Statistics 134: Concepts of Probability (Lugo)Homework 5AThis assignment is due on Friday, October 7, in class. In order to do it you should readSection 2.5.The first exam is on Wednesday, September 28. You should at least attempt the recom-mended problems here before the exam. Among the required problems, Pitman 2.5.6 is goodas exam preparation, as is P17(a-b); P16 and P17(c) are probably too hard to be exam-typeproblems.There will also be a “homework 5B” due on Friday, October 7. Hand in the requiredproblems of both assignments 5A and 5B as a single “Homework 5” on October 7.Recommended problemsFrom Pitman: 2.5.3, 2.5.7, 2.5.9, 2.5.12, 2.R.13, 2.R.15, 2.R.17, 2.R.23.2.5.12 asks you to find the probabilities of various poker hands. This is a recom-mended problem, even though it’s even-numbered, because the answer is very easily foundin any number of references. See http://en.wikipedia.org/wiki/List_of_poker_handsfor more complete definitions of the hands and http://en.wikipedia.org/wiki/Poker_probability for the solution.(Remember: you don’t need to hand these in.)Required problemsPitman 2.5.6 In a hand of 13 cards drawn randomly from a pack of 52, find the chance of:(a) no court cards (J, Q, K, A);(b) at least one ace but no other court cards;(c) at most one kind of court card.P16: Playoff probabilities.(a) I have 2n balls labeled 1, 2, . . . , 2n, and two boxes, East and West. I take n of theballs at random (without replacement) and place them in the East box; I take the remainingn balls and place them in the West box. I then take the smallest-numbered ball from eachbox. What is the probability that I end up with balls 1 and 2?For example, if n = 5 I might have 1, 3, 7, 8, 10 in East and 2, 4, 5, 6, 9 in West; then taking thesmallest-numbered ball from each box gives 1 and 2. On the other hand I could have 1, 2, 5, 6, 8 inEast and 3, 4, 7, 9, 10 in West, and then taking the smallest-numbered ball from each box gives 1and 3.(b) In the setup of (a), I take the two smallest-numbered balls from each box. What isthe probability that I end up with balls 1, 2, 3 and 4?Both of the examples above satisfy this. On the other hand I might have 1, 3, 4, 8, 9 in one boxand 2, 5 , 6, 7, 10 in the other; then taking the two smallest balls from each box gives 1, 2, 3, 5.1(c) I have 3n balls labeled 1, 2, . . . , 3n, and three boxes, East, Central, and West. I take nof the balls at random (without replacement) and place them in the East box; I take n of theremaining 2n balls at random (without replacement) and place them in the Central box; Itake the remaining n balls and put them in the West box. I then take the smallest-numberedball from each box, and then the smallest-numbered ball among the balls remaining. Whatis the probability that I end up with balls 1, 2, 3 and 4?For example, if n = 4 I might have 1, 5, 8, 12 in East, 2, 6, 9, 10 in Central, and 3, 4, 7, 11 inWest. The smallest balls in each box are 1, 2, 3 and the smallest ball not among these is 4, so Iget the balls 1, 2, 3, 4. On the other hand I could have 3, 8, 9, 10 in East, 5, 6, 7, 11 in Central, and1, 2, 4, 12 in West; then I get the balls 1, 2, 3, 5.As you might imagine from the title of this problem and the nomenclature I’ve chosen,this problem is intended to be a simple model of the probabilities that the teams with the bestregular-season records in a certain sports league make it to the post-season. In particularMajor League Baseball currently uses an arrangement very similar to (c), although there’sa slight complication from the fact that the divisions in MLB are not all of the same size.P17: Three-person birthday problem. In this problem, you will find an approxi-mation of the probability that, among n people, three of them will have the same birthday.(Note that in this problem we will make assumptions of independence that are not exactlycorrect.)(a) I pick three people at random. What is the probability that all three of them havethe same birthday (month and day)? Assume that all birthdays are equally likely and thatnobody is born on February 29. There are 365 days in a year.(b) Say we have n people in our population. What is the expected number of triples ofpeople that have the same birthday?(c) It is not true that “x, y, z share a birthday” and “u, v, w share a birthday” are inde-pendent whenever {x, y, z} 6= {u, v, w}. Why not?(d) Ignore (c), and assume that the events “x, y, z share a birthday” and “u, v, w share abirthday” are independent whenever we have two different triplets. What is the approximateprobability that no three people have the same birthday, as a function of n? Use a Poissonapproximation.(e) For what n is the function in (d) equal to 1/2?You should consider whether your analysis here would work in the case of 2 people sharing thesame birthday. It turns out that the approximation made in (d) becomes increasingly bad as thenumber of people intended to share a birthday