Math/Stat 394 Homework 5

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1 Math/Stat 394 Homework 5 1. If we select two black balls then X 4. This happens with probability ( 4 2). If we select two white balls then X 2. This happens with ( 14 probability (8 2). If we select two orange balls then X 0. This ( 14 happens with probability (2 2). If we select one black and one white ( 14 ball then X 1. This happens with probability 248. If we select one ( 12 black and one orange ball then X 2. This happens with probability 242. If we select one white and one orange ball then X 1. This ( 12 happens with probability 282 ( Five men and five women are ranked according to their scores on an examination. Assume no two scores are alike and all 10! possible rankings are equally likely. Let X denote the highest ranking achieved by a women (for instance, X 1 if the top-ranked person is female.) Find P ({X i}) for i 1, 2,..., 10. If the highest ranking woman has rank i then there are four women in the bottom 10 i. Thus ) for i 1,..., 6. P (X i) ( 10 i 4 ( 10 5 ) 5. Let X denote the difference between the number of heads and the number of tails obtained when a coin is tossed n times. What are the possible values of X? Let H be the number of heads and T be the number of tails. Then T n H and X H T 2H n. As H can take on all integer values between 0 and n X can take on all integer values between n and n which are of the same parity as n. 1

2 6. In problem 5, if the coin is assumed fair, for n 3 what are the probabilities associated with the values that X can take on? P (X 3) P (X 3) 1 8 and P (X 1) P (X 1) Admittedly sketchy proof. (a) 333/1000, 200/1000, 142/1000, /1000, 9/1000. The limit as k get large is 1/3, 1/5, 1/7, 1/15 and 1/105. (b) Let p k,n {j 1,..., n: j has no repeated prime factor of p 1,..., p k } n By the same reason as above p k lim n p k,n Taking the limit as k goes to we get k i1 p 2 i 1 p 2 i {j 1,..., n: j has no repeated prime factor} lim n n lim p k k 6 π A salesman has scheduled two appointments to sell encyclopedias. His first appointment will lead to a sale with probability.3, and his second will lead independently to a sale with probability.6. Any sale made is equally likley to be either the deluxe model, which costs $1000 or the standard model which cosets $500. Determine the probability mass function of X, the total dollar value of all sales. There possible outcomes are no sale (X 0), one deluxe (X 1000), one regular (X 500), two deluxe (X 2000) one regular and one deluxe (X 1500) or two regular(x 1000). No sales happen with probability (.7)(.4).28. One sale happens with probability.3(.4) + (.7)(.6).54. Thus one deluxe and one regular each happen with probability.27. Two sales happens with probability (.3)(.6).18. 2

3 So two deluxe and two regular happen with probability.045 and one regular and one deluxe happen with probability.09. Putting all this together we get P (X 0).28, P (X 500).27, P (X 1000) , P (X 1500).09 and P (X 2000) Player one wins no rounds if her card is lower than the card for player two. As all combinations are equally likely P (X 0) 1/2. Player one wins one round if her card is higher than the card for player two and lower than the card for player three. All six rankings of the first three players are equally likely so P (X 1) 1/6. Thus P (X 1) 1/6. Player one wins one round if her card is higher than the card for player two and three and lower than the card for player four. All twenty four rankings of the first four players are equally likely and two of them satisfy this condition (4123 and 4132). Thus P (X 2) 1/12. Player one wins three rounds if she has the second highest card and player 4 has the highest card. The probability that player 4 has the highest card is 1/5 and the conditional probability given that that player 1 has the next highest card is 1/4 thus P (X 3) 1/20. Player one wins four rounds if she has the highest card. As all players are equally likely to have the highest card P (X 4) 1/5 15. Let team i be the team with the ith worst record. They get balls in the lottery. Using notation from the next problem let Y 1 i if team i gets the first pick, Y 2 i if team i gets the second pick, and Y 3 i if team i gets the third pick. P (X 1) 11. By Bayes rule the probability that team 1 gets the second pick is P (X 2) 11 i2 11 i2 11 i2 P (X 2 and Y 1 i) P (X 2 Y 1 i)p (Y 1 i) 11 (). 3

4 P (X 3) P (X 4) P (X 3, Y 1 i and Y 2 j) P (Y 1 i)p (Y 2 j Y 1 i)p (X 3 Y 1 i and Y 2 j) () 11 () (). P (Y 1 i, Y 2 j and Y 3 k) P (Y 1 i)p (Y 2 j Y 1 i) P (Y 3 k Y 1 i and Y 2 j) () 12 k () (). 16. P (Y 1 m) 12 m. By Bayes rule the probability that team m gets the second pick is P (Y 2 m) P (Y 2 m and Y 1 i) P (X 2 Y 1 i)p (Y 1 i) 12 m (). P (Y 3 m) P (Y 3 m, Y 1 i and Y 2 j) P (Y 1 i)p (Y 2 j Y 1 i) P (Y 3 m Y 1 i and Y 2 j) () 12 m () (). 4

5 17. (a) We use the fact that P (X i) lim P (X i) P (X i ɛ) lim F (i) F (i ɛ) F (i) F (i ). Then we get P (X 1) F (1) F (1 ) 1/2 1/4 1/4. P (X 2) F (2) F (2 ) 11/12 3/4 1/6. P (X 3) F (3) F (3 ) 1 11/12 1/12. (b) We use the fact that P (X (i, j)) lim P (X j ɛ) P (X i) lim F (j ɛ) F (i) F (j ) F (i) So P (X (.5, 1.5)) 5/8 1/4 3/8. 5