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Unformatted text preview: C wins the second game. Solution : Let E be the event that A wins at least one game. Then E = { A 1 A 2 ,A 1 B 2 ,A 1 C 2 ,B 1 A 2 ,C 1 A 2 } and so P ( E ) = P ( A 1 A 2 ) + P ( A 1 B 2 ) + P ( A 1 C 2 ) + P ( B 1 A 2 ) + P ( C 1 A 2 ) = 5 / 9 Therefore, the required probability is given by P ( C 2  E ) = P ( C 2 ∩ E ) P ( E ) = P ( A 1 C 2 ) P ( E ) = 1 / 9 5 / 9 = 1 / 5 (b) Suppose instead that the probability that C wins a game is 5 / 7 , and that A and B are equally likely to win a game. Repeat part a). Solution : Here, P ( C ) = 5 / 7 , and P ( A ) = P ( B ) = 1 / 7 . Repeat part a) to obtain i. 0.0408 ii. 0.3846 5. (a) 0.56 (b) 0.857 6. Will be done in next week’s tutorial. 2...
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 Winter '12
 Javid
 Probability theory, Pallavolo Modena, Sisley Volley Treviso

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