# Probability and Statistics Midterm

Topics: Probability theory, Random variable, Probability distribution Pages: 11 (1611 words) Published: April 16, 2013

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...