assn2-solns - Math 361, Problem Set 2 September 17, 2010...

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Math 361, Problem Set 2 September 17, 2010 Due: 9/13/10 1. (1.3.11) A bowl contains 16 chips, of which 6 are red, 7 are white and 3 are blue. If four chips are taken at random and without replacement, Fnd the probability that (a) each of the 4 chips is red (b) none of the four chips is red (c) there is at least one chip of each color. Answer: ±or ( a ), there are ( 6 4 ) ways to make such a choice so p = ( 6 4 ) ( 16 4 ) ±or ( b ), the probability is ( 10 4 ) ( 16 4 ) (there are 10 non-red chips, of which I must choose 4). ±or ( c ), the probability is: p = ( 6 2 )( 7 1 )( 3 1 ) + ( 6 1 )( 7 2 )( 3 1 ) + ( 6 1 )( 7 1 )( 3 2 ) ( 16 4 ) Note that this is equivalent to 1 2 × 6 · 7 · 3 · 13 ( 16 4 ) . 2. (1.3.24) Consider three events C 1 ,C 2 3 . (a) Suppose C 1 2 3 are mutually exclusive events. If P ( C i )= p i , for i =1 , 2 , 3 what is the restriction on the sum p 1 ,p 2 3 . (b) In the notation of Part (a), if p 1 =4 / 10, p 2 =3 / 10, and p 3 =5 / 10 are C 1 2 and C 3 mutually exclusive? (c) Does it follow from your conclusion in part ( b ) that P ( C 1 C 2 C 3 ) > 0? Why or why not? 1
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Answer: For ( a ), p 1 + p 2 + p 3 1. For ( b ), this implies that the answer is no. However the answer to part ( c ) is also no - P ( C i C j ) 0 for some pair, but all three need not intersect. For instance it is possible for C 2 C 1 and C 1 C 2 = while obtaining those probabilities. 3. (1.4.7) A pair of 6-sided dice is cast until either the sum of seven or eight appears.
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assn2-solns - Math 361, Problem Set 2 September 17, 2010...

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