Discrete Probability Distributions [PDF]

Discrete Probability Distributions



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5.1

Introduction Most of us will have met the idea of a frequency distribution. We collected some data, subdivided it into classes and the frequency (or number of items) for each class was noted. Once a frequency distribution is constructed there are two particular statistics that characterise the distribution, the mean value and the variance. Because most engineering phenomena are subject to random influences, their ‘output’ is random in nature: more precisely, their output is a random variable. In order to model such phenomena, we use a probability distribution. Using the theoretical probability model we predict a corresponding frequency distribution which can then be compared with observation. In this block we examine discrete probability distributions where the values of the random variable may be written as a list. As with frequency distributions we find that a probability distribution can also be characterised by the mean and variance.

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Prerequisites

① understand the concepts of probability

Before starting this Block you should . . .

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Learning Outcomes After completing this Block you should be able to . . .

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Learning Style To achieve what is expected of you . . .

☞ allocate sufficient study time ✓ explain the term discrete probability distribution ✓ find the mean and variance of a discrete probability distribution

☞ briefly revise the prerequisite material ☞ attempt every guided exercise and most of the other exercises

1. Discrete Probability Distribution Consider the experiment of throwing two coins. With the usual notation the sample space for this experiment is: S = {HH, HT, T H, T T }. A random variable is a rule which assigns a number to each member of S and is denoted by a capital letter. We can choose this rule as we please. For example, (i) X: “number of heads which appear.” (ii) Y : “has the value 1 if both throws are the same and 0 otherwise.” and so on. With each value of the random variable we can associate a probability. To see how this is done we again consider the random variable X introduced above. This random variable has only three values: X = 0 (no heads appear), X = 1 (one head appears), X = 2 (both throws give a head). The probability that X has the value zero is written P (X = 0) and: P (X = 0) = P ({T T } occurs) =

1 4

Similarly 1 2 1 . P (X = 2) = P ({HH}) = 4

P (X = 1) = P ({HT } ∪ {T H}) =

We can record these values of X and the associated probabilities in tabular or in graphical form. See Figure 1. P (X= x) x P (X= x)

0 1 4

1 1 2

2 1 4

1 4

1 2

0

1

2

x

Figure 1 Either of these forms is referred to as the probability distribution of X. Try each part of this exercise Obtain the probability distribution of the random variable Y introduced above. Part (a) First find P (Y = 0) and P (Y = 1) Answer Part (b) Now construct the probability distribution Answer Engineering Mathematics: Open Learning Unit Level 1 5.1: Probability

2

In Figure 2 we show the probability distribution for the experiment of throwing a fair die once. Each of the six outcomes x has probability 16 . This is an example of a uniform distribution. P (X= x ) x P (X= x )

1

2

1 6

1 6

3 1 6

4 1 6

5 1 6

6 1 6

1 6 1 2 3 4 5 6

x

Figure 2 Now do this exercise A fair coin is thrown three times. List the equally likely outcomes. Calculate the probabilities of the events {no heads appear}, {exactly one head appears}, {two heads appear}, {three heads appear}. Answer You may have observed that the probabilities associated with a random variable satisfy the usual constraints imposed on probabilities: each is non-negative, none has a value greater than one and the sum of all the probabilities in the distribution is equal to one. Now do this exercise Consider the last guided exercise in which a fair coin is thrown three times. Let X be the random variable ”number of heads”. Display the probability distribution for the experiment above as (i) a table (ii) a graph. Answer Key Point Let X be a random variable associated with an experiment. Let the values of X be denoted by x1 , x2 , . . . , xn and let P (X = xi ) be the probability that xi occurs. We have two necessary conditions for a valid probability distribution • P (X = xi ) ≥ 0 for all xi •

n 

P (X = xi ) = 1

i=1

(These two statements are sufficient to guarantee that P (X = xi ) ≤ 1 for all xi )

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Engineering Mathematics: Open Learning Unit Level 1 5.1: Probability

Now do this exercise Explain why neither of the following represents a probability distribution P(X= x)

(i)

2 5

(ii)

xi 1 2 3 4 5 P (xi ) 0.3 0 0.4 −0.1 0.4

1 5 1 2 3 4

x

Answer

2. Mean and Variance of a discrete distribution One method of comparing the theoretical probability distribution with the experimental frequency distribution is to multiply the probabilities by the number of data values to create a set of expected frequencies. The following data was collected for the throwing of three coins 100 times. Number of heads 0 1 2 3 Frequency 11 37 39 13 The probability of obtaining no heads in three throws is 18 and in 100 throws we would expect on that basis 100× 18 = 12.5 heads. The probability of obtaining one head is 38 and in 100 throws we would expect 100 × 38 = 37.5. We can calculate the expected number of times of obtaining two heads and of obtaining three heads. We can therefore complete the table as follows Number of heads Observed Frequency Expected Frequency

0 1 2 3 11 37 39 13 12.5 37.5 37.5 12.5

We see that there is quite a good agreement between theory and experiment. A second way of comparing a theoretical probability distribution to an experimental frequency distribution is to compare their means and variances. Consider performing an experiment s in which distinct observations x1 , x2 , . . . , xs with frequencies f1 , f2 , . . . , fs are made. If n = i=1 fi , is the total number of observations then the quantity fni is called the relative frequency of the observation with value xi . Relative frequencies are akin to probabilities: informally, we would say that the chance of observing xk is fnk . This observation leads to the following keypoint: Key Point Relative frequencies and probabilities If a frequency distribution F is modelled by a probability distribution X the probabilities arising in X correspond to the relative frequencies of F Engineering Mathematics: Open Learning Unit Level 1 5.1: Probability

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For the frequency distribution the mean is defined by f1 x1 + f2 x2 + . . . + fs xs f1 + f2 + . . . + fs s  fi xi s 1 i=1 = = fi xi s  n i=1 fi

µ =

i=1

s    fi

=

n

i=1

xi

This leads us to define the corresponding quantity for a random variable. Key Point The expectation value of a random variable Let X be a random variable with values x1 , x2 , . . . , xn . Let the probability that X takes the value xi (i.e. P (X = xi )) be denoted by Pi . The expected value of X, which is written E(X) is defined as: n  E(X) = Pi x i = P 1 x 1 + P 2 x 2 + · · · + P n x n i=1

E(X) is sometimes denoted by the symbol µ. If X is the random variable “score observed when a die is thrown” then, as we have seen, its probability distribution is xi Pi Here E(X) =

6  i=1

Pi x i =

1

2

3

4

5

6

1 6

1 6

1 6

1 6

1 6

1 6

1 1 1 21 × 1 + × 2 + ... + × 6 = = 3.5 6 6 6 6

At first sight this might appear to be a strange result since no single throw of a die can produce 3.5. However, if we carry out the experiment a large number of times and average the score obtained this average should be close to 3.5 if the die is fair; also, the agreement between experimental average and the theoretical expectation value improves as the number of throws increases. Now do this exercise Find the expected value of the number of heads when three coins are thrown. Answer

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Engineering Mathematics: Open Learning Unit Level 1 5.1: Probability

The variance of a random variable Whereas the mean value of a frequency distribution is a measure of its centre, the variance is a measure of its spread. Once the variance is known the standard deviation σ is then found by taking the square root. (The standard deviation is sometimes used in preference to the variance since it has the same dimensions as the underlying data). The usual formula for calculating the variance, σ 2 , of a frequency distribution is: 1 σ = fi (xi − µ)2 n i=1 n

2

As a rough guide: the further away the data values are from the mean µ the larger will be the variance. The variance is often written in an alternative form which is obtained by expanding the square (xi − µ)2 and simplifying:  n   1 σ2 = fi x2i − µ2 n i=1 This is often quoted in words: The variance is equal to the mean of the squares minus the square of the mean. We now extend the concept of variance to a random variable. Key Point The variance of a random variable Let X be a random variable with values x1 , x2 , . . . , xn . The variance of X, which is written V (X) is defined by n  V (X) = Pi (xi − µ)2 i=1

where µ ≡ E(X). We note that V (X) can be written in the alternative form V (X) = E(X 2 ) − [E(X)]2  The standard deviation σ of a random variable is then V (X). Now do this exercise Find the variance and standard deviation of the number of heads in the three-coin experiment. Refer to the previous guided exercise for the probability distribution. Answer

Engineering Mathematics: Open Learning Unit Level 1 5.1: Probability

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More exercises for you to try 1. A random number generator produces sequences of independent digits, each of which is as likely to be any digit from 0 to 9 as any other. If X denotes any single digit find E(X). 2. A hand-held calculator has a clock cycle time of 100 nanoseconds; these are positions numbered 0, 1, . . . , 99. Assume a flag is set during a particular cycle at a random position. Thus, if X is the position number at which the flag is set. P (X = k) =

1 100

k = 0, 1, 2, . . . , 99.

Evaluate the average position number E(X), and σ, the standard deviation. (Hint: The sum of the first k integers is k(k + 1)/2 and the sum of their squares is: k(k + 1)(2k + 1)/6.) 3. Concentric circles of radius 1 cm and 3 cm are drawn on a circular target radius 5 cm. A darts player receives 10, 5 or 3 points for hitting the target inside the smaller circle, middle annular region and outer annular region respectively. The player has only a 50-50 chance of hitting the target at all but if he does hit it he is just as likely to hit any one point on it as any other. If X = ‘number of points scored on a single throw of a dart’ calculate the expected value of X. Answer

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Engineering Mathematics: Open Learning Unit Level 1 5.1: Probability

End of Block 5.1

Engineering Mathematics: Open Learning Unit Level 1 5.1: Probability

8

P (Y = 0) = 12 ,

P (Y = 1) = P ({HH} ∪ {T T }) =

1 2

Back to the theory

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Engineering Mathematics: Open Learning Unit Level 1 5.1: Probability

P (Y = y ) y P ( Y= y )

0

1

1 2

1 2

1 2 0 1

y

Back to the theory

Engineering Mathematics: Open Learning Unit Level 1 5.1: Probability

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The outcomes are {HHH, HHT, HT H, T HH, HT T, T HT, T T H, T T T } P (no heads)= 18 , P (one head) = 38 P (two heads)= 38 , P (three heads)= 18 Back to the theory

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Engineering Mathematics: Open Learning Unit Level 1 5.1: Probability

P(X= x ) 3 8

x

0

1

P(X= x )

1 8

3 8

2 3 8

3 1 8

1 8 0 1 2 3

x

Back to the theory

Engineering Mathematics: Open Learning Unit Level 1 5.1: Probability

12

(i)

4 

Pi = 1. The sum of four values P (xi ) is 65 .

i=1

(ii) In this case the sum of the probabilities is 1, but not all P (xi ) are non-negative Back to the theory

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Engineering Mathematics: Open Learning Unit Level 1 5.1: Probability

x P (X = x) E(X) =

1 8

× 0 + 38 × 1 + 38 × 2 + 18 × 3 =

12 8

0

1

2

3

1 8

3 8

3 8

1 8

= 1.5

Back to the theory

Engineering Mathematics: Open Learning Unit Level 1 5.1: Probability

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1 3 3 1 × 02 + × 12 + × 22 + × 32 8 8 8 8 1 3 3 1 = ×0+ ×1+ ×4+ ×9=3 8 8 8 8 3 V (X) = 3 − 2.25 = 0.75 = 4 √ 3 s.d. = 2 pi x2i =

Back to the theory

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Engineering Mathematics: Open Learning Unit Level 1 5.1: Probability

1.

x P (X = x)

1

0 /10

1

1 /10

1

2 /10

1

E(X) =

3 /10

1

4 /10

1

5 /10

1

6 /10

1

7 /10

1

8 /10

1

9 /10

1 {0 + 1 + 2 + 3 + . . . + 9} = 4.5 10

2. Same as Q.1 but with 100 positions



1 1 99(99 + 1)) E(X) = {0 + 1 + 2 + 3 + . . . + 99} = = 49.5 100 100 2

σ 2 = mean of squares - square of means 1 2 ∴ σ2 = [1 + 22 + . . . + 992 ] − (49.5)2 100 1 [99(100)(199)] = − 49.52 = 833.25 100 6 3. X can take 4 values 0, 3, 5 or 10 P (X = 0) = 0.5

[only 50/50 chance of hitting target.]

The probability that a number is hit is related to the areas of the annular regions which are, from the centre: π, (9π − π) = 8π, (25π − 9π) = 16π P (X = 3) = P [(3 is scored) ∩ (target is hit)] = P (3 is scored | target is hit).P (target is hit) =

8 16π 1 . = 25π 2 25

P (X = 5) = P (5 is scored | target is hit).P (target is hit) =

8π 1 4 . = 25π 2 25

P (X) = 10) = P (10 is scored | target is hit).P (target is hit) =

1 π 1 . = 25π 2 50

Engineering Mathematics: Open Learning Unit Level 1 5.1: Probability

16

x P (X = x)

25

0 /50

16

3 /50

8

5 /50

10 /50

1

∴

E(X) =

48+40+10 50

= 1.96.

Back to the theory

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Engineering Mathematics: Open Learning Unit Level 1 5.1: Probability