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Discussion 12 Worksheet—Solutions 1. a. Choose 2 of the 6 defects and 3 of the 9 non-defects. The answer is 6 9 2 3 15 5 . b. At most 2 means exactly 2 or exactly 1 or exactly 0 defects. The answer is 6 9 6 9 6 9 2 3 1 4 0 5 15 5 . c. The opposite of at least 2 defects is at most 1 defect (exactly 1 or exactly 0 defects). The answer is 6 9 6 9 1 4 0 5 1 15 5 2. 1 1 2 1 1 1 a. pE 1 2 2 2 2 2 4 1 1 3 b. pF 2 2 2 8 c. E F exactly one heads in first two tosses and exactly one heads in second two tosses. Since the second two tosses are independent of the first two, the probability of getting exactly one heads in the second two tosses is the same as our answer to a). So we get pE F . 1 1 2 2 d. pE | F 1 4 pE F 1 4 2 pF 38 3 e. pE pF 3 1 pE F , so the events are not independent. 16 4 3. a. There are 4 aces to choose from and we want 3 of them so 4 3 p3 aces 52 3 . b. There are four aces, four kings and four queens and we want one of each, in that order. So the desired probability is 444 . P 52 ,3 Note the permutation in the denominator reflects the fact that the hand is ordered. c. The opposite of at least one ace being drawn is: no aces are drawn. Thus the answer is 48 3 1 p no aces 1 52 3 . 4. If heads is three times as likely as tails then for a single coin toss pheads .75 and 10 ptails .25 . This is now a Bernoulli trials problem and the answer is 0.75 7 0.25 3 . 7 5. Define two events as follows: E all four tosses result in heads and F at least one toss in four results in heads. Then we must find pE | F . Now note that E F hence EF E pE | F and so pE F pE pE F 1 16 1 . pF 15 16 15 1 . 16 Also pF 1 pF 1 1 15 . 16 16 Thus 6. The expected value of the ping pong ball drawn is 1 2 1 1 3 27 . 1 2 3 4 5 8 8 8 8 8 8 7. Because the questions are answered at random, each answer is independent of all the others. The probability of getting a question correct is question wrong. (Just like tossing a fair coin.) 7 3 10 10 1 10 1 1 a. pexactly 7 correct answers . 7 2 2 7 2 1 , as is the probability of getting a 2 b. We need to have question #4 be one of the 7 correct answers plus six other correct 10 answers chosen from the 9 other questions. There are possible ways to get exactly 7 7 out 10 questions correct. So the desired probability is 9 9 1 6 6 . 10 10 7 7 c. Define the following events: E question #4 is answered correctly, and F exactly seven questions are answered correctly. Then we must find pE | F . Now E F #4 and exactly six other questions are answered correctly. Since the individual questions are 1 9 1 answered randomly, we have pE F 2 6 2 10 have pE | F 9 1 6 2 10 10 1 7 2 6 3 10 9 1 1 2 6 2 . Thus by part a) we 9 6 . 10 7 8. Let’s call the containers A and B. In general, the probability of going free is pgetting white | A chosen pA chosen pgetting white | B chosen pB chosen . Note that if the prisoner puts an equal number of white and black balls in A, then there necessarily will be an equal number of each type in B. (For instance, 2 of each in A implies 8 of each in B.) In any such case the probability of going free is 1 1 1 1 1 . 2 2 2 2 2 Can the prisoner do better? As we just saw, if there is 1 of each ball in A and 9 of each ball in B, the prisoner has equal odds of going free and not going free. What if we move the one black ball from A to B. Then there is 1 white ball in A and the other 19 balls are in B so the probability of going free is 1 1 9 1 0.737 . 2 2 19 As it turns out, this is the best possible configuration. With this setup, it is of no use to move any of the 9 white balls from B to A—this only lowers the probability of going free. Likewise, moving black balls from B to A also lowers the chances of going free.