1 2 also p l 10 a 1 2 p l 15 a p l 20 a 1 4 and p l

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) = 1 2 . Also, P ( L 10 | A ) = 1 / 2 , P ( L 15 | A ) = P ( L 20 | A ) = 1 / 4 , and P ( L 10 | B ) = P ( L 15 | B ) = P ( L 20 | B ) = 1 / 3 . (a) From Total Probability, we have P ( L 10 ) = P ( L 10 | A ) · P ( A ) + P ( L 10 | B ) · P ( B ) = 1 2 · 1 2 + 1 3 · 1 2 = 5 12 . P ( L 15 ) = P ( L 15 | A ) · P ( A ) + P ( L 15 | B ) · P ( B ) = 1 4 · 1 2 + 1 3 · 1 2 = 7 24 . P ( L 20 ) = P ( L 20 | A ) · P ( A ) + P ( L 20 | B ) · P ( B ) = 1 4 · 1 2 + 1 3 · 1 2 = 7 24 . (b) This corresponds to the event L 20 | (( L 10 L 15 ) c A ) , which can be written as L 20 | ( L 20 A ) . Thus P ( L 20 | L 20 A ) = P ( L 20 L 20 A ) P ( L 20 A ) = P ( L 20 A ) P ( L 20 A ) = 1 . 2
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(c) This corresponds to the event L 20 | ( L c 10 A ) . Thus P ( L 20 | L c 10 A ) = P ( L 20 L c 10 A ) P ( L c 10 A ) = P ( L 20 A ) P ( L c 10 A ) = P ( L 20 | A ) · P ( A ) P ( L c 10 | A ) · P ( A ) = 1 / 4 · 1 / 2 (1 - 1 / 2) · 1 / 2 = 1 2 . (d) This corresponds to the event A | L c 10 . From part (a), P ( L c 10 ) = 1 - P ( L 10 ) = 7 / 12 . By Bayes’ Rule, P ( A | L c 10 ) = P ( L c 10 | A ) · P ( A ) P ( L c 10 ) = (1 - 1 / 2) · 1 / 2 7 / 12 = 3 7 . (e) This corresponds to the event L 20 | L c 10 . Thus P ( L 20 | L c 10 ) = P ( L 20 L c 10 ) P ( L c 10 ) = P ( L 20 ) P ( L c 10 ) = 7 / 24 7 / 12 = 1 2 . 5. Yes. Since C is independent from A B we must have P ( C ( A B )) = P ( C ) P ( A B ) . However since A and B are pairwise independent then P ( A B ) = P ( A ) P ( B ) . Hence P ( C ( A B )) = P ( C ) P ( A ) P ( B ) . Finally since A , B and C are pairwise independent, we deduce that A , B and C are independent. 6. Even though in reality play ends as soon as the winning team is determined, for purposes of com- putation we can assume that the teams play all fives games. The probability that the See Knays win exactly k games out of 5 games can be written as ( 5 k ) (1 - p ) k p 5 - k . Since the See Knays will win the series if k = 3 , 4 , or 5 , P ( the See Knays win the series ) = 5 k =3 ( 5 k ) (1 - p ) k p 5 - k . 7. Let A be the event in which the factory A is chosen, B the factory B, and W be the event in which the selected jPhone works. P ( A ) = P ( B ) = 1 2 because we select a factory uniformly at random.
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