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Consider the question: what is the probability of getting a 6-card poker hand that contains a card from all four suits (order doesn't matter):

Each of the C(52, 6) hands is equally likely. Let E be the event that the hand selected contains 6 a card from all four suits. Then

Pr(E) = |E| / C(52, 6)

For |E|, first pick one heart, then one spade, then one diamond, then one clubs, and, finally, two additional cards. Therefore

|E| = 134 . C(48, 2) and hence Pr(E) = (134 . C(48, 2)) / C(52, 6)

Explain what is wrong with this solution. If there is over-counting in |E|, characterize all hands that are counted more than once, and how many times each such hand is counted.

What is the correct answer for Pr(E)?

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Statistics and Probability, Statistics

Category:- Statistics and Probability
Reference No.:- M91021673

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