_' C)_' ___ ' __________ I: ____ __' 9. Recall the birthday problem with 100 people. We saw that the probabil ity of nding at least two people with the same birthday is 0.9999997. We also pointed out that this is not so surprising if we think about pairs of pe0p1e: There are (130) = 4950 pairs where each pair has a probability of 1 / 365 of a match for birthdays. (a) Let X be the number of pairs with matching birthdays and com pute P(X > 0) using the binomial distribution with n = 4950 and p = 1 / 365. Compare with the answer we got in class. (b) The probability in (a) is approximate. The binomial distribution requires independence between trials which is not true here. Why not? Why is this still not a very big problem? (c) The probability of at least one triple with matching birthdays among 100 people is tricky to compute exactly, but the binomial approximation works well. Let Y be the number of such triples and compute P(Y > 0) with the binomial distribution. ((1) It can be shown that a binomial distribution with large n and small 33 can be approximated by a Poisson distribution with mean a 2 mo. Use such a Poisson approximation to compute P(X > 0) and P(Y > 0). (e) Compute the approximate probability that there is a least one quintuple with matching birthdays in the Swedish parliament, us ing a Poisson approximation