of a company’s television sets are manufactured at a plant in Portland and 60% at a plant in Houston. Of those made in Portland, 4% are defective, and of those made in Houston 6% are defective. (a) What percentage of their total production is defective? (b) If you buy a defective set from the company, what is the probability it was made in Houston?

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10. 95% of 18 year old males have their driver’s license, and 90% of females do. Half of all 18 year olds are male and half are female. (a) What percentage of 18 year olds have their driver’s license? (b) If you find the driver’s license of an 18 year old person lying on the sidewalk, what is the probability that it belongs to a female? 11. A man has 3 coins in his pocket. One is a fair coin, one has heads on both sides, and one has tails on both sides. The man pulls a coin from his pocket and tosses it. It comes up heads. What is the probability that the coin is the fair coin? 12. Two red and three green balls are in a bag. Balls are removed from the bag one at a time (without replacement) until a ball of each color has been obtained. Draw a tree diagram for this experiment. (a) What is the probability of drawing 2 green balls followed by a red ball? (b) What is the probability that the last ball drawn is red?

Section 5.5: Using Conditional Probabilities in Tree Diagrams

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5.6 Independence and Independent Trials Sometimes information simply isn’t relevant to the question at hand. For instance, suppose you say to a friend, “What is the probability that it will rain today?” And then your friend answers, “I’m not sure, but I think I will go to a movie tonight.” Your friend has given you a piece of information. But does it in any way influence the probability that it is going to rain? Example 5.33. A card is drawn from a deck of cards. Let’s compare these two events: A = an ace is drawn H = a heart is drawn P(H) = 13/52 = 1/4 since there are 13 hearts in the 52-card deck. But also P(H|A) =

P(H ↔ A) 1/52 1 = 4/52 = 4 . P(A)

(The 1/52 in the numerator is the probability of getting the ace of hearts.) The interesting thing about this situation is that P(H|A) = P(H). In other words, the conditional probability that H occurs given that A occurs is exactly the same as the original unconditional probability that H occurs given no information whatsoever. This in effect says that the information we are being presented has no effect on the probability of the event we are interested in. It’s just like your friend’s going to the movie has no effect on the weather. Notice also that P(A) = 4/52 = 1/13, and P(A|H) =

P(A ↔ H) 1/52 1 = 13/52 = 13 , P(H)

so it is also true the other way around: P(A|H) = P(A). And finally, there is still one additional way to look at the relation between A and H. Notice that 1 4 13 P(A ∩ H) = 52 = 52 × 52 = P(A) P(H). So the probability that both A and H occur is just the product of their individual probabilities. When the conditions of this example are met, the events are said to be independent events. The three equations in the following definition are said to be equivalent because they are either all true or all false. When they are all true we call the events independent events. To see that they are equivalent is easy. Suppose, for example, that equation (1) is true. Then

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P(A) = P(A|B) =

P(A ↔ B) , P(B)

and if we multiply both sides by P(B) this gives P(A) P(B) = P(A ∩ B), which says that equation (3) is true. By doing the same computation in reverse we can see that if equation (3) is true then equation (1) must be true. So (1) and (3) are equivalent. In exactly the same way we could check that (1) and (2) are equivalent.

Definition: Independent events Two events A and B with P(A) ≠ 0 and P(B) ≠ 0 are independent if P(A ∩ B) = P(A) P(B). Furthermore, itʼs easy to prove that the following conditions are all equivalent. (If one of them is true, all are true.) So you can use any one of the three conditions to check whether two events are independent. 1. P(A | B) = P(A) 2. P(B | A) = P(B) 3. P(A ∩ B) = P(A) P(B)

The important thing for you to remember in applications, however, is that you don’t have to check all three conditions. Checking one of them is enough, and it doesn’t matter which one you check. Example 5.34. Two cards are dealt from a 52-card deck. Consider the events A = event that the first card dealt is a heart B = event that the second card dealt is a heart In this case P(B|A) = 12/51, whereas P(B) = 1/4. If you’re in doubt that P(B) = 1/4, then you need to remember what the tree diagram for this experiment looks like: 1/17

13/68

13/68

19/34

H

c H

H

c H

12/51

39/51 13/51

H 13/52

c H

38/51

39/52

The two outcomes in the tree that show a heart on the second card have probabilities 1/17 and 13/68. The sum of these is 17/68 = 1/4. The way to understand this intuitively is to remember that P(B) means the probability that the second card is a heart given no information at all about the first card. If we don’t know anything about

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the first card, then we can think of the second card as equally likely to be any one of the 52 cards which includes 13 hearts, i.e. P(B) = 13/52 = 1/4. This way of thinking is correct in spite of the fact that when the second card is actually dealt there are only 51 cards in the deck. In this example A and B are not independent since P(B) ≠ P(B | A). Example 5.35. A card is dealt from a 52-card deck. The card is replaced in the deck, the deck is reshuffled, and a second card is dealt. Let A = event that the first card dealt is a heart B = event that the second card dealt is a heart. In this scenario A and B are independent events. The reason is that since the first card is replaced in the deck before the second is dealt, whether the first card is a heart has no effect on whether the second is. In other words P(B) = P(B|A) = 1/4. This is an example of what statisticians call sampling with replacement. Example 5.35 is a simple illustration of an important phenomena called an independent trials process. Imagine a situation such as this:

Definition: Independent trials process 1. An experiment is repeated a number of times. 2. Each time the result of the experiment is one of two outcomes which we will call success and failure. 3. The probability of success (which weʼll call p) remains the same every time we conduct the experiment, and the outcome of one experiment does not affect the outcome of another

Here is a tree diagram S F S F S F S F S F S F S F S F showing all possible results p q p q p q p q p q p q p q p q for four trials of an F F F S S S F S independent trials process. q p (The number of times the p q p q p q experiment is repeated is S F S F commonly referred to as p q p q the number of trials.) In F S the figure the failure probability for each trial is p q denoted by q. Of course it is necessary that p + q = 1. Now let’s think about this question: What is the probability that we get exactly 3

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successes in the 4 trials shown above? To determine this we first locate the outcomes that show exactly three successes occurring during the four trials. These outcomes correspond to the ones circled in the following picture. S

F

S

p

q p

F S

F S

q p

q p

F

S p

p

S

F

S

F S

F S

q

q

p

q p

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q

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q

p

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p

q

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p

q

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F

p

q

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F p

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S p

q

Notice that the probability of each one of these four outcomes is p3q. The reason this is true is that all these four outcomes involve 3 successes and 1 failure, and therefore we pick up p as a factor 3 times and q once. And notice why there are four circled outcomes in which exactly 3 successes occur. It’s because there are exactly 4 ways to get 3 successes in four trials, i.e. C(4,3) = 4. Summary: P(3 successes in 4 trials) = C(4,3) p3q.

Independent trials formula In an independent trials process where the probability of success is p and the probability of failure is q = 1 – p, the probability of exactly k successes in n trials is given by P(k successes in n trials) = C(n,k) pkqn–k. This is called the binomial probability distribution. It is useful any time you need to find the probability of a certain number of successes occurring in a fixed number of trials.

Example 5.36. Karen is a basketball player who is an 80% free throw shooter. If she attempts 6 free throws in a game, what is the probability she will hit exactly 4? What is the probability she will hit at least 4? Solution: Here we have n = number of trials, p = success probability = .8, q = failure probability = .2, and

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k = number of shots hit = 4. So P(4 hits in 6 attempts) = C(6,4) p4q2 ≈ .246. The probability of at least four hits in 6 shots is given by P(at least 4 hit) = P(4 hit) + P(5 hit) + P(6 hit) = C(6,4) p4q2 + C(6,5) p5q1 + C(6,6) p6q0 = 15 (.8)4 (.2)2 + 6 (.8)5(.2) + (.8)6 ≈ .881.

Problems 1.

Suppose A and B are independent events with P(A) = 1/3 and P(B) = 1/6. (a) Find P(A ∩ B) (b) Find P(A ∪ B) (c) Find P(A ∩ Bc)

2. A nickel and a dime are tossed. A, B, C, and D are the following events: A = event that the nickel comes up heads B = event that the dime comes up heads C = event that the nickel comes up tails D = event that the two coins show the same thing (heads or tails) (a) Evaluate P(C). (b) Evaluate P( C ∩ D). Answer TRUE or FALSE. (c) A and B are independent. (d) A and B are mutually exclusive. (e) A and C are independent. (f) A and C are mutually exclusive. (g) A and D are independent. (h) A and D are mutually exclusive. 3.

Lisa is taking a 10-question multiple choice test in which each question has 4 answers including just one correct answer. If Lisa answers every question by random guessing, what is the probability she will answer exactly 3 of the questions correctly?

4.

(a) If a coin is tossed 7 times, what is the probability of getting exactly 3 heads? (b) If a die is rolled 4 times, what is the probability of getting exactly 2 sixes? (c) If 3 cards are drawn from a standard 52-card deck, what is the probability of getting exactly 2 spades? [None of the cards are replaced in the deck.]

5.

Two cards are drawn one at a time from a 52-card deck. A = event that the first card drawn is red

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Chapter 5: Probability B = event that the second card drawn is red C = event that the first card drawn is a heart (a) Are A and B independent, mutually exclusive, or neither? (b) Are A and C independent, mutually exclusive, or neither? (c) Are B and C independent, mutually exclusive, or neither?

6.

If A and B are independent with P(A) = .4 and P(Bc) = .6, find P(A ∪ B).

7.

In a large city, 60% of the voters are Democrats. If a pollster talks to 8 voters, what is the probability that at least 6 of them are Democrats?

8.

Isaac is taking a 5-question multiple choice test in which each question has 4 answers including just one correct answer. He needs to answer at least 4 of the questions correctly in order to pass the test. What is the probability that he will pass?

9.

Dork Baker is a baseball player with a .300 batting average, i.e. he gets a hit 30% of the times he bats. (a) If he has 5 “at bats” in a game, what is the probability he will get at least one hit? (Treat this as an independent trials process.) (b) How many times would he have to come to bat in a game in order that he would have at least an 90% chance of getting a hit?

10. Under what conditions is it possible for two events to be independent and mutually exclusive? 11. Tom and Alice independently attempt to solve a certain problem. The probability that Tom solves it is 1/10 and the probability that Alice solves it is 1/3. (a) What is the probability that the problem will be solved? (b) If the problem is solved, what is the probability that Tom solved it? 12. Suppose that airplane engines operate independently and fail with probability 1/5. A plane can fly if at least half its engines work. (a) What is the probability that a plane with four engines will crash? (b) Would you be better off flying in a 2-engine plane? 13. 2/3 of the students at N.C. State do not get married during their undergraduate career. In a suite with 15 students, what is the probability that a majority of them will be single when they graduate? 14. An insurance salesman sells policies to his 5 brothers-in-law who are all 30 to 40 years old and in good health. Actuarial tables show that a man 30-40 years old has an 80% chance of living at least another 25 years. What is the probability that at least 3 of the 5 men will be alive in 25 years? 15. An wine store receives a shipment of 100 corkscrews from a certain manufacturer.

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On average, 10% of the corkscrews made by this manufacturer are defective. The manager of the wine store decides to test 8 of the corkscrews and to send back the whole shipment if any of the 8 are found to be defective. What is the probability she sends the corkscrews back? 16. In Problem 2 of Section 5.5, let A denote the event that 2 t-shirts are pulled out of the drawer, and let B be the event that the top drawer is the one chosen. Are A and B independent events? Why or why not? 17. A machine produces balloons that are good with probability .99 and defective with probability .01. What is the probability that a package of 12 balloons contains no more than two defective balloons? 18. A machine produces parts which are good 90% of the time and defective 10% of the time. Assuming that parts produced by the machine are independent trials, (a) what is the probability that a lot of 10 parts contains exactly one defective part? (b) what is the probability that a lot of 10 parts contains at most 2 defective parts?

5.7 Expected Value Everyone is familiar with various kinds of “averaging” processes. For example, your grade in this course will undoubtedly consist of some kind of “average” based on your grades throughout the semester. Expected value is a special kind of average that is important in probability. Example 5.37. When an ordinary six-sided die is rolled, the average of the numbers on the die is given by 1 21 6 × (1 + 2 + 3 + 4 + 5 + 6) = 6 = 3.5. Notice that this average can also be written as 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6). Here we have each of the possible outcomes {1,2,3,4,5,6} multiplied times its probability of occurring, namely 1/6 in each case. If you rolled the die a large number of times and each outcome appeared exactly 1/6 of the time, then the average for all the rolls would be exactly 3.5. This would be a case in which each outcome had appeared exactly as often as its theoretical probability predicts. Such exactness is not likely to occur in an actual experiment, but you should understand that 3.5 represents the theoretical average. Example 5.38. A card is dealt from a well-shuffled deck. Either the card is or is not a spade, so the number of spades dealt is 0 or 1. If we are interested in the question,

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Chapter 5: Probability

“How many spades have been dealt?”, the answer will have to be 0 or 1. Let’s consider this to be the sample space for this simple experiment: S = {0,1}. Notice that P(0 spades) = 3/4, and P(1 space) = 1/4 (since 1/4 of the cards in the deck are spades). In a large number of trials of this experiment, if the outcome “0 spades” occurred 3/4 of the time and the outcome “1 spade” occurred 1/4 of the time, then the average number of spades would be 0(3/4) + 1(1/4) = 1/4. This says that “on the average” you will get 1/4 spade per trial of this experiment.

Definition: Expected value Suppose an experiment is conducted in which the outcomes are the numbers {a1,a2,a3,...,an}, and suppose that the probability of outcome a1 is p1, the probability of a2 is p2, etc. Then the expected value for the experiment is E = a1p1 + a2p2 + a3p3 + ... + anpn .

Example 5.39. Michelle and Kim are playing a tennis match. The winner will be the first person to win two sets. Michelle is the better player, and in fact whenever they play a set the probability is 2/3 that Michelle will win the set. (a) What is the expected value for the number of sets Michelle will win? (b) What is the expected value for the number of sets that will be played in the match? Solution: The tree diagram for the tennis match was constructed in Example 5.30 and looks like this: 4/27

2/27

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12/27 3/27

1/3

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4/27

2/27

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(a) A glance at the tree (or a little common sense) tells us that the possibilities for the number of sets that Michelle wins are 0, 1, or 2. From the tree we get that

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P(Michelle wins 0 sets) = 3/27 P(Michelle wins 1 set) = 4/27 P(Michelle wins 2 sets) = 20/27 Therefore the expected value for the number of sets that Michelle will win is given by E = 0 × 3/27 + 1 × 4/27 + 2 × 20/27 = 44/27 ≈ 1.630 sets. (b) The tree shows two outcomes in which the match lasts only for two sets, and their probabilities are 12/27 and 3/27. Therefore P(match lasts 2 sets) = 15/27. And P(match lasts 3 sets) = 12/27. So the expected value for the number of sets played is given by E = 2 × 15/27 + 3 × 12/27 = 66/27 = 22/9 ≈ 2.444 sets. Example 5.40. Three good light bulbs and two defective ones have become mixed in a box. In order to find which are which, bulbs are going to be withdrawn from the box one at a time and tested until both bad ones or all three good ones have been found. (a) What is the expected value for the number of bulbs that will have to be tested? (b) What is the expected value for the number of good bulbs that will be tested? Solution: (a) The tree diagram for this testing process is found in Example 5.31. From the tree it is easily seen that either 2, 3, or 4 bulbs will be tested. Furthermore, from the tree we obtain that P(2 bulbs are tested) = 1/10 P(3 bulbs are tested) = 3/10 P(4 bulbs are tested) = 6/10. Therefore the expected value for the number of bulbs to be tested is given by E = 2 × 1/10 + 3 × 3/10 + 4 × 6/10 = 35/10 = 3.5. (b) The number of good bulbs found during the testing procedure will be either 0, 1, 2, or 3. The probabilities (obtained from the tree diagram) are as follows: P(0 good bulbs tested) = 1/10 P(1 good bulb tested) = 2/10 P(2 good bulbs tested) = 3/10 P(3 good bulbs tested) = 4/10. Therefore the expected value for the number of good bulbs tested is given by E = 0 × 1/10 + 1 × 2/10 + 2 × 3/10 + 3 × 4/10 = 20/10 = 2. Example 5.41. A coin is tossed 4 times. What is the expected value for the number of heads obtained? Solution: Let’s consider this question in light of the independent trials formula. If we consider a “heads” to be a success, then counting the number of heads in four tosses is equivalent to counting the number of successes in four trials with the probability being 1/2 for success in each trial. From the independent trials formula this means that

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Chapter 5: Probability P(0 heads) = C(4,0) (1/2)0 (1/2)4 = 1/16 P(1 head) = C(4,1) (1/2)1 (1/2)3 = 4/16 P(2 heads) = C(4,2) (1/2)2 (1/2)2 = 6/16 P(3 heads) = C(4,3) (1/2)3 (1/2)1 = 4/16 P(4 heads) = C(4,0) (1/2)4 (1/2)0 = 1/16. Therefore the expected value for the number of heads in the four tosses is given by E = 0 × 1/16 + 1 × 4/16 + 2 × 6/16 + 3 × 4/16 + 4 × 1/16 = 32/16 = 2.

This doesn’t seem very surprising, does it? Surely the “average” number of heads in four coin tosses should be equal to two. Isn’t there some simpler arithmetic that would lead to this answer? The answer is yes, and the relevant fact is the following:

Expected Value and Independent Trials If n independent trials are conducted of an experiment for which the success probability is p on each trial, then the expected number of successes to occur during the n trials is given by E = np.

Example 5.42. Remember the basketball player named Karen who is an 80% free throw shooter and who attempted six shots in a game? (See Example 5.36.) What would be the expected value for the number of shots she hits in the six attempts? Solution: Here we have an independent trials process with n = 6 and p = .8. So the expected number of shots that she would hit would be E = np = 6 × .8 = 4.8.

Problems 1.

Edith is throwing darts at a target. On each throw she has a 40% chance of hitting the target. (a) What is the probability she hits exactly 2 of the 5 throws? (b) In 5 throws, what is the expected number of times she will hit?

2.

Mark is a 70% free throw shooter in basketball. If he goes to the foul line to shoot a “one-and-one”, (a) what is the expected number of points he will score?

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(b) what is the expected number of shots he will take? [Note: A one-and-one means that he shoots once and if he hits he gets another shot.] 3.

Three male and three female patients are seated in a doctor’s waiting room. Patients are randomly called in to see the doctor. As soon as she has treated two of the female patients, the doctor plans to take a coffee break. (a) Draw a tree diagram to represent the possibilities for the patients treated before the doctor’s coffee break. (b) What is the expected number of patients the doctor will treat before her coffee break? (c) What is the expected number of male patients the doctor will treat before her coffee break? [Can you see the reason for the relationship between this answer and the answer to part (b)?]

4.

Four eggs are left in a carton. Two of them are rotten. You start cracking the eggs one at a time until you have cracked both rotten eggs or both good eggs. (a) Draw a tree diagram to represent this process. (b) What is the probability you have to crack 3 eggs? (c) What is the expected value for the number of eggs you will crack? (d) What is the expected value for the number of rotten eggs that will be cracked?

5.

Each day that Sue drives to NCSU, there is a 60% chance that she will have to stop at the Western Boulevard stoplight. If she drives to the university 5 days in a given week, (a) what is the probability she will get caught by the light exactly 4 times? (b) what is the probability she will get caught by the light at least 4 times? (c) what is the expected value for the number of times she will have to stop?

6.

Sam buys a raffle ticket for $1. 1000 of the tickets are to be sold. First prize is a TV valued at $350, second prize is a tape deck worth $115, and third prize is a camera worth $40. (a) What is the expected value of the raffle ticket? (b) If Sam buys 5 tickets, what is the probability he wins a prize?

7.

The number of traffic accidents in Fuquay-Varina during rush hour on Friday afternoons is 0, 1, 2, or 3 with probabilities .94, .03, .02, and .01 respectively. (a) Find the expected number of accidents during the Friday afternoon rush hour. (b) How many accidents on Friday afternoons would you expect during a year? (Hint: How many Friday afternoons are there during a year?)

8.

Bill’s dresser drawer contains 7 blue socks and 3 white ones. He reaches into the drawer and pulls out socks one at a time until he has two that match. (a) Draw a tree diagram for this process. (b) What is the probability he gets a pair of white socks? (c) What is the probability he will get a match with the first two socks? (d) What is the expected number of socks he will pull from the drawer?

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9. A student is trying to pass a competency exam in French. Each time she takes the exam she has a 25% chance of passing, and she is allowed a maximum of three attempts. (a) Draw a tree diagram to represent her attempts to pass the exam. (b) What is the probability she will eventually pass the exam? (c) What is the probability she will take the exam three times? (d) What is the expected number of times she will take the exam? (e) What is the expected number of times she will fail the exam? 10. If 2 cards are dealt from a standard deck of 52 cards, what is the expected number of spades dealt? 11. 3 men and 4 women are at a small party at which 3 door prizes will be awarded. Nobody will receive more than one prize. (a) Find the probability that none of the prizes is won by a man. (b) Find the probability that exactly one of the prizes is won by a man. (c) Find the probability that exactly two of the prizes are won by men. (d) Find the probability that all three prizes are won by men. (e) What is the expected value for the number of prizes to be won by men? 12. In Problem 13 of Section 5.4, (a) what is the expected value for the number of mints chosen? (b) what is the expected value for the number of candies chosen that are wrapped in red foil? 13. Paul has three $5 bills and three $10 bills in his pocket. Nick has four $1 bills and two $20 bills in his pocket. Each person pulls out a bill from his own pocket. What is the expected value for the amount of money that Paul will receive in each of the following situations? (a) The person that pulls out the lower amount pays the other person $5. (b) The person that pulls out the lower amount pays the other person a number of dollars equal to the difference between the two amounts. (c) The person that pulls out the lower amount pays the other person the amount of money shown on the bill pulled out by the other person. 14. Tanya wants to pass a qualifying exam, and she is allowed 3 attempts at passing the exam if necessary. Each time she takes the exam there is a 40% chance she will pass. What is the expected number of times she will take the exam? [Hint: Draw a tree.]

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Chapter 5 Review Problems 1.

John, Ed, Ann, and Barbara are lining up in a random order to buy tickets to a movie. What is the probability that they will line up in alphabetical order?

2.

John and Jerry are 2 applicants among 40 applicants at a company that plans to hire 10 people. If the people chosen are randomly selected from among the 40 applicants, (a) what is the probability that John will get a job? (b) what is the probability that both John and Jerry will get jobs? (c) what is the probability that neither will get a job?

3.

The board of directors authorizes the bank where Sam works to promote 3 of their 10 junior executives to the rank of vice-president. The promotions will be random. (a) What is the probability that Sam will be promoted? (b) What is the probability that Sam won’t be promoted but that his arch rival Carl will be?

4.

To play a particular card game, a player is dealt 2 cards from an ordinary 52-card deck. (a) Disregarding the order in which the cards are dealt, how many different hands are possible? (b) What is the probability that a given hand (of 2 cards) contains 2 spades? (c) What is the probability that the player will be dealt two cards of the same suit?

5.

There are 4 seasons (spring, summer, winter, fall). Assume that people are equally likely to be born in each of the seasons. Find the probability that in a group of 3 people, at least 2 were born in the same season.

6.

Four salesmen play “odd man out” to see who pays for lunch. They each flip a coin, and if there is a salesman who doesn’t match the others he pays for lunch. (For instance, he might get “heads” while the other three get “tails.”) What is the probability that there is an “odd man” the first time they flip?

7.

An ordinary die is rolled twice. (a) What is the probability that the sum of the two numbers obtained is equal to 9? (b) What is the probability the sum is 9 if you know that the first roll showed a 5? (c) What is the probability the sum is 9 if you know that the second roll showed a 5? (d) What is the probability the sum is 9 if you know that at least one of the rolls showed a 5? (e) What is the probability the sum is 9 if you know that the first roll showed a 2? (f) What is the expected value for the sum?

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8.

Of a group of students applying to a particular college, 55% were female and 45% were male. Of the female applicants 40% were accepted, and of the male applicants 35% were accepted. (a) What percentage of applicants were accepted? (b) If someone tells you that her cousin applied and was accepted, what is the probability that her cousin is female?

9.

A submarine is firing torpedoes at a ship. Each has probability .6 of hitting the ship. How many must be fired in order to be 95% certain of scoring a hit?

10. A basketball player hits 70% of his free throws. If his attempts are considered independent trials and he makes 5 attempts in a game, (a) what is the probability he hits exactly 4 shots? (b) what is the probability he hits at least 4 shots? (c) what is the expected value for the number of shots he hits? 11. In a box containing 5 pieces of chocolate candy that are all wrapped alike, Kay knows that there are 2 pieces that she likes. She is determined to keep opening pieces of candy until she finds these 2 pieces. (a) Draw a tree for this process of opening pieces of candy. (b) What is the probability that she opens exactly 4 pieces of candy? (c) What is the probability that she opens exactly 4 pieces if we know that she didn’t like the first piece she opened? (d) If she opened exactly 3 pieces to find the 2 that she liked, what is the probability that she liked the first piece she opened? (e) What is the expected number of pieces for her to open? 12. Two cards are drawn without replacement from a deck of cards. probability (a) both cards are face cards (kings, queens, and jacks)? (b) both cards are face cards given that at least one is a face card? (c) both are face cards given that at least one of the cards is a king?

What is the

13. Fred and Jerry like to go to trap shoots together. There are 3 events at a trap shoot, the 16-yard event, handicap, and doubles. Fred beats Jerry half the time in the 16yard event, 2/3 of the time in the handicap event, and 3/5 of the time in the doubles event. If the results in the 3 events are independent, what is the probability that at a particular trap shoot (a) Fred beats Jerry in all three events? (b) Fred beats Jerry in at least two events? (c) Fred beats Jerry in at least two events given that Fred beats Jerry in the 16-yard event? 14. Susan has 2 black pens and 3 red pens in her purse. She pulls out pens one at a time without replacement until she has a pen of each color.

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(a) Draw the probability tree diagram. (b) What is the probability she pulls out at least 3 pens? (c) What is the probability she pulls out at least 3 pens given the first pen she chooses is red? (d) What is the probability the first pen she chooses is red given that she pulls out at least 3 pens? (e) Let A be the event “the first pen chosen is red” and let B be the event “at least 3 pens are chosen”. Are A and B independent? Why or why not? (f) What is the expected number of pens she chooses? (g) What is the expected number of red pens she chooses? 15. You have to answer 5 multiple-choice questions. Each question has 4 different answers, only one of which is correct. If you answer them by randomly guessing, (a) what is the probability you will get all 5 correct? (c) what is the probability you will get at least 4 correct? (d) what is the expected value for the number you will get correct?