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Unformatted text preview: Solutions to Homework 7, Math 55 1. The probability that Beatrix starts with the crown and throws it to Andrew is 1 3 1 2 = 1 6 ; similarly, the probability that Charles starts with the crown and throws it to Andrew is 1 6 . Therefore, the probability that Andrew has the crown after one turn is 1 6 + 1 6 = 1 3 . Similarly, the probability that Charles has the crown after one turn is 1 3 from Andrew, and 1 3 1 2 = 1 6 from Beatrix, so the total probability is 1 3 + 1 6 = 1 2 . On the other hand, if Beatrix gets the crown, it must be from Charles; therefore, the probability that Beatrix has the crown after one turn is 1 3 1 2 = 1 6 . 2. We have p ( A ) = 4 52 = 1 13 ; p ( B ) = 13 52 = 1 4 ; and p ( C ) = 26 52 = 1 2 . Now p ( A B ) = 1 52 = p ( A ) p ( B ), and p ( A C ) = 2 52 = p ( A ) p ( C ), but p ( B C ) = 13 52 6 = p ( B ) p ( C ). Thus, A and B are independent, and so are A and C , but B and C are not independent. If we add a joker to the deck, then p ( A ) = 4 53 ; p ( B ) = 13 53 ; and p ( C ) = 26 53 . However, p ( A B ) = 1 53 6 = p ( A ) p ( B ); p ( A C ) = 2 53 6 = p ( A ) p ( C ); and p ( B C ) = 13 53 6 = p ( B ) p ( C ). Thus, in this case, no pair of events from A, B, C is independent. 3. We have p ( A B ) = p ( S ( A B )) = 1 p ( A B ) = 1 ( p ( A ) + p ( B ) p ( A B )) = 1 p ( A ) p ( B ) + p ( A ) p ( B ) = (1 p ( A ))(1 p ( B )) = p ( A ) p ( B ). Therefore, A and B are independent. Similarly, p ( A B ) = p ( A ( A B )) = p ( A ) p ( A B ) = p ( A ) p ( A ) p ( B ) = p ( A )(1 p ( B )) = p ( A ) p ( B ), so A and B are independent. (Note that in the step p ( A ( A B )) = p ( A ) p ( A B ), we need to use the fact that A B A .) However, if A and A are independent, then p ( A ) p ( A ) = p ( A A ) = p ( ) = 0, which implies p ( A ) = 0 or p ( A ) = 0, and in the latter case, p ( A ) = 1 p ( A ) = 1. Thus, A and A are not independent unless p ( A ) = 0 or p ( A ) = 1....
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 Spring '08
 STRAIN
 Math, Probability

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