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1 Name Student ID # Instructor: SOLUTION Sergey Kirshner STAT 516 Fall 09 Practice Midterm #1 January 31, 2010 You are not allowed to use books or notes. Non-programmable non-graphic calculators are permitted. Please read the directions carefully. There are 8 problems worth 20 points each. The exam is graded out of 150 points, so there is a possibility of additional 10 points. Full credit is given for complete correct solutions. Partial credit is limited and awarded at the discretion of the instructor. You have 50 minutes to complete the exam. Please show all your work. Problem # Number of points 1 /20 2 /20 3 /20 4 /20 5 /20 6 /20 7 /20 8 /20 Total /150 1

2 1. (For all parts, write down the expression without evaluating it.) A bridge hand (13 cards) is drawn at random from a standard deck of 52 cards. (a) (5pts) How many different hands are there? Solution: ( 52 ) 13 = 52!. 13!39! (b) (5pts) What is the probability that the hand contains 8 red cards (hearts or diamonds) and 5 black cards (clubs or spades)? Solution: ( 26 ( 52 13) 8 )( 26 5 ) = 26! 18!8! 26! 21!5! 52! 39!13!. (c) (10pts) What is the probability that the hand contains an ace or a king? Solution: Let A denote the event that the drawn hand contains and ace, and let K denote the event that the hand contains a king. We are looking for P (A K) = 1 P ((A K) c ) = 1 P (A c K c ). A c K c is the event of not having either a king or an ace in a hand. Since all hands are equiprobable, P (A c K c ) = (44 44! 13!31! 52! 13!39! = 44!39! 44!39!. Thus P (A K) = 1. 31!52! 31!52! 13) ( 52 13) = 2

3 2. (20 points) An urn contains 5 blue balls, 3 green balls, and an unknown number of red balls. 2 balls are drawn from the urn (without replacement). If the probability that the draw consists of 2 blue balls is 2, how many red balls are in the urn? 9 Solution: Let x be the number of red balls in the urn. We can assume that each draw of 2 balls is equally likely, and there are ( ) 5+3+x 2 = (x+8)! = (x+8)(x+7) of them. 2!(x+6)! 2 The number of draws consisting of 2 blue balls is ( ) 5 2 = 5! = 10. Thus 2!3! 10 (x+8)(x+7) 2 = 2 9 (x + 8) (x + 7) = 90 x2 + 15x 34 = 0. Thus x = 15± = 15±19, so either x = 2 or x = 17. Since the number of 2 2 balls cannot be negative, the number of red balls in the urn is 2. 3

4 3. (20 points) Among the 100 patrons of a bar, 42 are fans of basketball, 41 are fans of baseball, 65 are fans of either soccer or basketball, 60 are of either soccer or baseball, and 75 are of either basketball or baseball. If only 1 patron is a fan of all three sports, and if 10 fans don t care about any of the three sports, how many patrons are soccer fans? Solution: Let S be the set of all patrons, and let A, B, and C be subsets of S corresponding to the fans of soccer, basketball, and baseball. We are given that S = 100, B = 42, C = 41, A B = 65, A C = 60, B C = 75, A B C = 1, and (A B C) c = 10. We are asked to find A, and we will do so using the inclusion-exclusion principle: A B C = A + B + C A B A C B C + A B C = A + B + C ( A + B A B ) ( A + C A C ) ( B + C B C ) + A B C = A B C + A B + A C + B C + A B C, so A = A B C B C + A B + A C + B C + A B C. A B C = S \ (A B C) c, so A B C = S (A B C) c = = 90, and A = = 28. 4

5 4. (20 points) A student is taking midterms in Linear Algebra, Probability, and Analysis. The probabilities of getting an A in Linear Algebra, Probability, and Analysis are 0.3, 0.5, and 0.25, respectively. Given that grades on the midterms in these courses are mutually independent, the probability that a student will get at least one A. Solution: Let LA denote the event of getting an A on the Linear Algebra test, P on the Probability test, and A on the Analysis test, respectively. Using DeMorgan s Law and taking independence into account, P (LA P A) = 1 P ((LA P A) c ) = 1 P (LA c P c A c ) = 1 P (LA c ) P (P c ) P (A c ) = 1 (1 P (LA)) (1 P (P )) (1 P (A)) = 1 (1 0.3) (1 0.5) (1 0.25) =

6 5. (20 points) A child is playing with wooden blocks among which there are 8 blue, 5 yellow, and 2 red. The blocks are identical except for their color. A child arranges all of them randomly in a straight line. What is the probability that the first five blocks are blue given that last 4 blocks are yellow? Solution: There are a total of = 15 blocks. Let S denote the set of all possible block arrangements. Let B denote the set of all block arrangements such that the first five are blue, and let Y denote the set of arrangements such that the last four are yellow. We are looking to find P (B Y ) = P (B Y ) /P (Y ). B Y corresponds to the set of all block arrangements such that the first five are blue, and the last four are yellow. Since all of the outcomes are equally likely, P (B Y ) = P (B Y ) P (Y ) = B Y S Y S = B Y. Y To find Y, notice that when the last four blocks are yellow, there are 8 blue, 1 yellow, and 2 red blocks left to distribute among 11 slots, and there are ( 11 8,1,2) ways to do so. Similarly, for B Y, there are 3 blue, 1 yellow, and 2 red blocks left to distribute among 6 slots, total of ( 6 3,1,2) ways to do so. Thus P (B Y ) = B Y Y = ( 6 3,1,2 ( 11 8,1,2 ) ) = 6! 3!1!2! 11! 8!1!2! = 8!6! 11!3! =

7 6. (20 points) Biased coins C 1 and C 2 are flipped once independently of each other. The probability that two tails are observed is twice the probability that two heads are observed. Given that the probability the coin C 1 would show heads and coin C 2 would show tails is 0.1, what is the probability of the coin C 1 showing heads? Solution: Let H 1 denote the event of C 1 showing heads, and H 2 of C 2 showing heads, respectively. The taking independence into account 0.1 = P (H 1 H c 2) = P (H 1 ) P (H c 2) = (P (H 1 H 2 ) + P (H 1 H c 2)) (P (H 1 H c 2) + P (H c 1 H c 2)) = (P (H 1 H 2 ) + 0.1) ( P (H 1 H 2 )) ; 0 = 2P (H 1 H 2 ) P (H 1 H 2 ) 0.09; P (H 1 H 2 ) = 0.3 ± ( 0.09) 2 2 = 0.3 ± Since the P (H 1 H 2 ) is non-negative, only one answer is possible, P (H 1 H 2 ) = P (H 1 ) = P (H 1 H 2 ) + P (H 1 H c 2) = =

8 7. (20 points) Three identically looking coins lie on the table. Coin I is fair, coin II comes up tails 70% of the time, and coin III comes up tails 40% of the time. A coin is picked at random and flipped once. The outcome of the flip was heads. What is the probability that coin I was chosen? Solution: Let I, II, III denote the events that the coin I, II, and III were picked, respectively. Let H denote the event that the picked coin landed heads. Then P (I) = P (II) = P (III) = 1, P (H I) = 1, P (H II) = 3 6, and P (H III) = = 3. Using Bayes Rule, P (I H) = P (I) P (H I), P (H) P (H) = P (H, I) + P (H, II) + P (H, III) = P (I) P (H I) + P (II) P (H II) + P (III) P (H III) = = 7 15, P (I H) = =

9 8. (20 points) 50 distinct numbers are chosen at random from {1,..., 100}. What is the probability that the smallest number among the chosen is k? What is the range of value for k? Solution: k cannot be smaller than 1 or larger than 51 (otherwise, there would be less than = 50 numbers chosen as the chosen numbers are distinct). If k is the minimum among chosen numbers, it means that none of the 1,..., k 1 were chosen, and k was. Let A k denote the event that the minimum of the chosen numbers is k, P (A k ) = P (1,..., k 1 not chosen, k is chosen) = P (1,..., k 1 not chosen) P (k is chosen 1,..., k 1 not chosen) ) ) = ( 100 (k 1) 50 ( ) ( 100 k 49 ( 100 (k 1) 50 ) = (100 k 49 ) ( ), k = 1,..., 51. 9

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