SOLUTION: Suppose that seven playing cards are to be randomly selected from a full deck of 52. A. Find the probability of getting "three of a kind" and nothing better (i.e. three cards o

Question 819312: Suppose that seven playing cards are to be randomly selected from a full deck of 52.
A. Find the probability of getting "three of a kind" and nothing better (i.e. three cards of the same denomination, with no more than one of any other denomination.)
B. Find the probability of obtaining a "flush", ie. seven cards that are all in the same suit.
C. Find the odds of the event described in part (b).

Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Suppose that seven playing cards are to be randomly selected from a full deck of 52.
A. Find the probability of getting "three of a kind" and nothing better (i.e. three cards of the same denomination, with no more than one of any other denomination.)
Pick a denomination: 13 ways
Pick 3 cards of that denomination: 4C3 = 4 ways
Pick another denomination: 12 ways
Pick a card of that denomination: 4 ways
Pick a 3rd denomination: 11 ways: 11 ways
Pick a card of that denomination: 4 ways
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# of ways to succeed: (13*4*12*4*11*4)
# of possible outcomes: 52C5
Ans: [13P3*4^3]/52C5
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B. Find the probability of obtaining a "flush", ie. seven cards that are all in the same suit.
Pick a suit: 4 ways
Pick 7 cards from that suit: 13C7
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Ans: (4*13C7)/52C7 = 6864/52C7
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C. Find the odds of the event described in part (b).
Ans: 6864:[52C7-6864] = 6864/133777696
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Cheers,
Stan H.
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Answer by Edwin McCravy(20059) (Show Source): You can put this solution on YOUR website!

I disagree with the other tutor on A: Pick a denomination: 13C1 = 13 ways Pick 3 cards from that denomination: 4C3 = 4 ways Pick four other different denominations: 12C4 = 495 ways Pick one card from the lowest of those 4 denominations: 4C1 = 4 ways Pick one card from the second lowest of those 4 denominations: 4C1 = 4 ways Pick one card from the next to the highest of those 4 denominations: 4C1 = 4 ways Pick one card from the highest of those 4 denominations: 4C1 = 4 ways. Number of successful hands = (13C1)(4C3)(12C4)(4C1)(4C1)(4C1)(4C1) = 13*4*495*4*4*4*4 = 6589440 Number of possible hands = 52C7 = 133784560 Probability of three of a kind = 6589440/133784560 = 0.0492541142 Edwin

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