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MT1_sol - Feb 7 2008 NAME Math 425 SECTION 8 MIDTERM 1...

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Feb 7, 2008 NAME: Math 425, SECTION 8 MIDTERM 1 – SOLUTIONS Problem 1 [20 pts] A poker hand consisting of 5 cards is dealt from a standard deck of 52 cards. Assume that all combinations are equally likely to occur. What is the probability that the hand is: (a) a flush (that is, all five cards are of the same suit)? Answer: P (flush) = 4 ( 13 5 ) ( 52 5 ) (b) a pair (that is, there are two cards of the same denomination, and the rest are of different denominations) ? Answer: P (pair) = 13 ( 4 2 )( 12 3 ) 4 3 ( 52 5 ) Our sample space S consists of ( 52 5 ) poker hands. Suppose we have a pair. There are 13 ways to choose the denomination in the pair, ( 4 2 ) ways to choose for the suits in the pair, ( 12 3 ) ways to choose the denominations of the remaining three cards and 4 3 ways to choose their suits. Note. It is tempting to write 12 × 11 × 10 instead of ( 12 3 ) in the formula above. By doing so, we take into account the order of the remaining three cards, so we must divide by 3!. Problem 2 [20 points] (a) Ten students are randomly divided into Team A and Team B, with 5 students each. How many divisions are possible? Solution. There are ( 10 5 ) ways to choose the five students in Team A. But now, we also know who is in Team B. So, the answer is ( 10 5 ) = 252. (b) Among the ten students, there are two brothers: John and Fred. What is the probability that they are both in the same team? Please state clearly the assumptions you are making.

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MT1_sol - Feb 7 2008 NAME Math 425 SECTION 8 MIDTERM 1...

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