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Exercise_STAT515Fall2011

Exercise_STAT515Fall2011 - Answers to some exercises in...

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Answers to some exercises in chapter 3 and 4 (STAT 515, Fall 2011) 3.66. a. Because A and B are mutually exclusive, P ( A B ) = P ( A ) + P ( B ) = 0 . 85 . b. By the definition of mutually exclusive events, P ( A B ) = 0 . c. P ( A | B ) = P ( A B ) /P ( B ) = 0 , since the numerator is zero from part b . d. Because B and C are mutually exclusive, P ( B C ) = P ( B ) + P ( C ) = 0 . 70 . 3.70. a. Because the Venn diagram shows that P ( A C ) = P ( B C ) = 0 , so A and C are mutually exclusive, also B and C are mutually exclusive. b. According to part a , we know the following two pairs, ( A, C ) and ( B, C ) , are not inde- pendent. Note that mutually exclusive events cannot be independent events. Next check if A and B are independent. Note that P ( A B ) = P (3) = 0 . 3. Because P ( A ) = P (1) + P (2) + P (3) = 0 . 2 + 0 . 05 + 0 . 3 = 0 . 55 , and P ( B ) = P (3) + P (4) = 0 . 3 + 0 . 1 = 0 . 4, therefore P ( A B ) = P ( A ) P ( B ) . This implies that A and B are not inde- pendent. In summary none of the pairs are independent. c. P ( A B ) = P (1) + P (2) + P (3) + P (4) = 0 . 20 + 0 . 05 + 0 . 30 + 0 . 10 = 0 . 65, Using the additive rule, P ( A B ) = P ( A ) + P ( B ) - P ( A B ) = 0 . 55 + 0 . 4 - 0 . 3 = 0 . 65 .
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