CEE 202 Homework 2 Solutions

# 108104000144 0077 you could have done this much more

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Unformatted text preview: these two cards of one kind, they can be 1,2,3,..13 and each of them has 4 ȹ 4 ȹ 4! suits, so there are n1 = 13 × ȹ ȹ = 13 × = 78 combinations 2!2! ȹ 2 Ⱥ For the rest three cards (of one kind), they can be any number except the one appears on those two cards, the combinations are ȹ 4 ȹ 4! n2 = (13 − 1) × ȹ ȹ = 12 × = 48 3!1! ȹ 3 Ⱥ For this event n=n1*n2=78*48=3744 The probability of full house is P(full house)=3744/2598960=0.00144 b) If there are exactly two aces, these two cards of one kind can only come from ȹ 4 ȹ 4! the 4 suits of aces, then the combination is n1 = ȹ ȹ = = 6 ȹ 2 Ⱥ 2!2! And for the rest three cards, the combination will be the same as that in part (a), n2=48 For this even n=n1*n2=16*48=288 The probability of full house which contains exactly two aces is P (full house⋂ twoaces) =288/2598960=1.108*10 ­4 Given the condition of full hou...
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