SOLUTION: Please help me to solve this. In a poker hand consisting of 5 cards, what is the probability of holding 2 Aces and 3 Kings?

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Question 1101680: Please help me to solve this.
In a poker hand consisting of 5 cards, what is the probability of holding 2 Aces and 3 Kings?
Found 2 solutions by Edwin McCravy, Theo:
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
(4 aces choose 2) AND (4 kings choose 3) = (4C2)(4C3) = 6×4 = 24 ways

(52 cards choose 5) = 52C5 = 2698960

So the probability is 24 ways out of 2598960 or

24/2598960 which reduces to 1/108290

That 1 way out of 108,290 ways.

Not very likely!

Edwin
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
p(first ace) = 4/52
p(second ace) = 3/51
p(first king) = 4/50
p(second king) = 3/49
p(third king) = 2/48

p(2 aces and 3 kings) = (4*3*4*3*2) / (52*51*50*49*48)

this is equal to 288 / 311875200.

the scientific notation equivalent of this is 9.234463016 * 10^-7.

the decimal equivalent of this is .0000009234463016.


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