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Introduction to Probability - Solutions Manual

Introduction to Probability - Solutions Manual - Charles M...

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Charles M. Grinstead and J. Laurie Snell: INTRODUCTION to PROBABILITY Published by AMS Solutions to the exercises SECTION 1.1 1. As n increases, the proportion of heads gets closer to 1/2, but the di ff erence between the number of heads and half the number of flips tends to increase (although it will occasionally be 0). 3. (b) If one simulates a su ffi ciently large number of rolls, one should be able to conclude that the gamblers were correct. 5. The smallest n should be about 150. 7. The graph of winnings for betting on a color is much smoother (i.e. has smaller fluctuations) than the graph for betting on a number. 9. Each time you win, you either win an amount that you have already lost or one of the original numbers 1,2,3,4, and hence your net winning is just the sum of these four numbers. This is not a foolproof system, since you may reach a point where you have to bet more money than you have. If you and the bank had unlimited resources it would be foolproof. 11. For two tosses, the probabilities that Peter wins 0 and 2 are 1/2 and 1/4, respectively. For four tosses, the probabilities that Peter wins 0, 2, and 4 are 3/8, 1/4, and 1/16, respectively. 13. Your simulation should result in about 25 days in a year having more than 60 percent boys in the large hospital and about 55 days in a year having more than 60 percent boys in the small hospital. 15. In about 25 percent of the games the player will have a streak of five. SECTION 1.2 1. P ( { a,b,c } ) = 1 P ( { a } ) = 1 / 2 P ( { a,b } ) = 5 / 6 P ( { b } ) = 1 / 3 P ( { b,c } ) = 1 / 2 P ( { c } ) = 1 / 6 P ( { a,c } ) = 2 / 3 P ( φ ) = 0 3. (b), (d) 5. (a) 1/2 (b) 1/4 (c) 3/8 (d) 7/8 7. 11/12 9. 3 / 4 , 1 11. 1 : 12 , 1 : 3 , 1 : 35 13. 11:4 15. Let the sample space be: ω 1 = { A,A } ω 4 = { B,A } ω 7 = { C,A } 1
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ω 2 = { A,B } ω 5 = { B,B } ω 8 = { C,B } ω 3 = { A,C } ω 6 = { B,C } ω 9 = { C,C } where the first grade is John’s and the second is Mary’s. You are given that P ( ω 4 ) + P ( ω 5 ) + P ( ω 6 ) = . 3 , P ( ω 2 ) + P ( ω 5 ) + P ( ω 8 ) = . 4 , P ( ω 5 ) + P ( ω 6 ) + P ( ω 8 ) = . 1 . Adding the first two equations and subtracting the third, we obtain the desired probability as P ( ω 2 ) + P ( ω 4 ) + P ( ω 5 ) = . 6 . 17. The sample space for a sequence of m experiments is the set of m -tuples of S ’s and F ’s, where S represents a success and F a failure. The probability assigned to a sample point with k successes and m - k failures is 1 n k n - 1 n m - k . (a) Let k = 0 in the above expression. (b) If m = n log2, then lim n →∞ 1 - 1 n m = lim n →∞ 1 - 1 n n log2 = lim n →∞ ( 1 - 1 n n log2 = e - 1 log2 = 1 2 . (c) Probably, since 6log2 4 . 159 and 36log2 24 . 953. 19. The left-side is the sum of the probabilities of all elements in one of the three sets. For the right side, if an outcome is in all three sets its probability is added three times, then subtracted three times, then added once, so in the final sum it is counted just once. An element that is in exactly two sets is added twice, then subtracted once, and so it is counted correctly. Finally, an element in exactly one set is counted only once by the right side.
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