hw2 solutions

# 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 ) , so the probability is 1 - 0 . 65 = 0 . 35. c The event is A B , 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 ) ( A B ). There are two ways to find the probability of this event. One is to notice that A B = ( A B ) ( A B ) ( A B ), and all three of those are mutually exclusive, so P ( A B ) = P ( A B ) + P ( A B ) + P ( A B ) P ( A B ) - P ( A B ) = 0 . 9 - 0 . 6 = 0 . 3 = P ( A B ) + P ( A B ) The other is to work out P ( A B ) and P ( A B ) directly. A = ( A B ) ( A B ) P ( A ) = P ( A B ) + P ( A B ) P ( A B ) = 0 . 8 - 0 . 6 = 0 . 2 B = ( A B ) ( A B ) P ( B ) = P ( A B ) + P ( A B ) P ( A B ) = 0 . 7 - 0 . 6 = 0 . 1 P ( A B ) + P ( A 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 ). Since A B and A B are disjoint, we can say that P ( A ) = P ( A B ) + P ( A B ), so P ( A B ) = 0 . 4 - 0 . 3 = 0 . 1. c Stopping at exactly one signal is the event ( A B ) ( A B ). These are disjoint, so the probability is P ( A B )+ P ( A B ). We’ve just gotten the first probability. We can find the second the same way: P ( A 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 ) P ( B ) = P ( A ) + P ( B A ) 2
Since P ( C ) 0 for any event C , we know P ( B A ) 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 1 ) = 1 - P ( A 1 ) = 1 - 0 . 12 = 0 . 88. b We want P ( A 1 A 2 ). We are given P ( A 1 ), P ( A 2 ) and P ( A 1 A 2 ). We can relate all four of these: P ( A 1 A 2 ) = P ( A 1 ) + P ( A 2 ) - P ( A 1 A 2 ) = 0 . 12 + 0 . 07 - 0 . 13 = 0 . 06 c We want the event A 1 A 2 A 3 . Let’s treat this as ( A 1 A 2 ) A 3 , since we already know the probability of A 1 A 2 . We can write A 1 A 2 = (( A 1 A 2 ) A 3 ) (( A 1 A 2 ) A 3 ), since if it has type 1 and type 2 defects, it can either have type 3 defects or not. Since A 1 A 2 A 3 and A 1 A 2 A 3 are mutually exclusive, we’ve got P ( A 1 A 2 ) = P ( A 1 A 2 A 3 ) + P ( A 1 A 2 A 3 ) We’re given P ( A 1 A 2 A 3 ) = 0 . 01, and we’ve just found P ( A 1 A 2 ) = 0 . 06, so P ( A 1 A 2 A 3 ) = 0 . 05.

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• Fall '05
• Shalizi
• Statistics, Zagreb, ways, Shift work, G DG D G DS D S DS SG DG S D G

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