sec_2_5_sol

# sec_2_5_sol - APPM 4/5570 Solutions to Problems from...

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Unformatted text preview: APPM 4/5570 Solutions to Problems from Section 2.5 71. a. P ( B | A ) = P ( B ) = 1- . 7 = 0 . 3 If A and B are independent events then so are “not A ” and “not B ”. b. P ( A ∪ B ) = P ( A ) + P ( B )- P ( A ∩ B ) = . 4 + 0 . 7- (0 . 4)(0 . 7) = 0 . 82 (Note that P ( A ∩ B ) = P ( A ) P ( B ) by independence of A and B .) c. P ( A ∩ B | A ∪ B ) = P (( A ∩ B ) ∩ ( A ∪ B )) P ( A ∪ B ) With the help of a Venn diagram, we can see that ( A ∩ B ) ∩ ( A ∪ B ) = A ∩ B So, we have P ( A ∩ B | A ∪ B ) = P ( A ∩ B ) P ( A ∪ B ) We have already computed the denominator. As for the numerator, we use the fact that A and B being independent implies that A and B are independent, so P ( A ∩ B ) = P ( A ) · P ( B ) = (0 . 4)(0 . 3) = 0 . 12 Hence, the answer is P ( A ∩ B | A ∪ B ) = . 12 . 82 ≈ . 1463 . 74. a. P ( both O ) = P ( first is O AND second is O ) indep = P ( first is O ) · P ( second is O ) = (0 . 44)(0 . 44) = 0 . 1936 . b....
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## This note was uploaded on 01/23/2011 for the course APPM 5440 at Colorado.

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sec_2_5_sol - APPM 4/5570 Solutions to Problems from...

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