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Mathematics 13 Online
OpenStudy (anonymous):

If a poker hand (five cards) is known to contain at least three aces, what is the probability that it contains all four aces?

OpenStudy (zarkon):

I think you should use the definition of conditional probability.

OpenStudy (zarkon):

P(have all four aces| hand contains at least three aces) =P(have all four aces,hand contains at least three aces)/P(hand contains at least three aces) =P(have all four aces)/P(hand contains at least three aces)

OpenStudy (zarkon):

let A=#aces =P(A=4)/P(A>=3)=P(A=4)/[P(A=3)+P(A=4)]

OpenStudy (dumbcow):

Zarkon, you are correct i thought i could take a shortcut but you get a different answer i believe you should get 48/4560 = 1/95

OpenStudy (zarkon):

\[P(A=4)=\frac{_{4}C_4\cdot_{48}C_{1}}{_{52}C_5}\] \[P(A=3)=\frac{ _{4}C_3\cdot_{48}C_{2}}{_{52}C_5}\]

OpenStudy (zarkon):

1/95 is correct

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