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Suppose the probability for A to win game B 0.4. If A has an option of playing either a" best of 3 game" of a "best of 5 game" match against B, which option should be choose so that the probability of his winning the match is higher? |
Answer» The probability p1 (say) of winning the best of three games is = the prob. of winning two games + the prob. of winning three games. = 3C2 (0.6) (0.4)2 + 3C3 (0.4)3 [Using Binomial distribution] Similarly the probability of winning the best five games is p2 (say) = the prob. of winning three games + the prob. of winning 5 games. = 5C3 (0.6)2 (0.4)3 + 5C3 (0.6) (0.4)3 + 5C5 (0.4)5 We have p1 = 0.288 + 0.064 = 0.352 And p2 = 0.2304 + 0.0768 + 0.01024 = 0.31744 As p1 > p2 ∴ A must choose the first offer i.e. best of three games. |
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