Idea Transcript
N.
Name: MATH1215: Mathematical Thinking Sec. 08 Spring 2011 Worksheet 5: March 12, 2011
Problem 1 Write the following rules: (a) Union rule for probability: P(E ∪ F ) = P(E) + P(F ) − P(E ∩ F ) (b) Complement rule for probability: P(E 0 ) = 1 − P(E) (c) Product rule for probability (use conditional probability): P(E ∩ F ) = P(E)P(F |E) = P(F )P(E|F ) (d) Product rule for independent events: P(E ∩ F ) = P(E)P(F ) Problem 2 Complete the following definitions: (a) Events E and F are mutually exclusive events if E ∩ F = 0, that is, P(E|F ) = 0 (equivalently, P(F |E) = 0). (b) Events E and F are independent if P(E|F ) = P(E) (equivalently P(F |E) = P(F )). Problem 3 Circle the correct answer: (1) Let E and F be mutually exclusive events. Which of the following are true? (i) E and F are independent events. (ii) P(E ∩ F ) = 0. (iii) P(E ∪ F ) = P(E) + P(F ). A. (i)
B. (i) and (ii)
C. (ii) and (iii)
D. (ii) only
E. All of them.
(2) P(E|F ) =? A.
P(E) P(F )
B.
P(F ) P(E)
C.
P(E ∩ F ) P(F )
D.
P(E ∩ F ) P(E)
E. P(E ∩ F )P(E)
(3) Suppose that E and F are independent and P(E) = 0.3, P(F ) = 0.5. What is P(E|F )?
A. 0.3
B. 0
C. 0.15
D. 0.6
E. Insufficient information
(4) Suppose that E and F are two independent events and P(E) = A.
4 10
B.
7 10
C.
2 10
D.
4 10 ,
P(F ) =
9 10
E.
5 10 .
What is P(E ∪ F )?
8 10
Indeed, using union rule and the fact that E and F are independent, P(E ∪ F ) = P(E) + P(F ) − P(E)P(F ) =
4 5 20 7 + − = 10 10 100 10
(5) If the odds in favor of an event E are 2 to 3 then P(E 0 ) is equal to: 1
A.
2 3
B.
2 5
D.
C. 1
3 5
E.
1 3
Indeed, since m = 2, n = 3, P(E 0 ) =
n 3 3 = = m+n 2+3 5
Problem 4 For two events A and B, we know that (i) P(A) =
7 10
(ii) P(B 0 ) = 0.3 (iii) P(A0 ∩ B) =
1 5
Using a Venn diagram, find the following probabilities: (a) P(A ∪ B) =
9 10
(b) P(A0 ∪ B 0 ) = (c) P(A|B 0 ) =
1 2
P(A∩B 0 ) P(B 0 )
=
2 10 3 10
=
2 3
Indeed, the Venn diagram is given by: t
U A y
x
B z
The information we have correspond to: (i) x + y = 0.7 (ii) P(B) = 0.7, then x + z = 0.7 (iii) z = 0.2 So, we get x = 0.5, y = 0.2 and t = 1 − (x + y + z) = 1 − 0.9 = 0.1. 0.1
U A 0.2
0.5
B 0.2
From this diagram we can easily pick up the values for (a) − (b) − (c). Problem 5 For two events X, Y such that P(Y ) = 0.7, P(X 0 ) = 53 , P(Y |X) = 12 . (a) P(X) = 1 − P(X 0 ) =
2 5
=
4 10
(b) P(X ∩ Y ) = P(X)P(Y |X) =
4 10
·
5 10
=
20 100
= 0.2
(c) Complete the Venn diagram with probabilities:
2
0.1
U
X
Y 0.2
0.2
(d) P(X|Y ) =
P(X∩Y ) P(Y )
=
2 10 7 10
=
0.5
2 7
(e) Are X and Y independent? Explain. Since P(X|Y ) 6= P(X), the two events are not independent. Problem 6 Two dice are rolled. Let E be the event that the first die shows an even number and F be the event that the second die shows 6. (a) Are E, F independent events? Since P(E) =
3 6
= 0.5 and P(E|F ) =
3 6
= 0.5, we have P(E|F ) = P(E)
that is, the two events are independent. (Indeed, if the second die shows 6, the probability that the first die shows an even number still remains the same). (b) Are E, F mutually exclusive events? Of course not. For example, (4, 6) is an event in E ∩ F . (c) P(E|F ) =
1 2
(d) P(E ∩ F ) = P(F )P(E|F ) =
1 6
·
1 2
=
1 12
Problem 7 Two cards are drawn without replacement from a standard deck. (a) Find the probability of S2 given R1 , that is, the probability that the second card is a spade, given that the first card is red. P(S2 |R1 ) =
13 51 ,
since spades are not red.
(b) Find the probability that both cards are diamonds. P(D1 ∩ D2 ) = P(D1 )P(D2 |D1 ) =
13 52
·
12 51
=
3 51
Problem 8 Three cards are drawn without replacement from a standard deck. (a) Find the probability of a tris of aces, that is: P(A1 ∩ A2 ∩ A3 ) = P(A1 )P(A2 |A1 )P(A3 |A2 ∩ A1 ) = (b) Find the probability of a tris.
3
4 52
·
3 51
·
2 50
In a standard deck, we have 13 possible cards (up to a suit). So, the event T = {tris} is the union of thirteen events, each of one corresponds to one of those cards, that is: T = {AAA, 111, 222, 333, 444, 555, . . .} It’s easy to see that these events have the same probability, so P(T ) = 13 · P(AAA) = 13 ·
3 4 3 2 · · = ≈ 0.0023 52 51 50 25 · 51
Problem 9 Consider the following diagram: 7B ppp
0.8 ppp
p ppp ppp @ A NNNN NNN NN 0.2 NNN N' 0.6 B0 == == == == == 7B = 0.4 == ppp 0.1 ppp == == ppp pppp A0 NNN NNN NNN 0.9 NNN N' B0 (a) P(B|A0 ) = 0.1 (b) P(A ∩ B) = P(A)P(B|A) =
6 10
·
8 10
=
48 100
(c) P(B) = P(B ∩ A0 ) + P(B ∩ A) = P(A0 )P(B|A0 ) + P(B ∩ A) =
4 10
·
1 10
+
48 100
=
52 100
Problem 10 According to a survey of a group of college students, 40% of the students have heard Tchaikovsky’s Piano Concert No.1 (T), of those 50% have read a novel by Dostoyevsky (D). Of those who have not heard the concerto, 25% have read a novel by Dostoyevsky. (a) Draw a tree diagram to summarize the results of the survey.
4
7D ppp
0.5 ppp
p ppp ppp T NNN @ NNN NNN 0.5 NNN N' 0.4 D0