Probability and Statistics for Engineering and the Sciences (with CD-ROM and InfoTrac )

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Solutions to Homework Assignment #2 Statistics 220, Fall 2005 Due September 14, 2005 The following formula is true for any two events A and B , and is useful in several problems: P ( A B ) = P ( A ) + P ( B ) - P ( A B ) It follows from the formula for unions, P ( A B ) = P ( A ) + P ( B ) - P ( A B ). 2.6 a The possible outcomes are: 123, 124, 125, 213, 214, 215, 13, 14, 15, 23, 24, 25, 3, 4, 5 b The possible outcomes in A are just 3, 4 and 5. c B = { 125 , 215 , 15 , 25 , 5 } . d C = { 23 , 24 , 25 , 3 , 4 , 5 } . 2.12 a P ( A B ) = P ( A ) + P ( B ) - P ( A B ) = 0 . 5 + 0 . 4 - 0 . 25 = 0 . 65. b Having neither card is the event ( A B ) 0 , so the probability is 1 - 0 . 65 = 0 . 35. c The event is A B 0 , and the probability of that is P ( A ) - P ( A B ) = 0 . 5 - 0 . 25 = 0 . 25. 1
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2.14 a P ( A B ) = P ( A ) + P ( B ) - P ( A B ) = 0 . 8 + 0 . 7 - 0 . 9 = 0 . 6. b The event in question is ( A B 0 ) ( A 0 B ). There are two ways to find the probability of this event. One is to notice that A B = ( A B ) ( A B 0 ) ( A 0 B ), and all three of those are mutually exclusive, so P ( A B ) = P ( A B ) + P ( A B 0 ) + P ( A 0 B ) P ( A B ) - P ( A B ) = 0 . 9 - 0 . 6 = 0 . 3 = P ( A B 0 ) + P ( A 0 B ) The other is to work out P ( A B 0 ) and P ( A 0 B ) directly. A = ( A B ) ( A B 0 ) P ( A ) = P ( A B ) + P ( A B 0 ) P ( A B 0 ) = 0 . 8 - 0 . 6 = 0 . 2 B = ( A B ) ( A 0 B ) P ( B ) = P ( A B ) + P ( A 0 B ) P ( A 0 B ) = 0 . 7 - 0 . 6 = 0 . 1 P ( A B 0 ) + P ( A 0 B ) = 0 . 3 2.22 Let A = stopping at the first signal and B = stopping at the second signal. a P ( A B ) = P ( A ) + P ( B ) - P ( A B ) = 0 . 4 + 0 . 5 - 0 . 6 = 0 . 3. b This asks for P ( A B 0 ). Since A B and A B 0 are disjoint, we can say that P ( A ) = P ( A B ) + P ( A B 0 ), so P ( A B 0 ) = 0 . 4 - 0 . 3 = 0 . 1. c Stopping at exactly one signal is the event ( A B 0 ) ( A 0 B ). These are disjoint, so the probability is P ( A B 0 )+ P ( A 0 B ). We’ve just gotten the first probability. We can find the second the same way: P ( A 0 B ) = P ( B ) - P ( A B ) = 0 . 5 - 0 . 3 = 0 . 2. So the probability of stopping at exactly one intersection is 0 . 1 + 0 . 2 = 0 . 3. 2.24 When they give us a hint, use it! B = A ( B A 0 ) P ( B ) = P ( A ) + P ( B A 0 ) 2
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Since P ( C ) 0 for any event C , we know P ( B A 0 ) 0. So P ( B ) P ( A ), q.e.d. As for what this implies for general A and B , we know that A B is always a subset of A , which is always a subset of A B , so P ( A B ) P ( A ) P ( A B ). 2.26 a P ( A 0 1 ) = 1 - P ( A 1 ) = 1 - 0 . 12 = 0 . 88. b
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This homework help was uploaded on 01/31/2008 for the course STAT 220 taught by Professor Shalizi during the Fall '05 term at Carnegie Mellon.

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hw2 solutions - Solutions to Homework Assignment #2...

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