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Unformatted text preview: MATH242/STAT242/STAT542 Theory of Statistics Solution to Problem Set 1 prepared by Ankur Sharma 1 (Chapter 1 Problem 60) Let E i be the event that an item is produced in the i th shift. Let D be the event that the item is defective. We are given P ( E 1 ) = P ( E 2 ) = P ( E 3 ) = 1 / 3 We are also given P ( D  E 1 ) = 0 . 01, P ( D  E 2 ) = 0 . 02, P ( D  E 3 ) = 0 . 05 So, P ( D ) = ∑ P ( D  E i ) P ( E i ) = (1 / 3) * (0 . 01 + 0 . 02 + 0 . 05) = 0 . 08 / 3 i.e., 8 3 % of items are defective. Also, probability that an item was produced in the 3rd shift given that it was defective is P ( E 3  D ) = P ( E 3 ∩ D ) P ( D ) = P ( D  E 3 ) * P ( E 3 ) P ( D ) = . 05 * (1 / 3) (0 . 08 / 3) = 5 / 8 2 (Chapter 1 Problem 77) Probability of hitting the bull’s eye for the player in any trial = p = 0 . 05 So probability of not hitting in any trial =1 p = 0 . 95 Thus, probability of never hitting in n trials = (1 p ) n = 0 . 95 n and the probability of hitting at least once in n trials = 1 probability of never hitting = 1 . 95 n If this is 0 . 5 then, 1 . 95 n = 0 . 5 or n = log 0 . 5 log . 95 = 13 . 51 n has to be an integer, so for n = 14, the probability of hitting atleast once crosses 0 . 5 3 (Chapter 2 Problem 25) Probability of being dealt royal straight flush = p = 1 . 3 * 10 8 total number of hands seen = n = 100 * 52 * 20 = 104000 So N , the number of royal stright flushes dealt, follows a Binomial(n,p) distribution...
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 Spring '11
 LisaZhang
 Statistics, Normal Distribution, Standard Deviation, Probability theory, Trigraph, probability density function, Cumulative distribution function

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