Exam 1 Solutions. Problem Points Score Total 145

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1 Exam Solutions Read each question carefully and answer all to the best of your ability. Show work to receive as much credit as possible. At the end of the exam, please sign the box below. Problem Points Score Total 45 Name: I have adhered to the Duke Community Standard in completing this test.

2 Question. (0 points Suppose that the events A and B are independent with respective probabilities /5 and /4. Find: and B a P (A c B. SOLUTION. By independence of A and B, we know that A c and B are also independent. Hence b P (A B c. P (A c B = P (A c P (B = ( P (AP (B = = 5. SOLUTION. Using inclusion/exclusion and independence of A and B c via independence of A c P (exactly one of A or B occurs P (A B c = P (A + P (B c P (AB c = P (AP (Bc = = 4 5. SOLUTION. Using independence for the third time, note that the probability of this event is precisely P (A(AB c B(AB c = P (A + P (B P (AP (B = = 7 0.

3 Question. (0 points A population of 70 registered voters contains 40 in favor of Proposition 34 and 30 opposed. An opinion survey selects a random sample without replacement of 0 voters from this population. a What is the probability that there will be no one in favor of Proposition 34 in the sample? SOLUTION. For this problem, use the hypergeometric distribution with G = 40, B = 30, N = 70, and n = 0. Therefore, the probability that no one is in favor is ( 40 ( 30 ( =. b What is the probability that there will be at least one person in favor? SOLUTION. 0 c What is the probability that exactly one pro 34 person will appear in the sample. SOLUTION. ( 40 ( ( = 40(. 0 3

4 Question 3. (0 points Given that there were 4 heads in 0 independent fair coin tosses, calculate: a the chance that the first toss landed heads; SOLUTION. Let B be the event that there are 4 heads in 0 tosses and A be the event that the first toss is heads. Using Bayes P (A B = P (B P (B A P (A = b the chance that the first two tosses landed heads; SOLUTION. Let C be the event that the first two tosses landed heads. Then by Bayes again: P (C B = ( 8 P (B CP (C =. P (B ( 9 3 ( 0 4 c the chance that at least two of the first five tosses landed heads. SOLUTION. Let D j denote the event that j of the first five tosses lands heads and D be the event that at least two of the first five tosses land heads. Then by the rules of probability and Bayes P (D B = P (D 0 B P (D B = P (B P (B D 0P (D 0 = ( 0 4 [( ( ] 5. 3 ( 0 4. P (B P (B D P (D 4

5 Question 4. (0 points There are two boxes: Box and Box. Box contains 4 red balls and 5 black balls. Box has red balls and 5 black balls. One of the two boxes is picked at random, and then a ball is picked at random from that box. a Is the color of the ball independent of which box is chosen? Explain. SOLUTION. Yes. Let R be the event that a red ball is selected, B be the event that a black ball is selected, and B i, i =, be the event that box i is selected. Then P (R B = 4 9 = P (R = (4/9 + 4/9 = P (R B P (B + P (R B P (B P (B B = 5 9 = P (B = (5/9 + 5/9 = P (B B P (B + P (B B P (B, so color is independent of the Box. b What if instead there were red balls and 4 black balls in Box? Explain. SOLUTION. No. P (R B = 4/9 P (R = (4/9 + 6/3. c Suppose again that Box contains 4 red balls and 5 black balls and Box has red balls and 4 black balls. Given that a red ball is selected, what is the chance that it came from Box? SOLUTION. Apply Bayes again: P (B R = P (R P (R B P (B = 4 9 = (4/9 + 6/ /9 + 6/

6 Question 5. (5 points A pollster is interested in determining the proportion p of Durhamites who will vote for Mitt Romney in the 0 Presidential election. How many people must the pollster survey to be 97.% confident that the observed frequency ˆp of persons who will vote for Mitt Romney is such that p ˆp 0.0? SOLUTION. We search for a z score such that in other words, we search for z such that.97 = Φ(z Φ( z = Φ(z, Φ(z =.97/ =.986. From the table in you Appendix, we see that z =.. Since we do not know p, we need to survey enough people n such that which implies n = n =.0 6

7 Question 6. (5 points Suppose that each week you buy a lottery ticket which has a chance of /000 of a win. You do this each week for three years. a Using the appropriate approximation scheme, roughly calculate the chance you will get k = 0,, wins during the three years. SOLUTION. Here, we ll use the Poisson approximation since p is small with µ = np =.56 P (0 e P (.56e P (.56 e b Find exactly (not approximately the chance of k = 0,, (this is THREE events wins in the three years. SOLUTION. P (0 = (999/ P ( = 56(/000 (999/ ( 56 P ( = (/000 (999/

8 Question 7. (5 points You are out one night with a friend and a shadowy figure approaches you and says: Would you like to play a game? Here s how it works: I ll toss three coins and look at them. If they are all tails, then I ll show you them and flip again. Otherwise, I ll show you one coin. If it s heads, I ll give you n dollars. If it s tails, you give me k dollars. When should you play the game? Explain your answer in terms of n and k. SOLUTION. If you play repeatedly, your expected earnings would be n 7 k 6 7. In other words, to be worth your while (assuming he wants to play repeatedly you would need n > 6k. 8

9 Question 8. (0 points At a winter party there are n people, all of which have put their n heavy coats in a closet. During the party, the electricity goes off and the house gets cold, so all people decide to retrieve their coats. Because it is dark, they cannot see and hence each person picks a coat uniformly at random from the closet. a Find the probability that at least one person gets their coat from the dark closet. [Hint: Use the inclusion-exclusion formula for n events A, A,..., A n : P ( n n A j = P (A j P (A i A i + P (A i A i A i3 +( n+ P (A A A n.] j= j= i <i n i <i <i 3 n SOLUTION. Let A j be the event that the jth person gets his coat. Then P (at least one person get his coat = P ( n A j = n P (A j j= First note that i <i n P (A i A i + j= i <i <i 3 n P (A j = (n! n! P (A i A i A i3 + ( n+ P (A A A n. since there are n! different ways to distribute the coats, (n! have person j with his coat. Then note that for i j P (A i A j = P (A i A j P (A j = Doing this inductively the sum becomes! + 3! + ( n+ n!. ( n (n! (n! = (n! n! n(n + b What is the limit of this probability as n? ( n 3 SOLUTION. As n, we see that this probability becomes n(n. n(n (n + ( n+ n! ( k+ k= k! ( k = k! k= = e. 9