Probability and Statistics Midterm

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Probability and Statistics, Midterm Nov. 3, 2011 Instructions. The exam is from 3:40-5:30. Simplify your answer as much as you can, for instance, x 1 x+1 should be simpli…ed to x 1: But the answer does not need to be numbers if the caculation is complicated, e5 for example, you can leave 20! in your answer. 1. (30 points) (a) One of three students, A, B and C, will get the prize for being the "best student in Probability and Statistics". The night before the prize is announced, student A …nds the professor and asks, "No matter I get the prize or not, we know that at least one of B and C does not get it. Would you please tell me one of them who does not win it? I’ like to prepare a gift for her." The professor refuses, d "No, I cannot give you any hint about the probability that you win the prize." But student A insists, "Think about it again, professor. Telling me one of B and C who does not win the prize won’ give me t any clue about my chances of winning it." Do you agree with A’ claim? Why? s Answer: His claim is correct under certain assumptions. Suppose that the prior probability of each one winning the prize is 1=3; then the question is exactly the same as the "two sheep, one car" game that we introduced in the class. In fact, if the professor tells A that B does not win, the posterior can be caculated as following: Prior A wins B wins C wins 1=3 1=3 1=3 Conditional probability that the professor says "B" 1=2 0 1 Posterior (1=3) (1=2) = 1=3 (1=3) (1=2) + (1=3) 1 0 (1=3) 1 = 2=3 (1=3) (1=2) + (1=3) 1 2

Thus the conditional probability that A wins is equal to the unconditional probability. (b) There are two villages. In the …rst village, each couple is allowed to have one child at most. In the other one there is no such a constraint. Suppose all parents want boys, so that in the second village each couple will keep trying until they get a boy. In the end, will the Boy/Girl ratios in the two villages be extremely di¤erent? Reason your answer. Answer: Intuitively, if each child independently has probability to be a boy and 1 probability to be a girl, then no matter how many children a couple already have, the probability of the next child being a boy is still : Comparing the two villages, although the numbers of children (per family) may be very di¤erent, the probabilities of each child being a boy are both . Thus the percentage of boys is approximately in each village. The boy/girl ratio should be approximately = (1 ): Formally, let random variable X be the number of girls in a family in the second village. Then under conventional assumptions, X has a geometric distribution (a special negative binomial distribution with r = 0). The p.d.f. of X is x f (x) = (1 ) and the expectation of X is E (X) = 1 X

x (1 (1

)

x

x=0

) E (X) = =
1 X

) E (X) = )
x 1 X

x=0

1 X

x (1 )

)
x

x

x=0

1 X

x (1

)

x+1

x (1 (1 )

(x

1) (1 )
1

) E (X) = (1

x=0 1 X

x=1 x

=

x=1

1 (1 1 (1

)

(1

)

=1

)= 1

Suppose that there are n families in the second village and let Xi be the number of girls in family i. Then Xi are independent and have the same distribution as X: Since the number of boys per family in the second village is 1; the boy/girl ratio of second village is Pn 1 1 Pni=1 = Pn Xi i=1 i=1 Xi n Using the law of large numbers (which we’ learn soon), the ratio converges to ll 1

almost surely.

(c) 11 cards, one letter on each, compose the word "probability". Now randomly pick 7 cards one by one and without replacement. What is the probability that the 7 letters on the 7 cards sequentially form the word "ability"? What is the probability that the 7 cards can be used (by reordering) to form the word "ability". Answer: (i) In total there are "ability", there are 11! (11 7)!

possible ordered sequences of the 7 cards, and for the them to be 1 2 2 1 (2 1) 1 1 = 4

possible cases, thus the probability is

4 4! 11!

(ii) In total there are 11...
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