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Then P(at least one 4 comes up) P(at least one 4 comes up) 1 1 1 1
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P(no 4 comes up) P(no 4 comes up)
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1 1 6 6
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P(no 4 on 1st toss and no 4 on 2nd toss) P(Ar > Ar ) 1 2 5 5 6 6 1 11 36 P(Ar ) P(Ar2 ) 1
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Total number of equally likely ways in which both dice can fall 6 6 36. Also Number of ways in which A1 occurs but not A2 5 Number of ways in which A2 occurs but not A1 5 Number of ways in which both A1 and A2 occur 1 Then the number of ways in which at least one of the events A1 or A2 occurs P(A1 < A2) 11 > 36. 5 5 1 11. Therefore,
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1.13. One bag contains 4 white balls and 2 black balls; another contains 3 white balls and 5 black balls. If one ball is drawn from each bag, find the probability that (a) both are white, (b) both are black, (c) one is white and one is black.
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Let W1 event white ball from first bag, W2 P(W1) P(W2 u W1) P(Wr ) P(Wr u Wr ) 1 2 1 P(W1) P(W2) P(Wr ) P(Wr ) 1 2 event white ball from second bag. 4 4 2 2 4 2 3 3 5 5 3 1 4 5 24
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(a) P(W1 > W2) (b) P(Wr > Wr ) 1 2
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(c) The required probability is 1 P(W1 > W2)
P(Wr > Wr2 ) 1
5 24
13 24
1.14. Prove Theorem 1-10, page 7.
We prove the theorem for the case n 2. Extensions to larger values of n are easily made. If event A must result in one of the two mutually exclusive events A1, A2, then A (A > A1) < (A > A2)
CHAPTER 1 Basic Probability
But A > A1 and A > A2 are mutually exclusive since A1 and A2 are. Therefore, by Axiom 3, P(A) P(A > A1) P(A > A2) P(A2) P(A u A2)
P(A1) P(A u A1) using (18), page 7.
1.15. Box I contains 3 red and 2 blue marbles while Box II contains 2 red and 8 blue marbles. A fair coin is tossed. If the coin turns up heads, a marble is chosen from Box I; if it turns up tails, a marble is chosen from Box II. Find the probability that a red marble is chosen.
Let R denote the event a red marble is chosen while I and II denote the events that Box I and Box II are chosen, respectively. Since a red marble can result by choosing either Box I or II, we can use the results of Problem 1.14 with A R, A1 I, A2 II. Therefore, the probability of choosing a red marble is P(R) P(I) P(R u I) P(II) P(R u II) 1 3 2 3 2 1 2 2 2 8 2 5
Bayes theorem 1.16. Prove Bayes theorem (Theorem 1-11, page 8).
Since A results in one of the mutually exclusive events A1, A2, c , An, we have by Theorem 1-10 (Problem 1.14), P(A) P(A1) P(A u A1) c
P(An) P(A u An)
a P(Aj ) P(A u Aj )
Therefore,
P(Ak u A)
P(Ak > A) P(A)
P(Ak) P(A u Ak )
a P(Aj) P(A u Aj )
1.17. Suppose in Problem 1.15 that the one who tosses the coin does not reveal whether it has turned up heads or tails (so that the box from which a marble was chosen is not revealed) but does reveal that a red marble was chosen. What is the probability that Box I was chosen (i.e., the coin turned up heads)
Let us use the same terminology as in Problem 1.15, i.e., A R, A1 I, A2 II. We seek the probability that Box I was chosen given that a red marble is known to have been chosen. Using Bayes rule with n 2, this probability is given by 1 3 2 3 2
P(I u R)
P(I ) P(R u I ) P(I ) P(R u I ) P(II ) P(R u II )
Combinational analysis, counting, and tree diagrams 1.18. A committee of 3 members is to be formed consisting of one representative each from labor, management, and the public. If there are 3 possible representatives from labor, 2 from management, and 4 from the public, determine how many different committees can be formed using (a) the fundamental principle of counting and (b) a tree diagram.
(a) We can choose a labor representative in 3 different ways, and after this a management representative in 2 different ways. Then there are 3 2 6 different ways of choosing a labor and management representative. With each of these ways we can choose a public representative in 4 different ways. Therefore, the number of different committees that can be formed is 3 2 4 24.
1 3 2 3 2
1 2 2 2 8
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