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Unformatted text preview: Solutions to Homework # 1 1. First, we note that there are ( 52 5 ) ways to select a poker hand of 5 cards. a) There are 4 ways to choose a suit and then there are ( 13 5 ) ways to choose a hand of this suit. Hence the probability to get a flush is 4 parenleftbigg 13 5 parenrightbigg / parenleftbigg 52 5 parenrightbigg = 4 13! 5!8! / 52! 5!47! = 4 13!47! 8!52! = 33 16660 ≈ . 001980 . Answer: the probability of a flush is 4 13!47! 8!52! . b) We choose a pair of denominations a and b in ( 13 2 ) ways, after which we choose the remaining denomination c in 11 ways. Next, we choose a pair of cards of denomination a in ( 4 2 ) = 6 ways, a pair of cards of denomination b in ( 4 2 ) = 6 ways and a card of denomination c in 4 ways. Hence the probability to have two pairs is parenleftbigg 13 2 parenrightbigg · 11 · 6 · 6 · 4 / parenleftbigg 52 5 parenrightbigg = 13! · 11 · 6 · 6 · 4 · 5! · 47! 2!11!52! = 198 4165 ≈ . 0475 . Answer: the probability of two pairs is 13! · 11 · 6 · 6 · 4 · 5! · 47!...
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 Winter '12
 AlexanderBarvinok
 Probability, #, Prime number, i=1

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