SOLUTION: I'm working on a homework assignment from my Topics in Mathematical Reasoning class and I've got an answer to this question, but there's no way for me to check the answer I have.

Question 382737: I'm working on a homework assignment from my Topics in Mathematical Reasoning class and I've got an answer to this question, but there's no way for me to check the answer I have.
The question is:
A full house in poker consists of three of one kind and two of another kind in a five-card hand. For example, if a hand contains three Kings and two 5's, it is a full house. If 5 cards are dealt at random from a standard deck of 52 cards, without replacement, determine the probability of getting three Kings and two 5's.
My answer is 1/108,290
Found 2 solutions by richard1234, stanbon:
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
We should first compute the number of five-card full houses with three Kings and two 5's. This number is just 4C3*4C2 = 4*6 = 24.
There are 52C5 possible hands in poker, so the probability is 24/(52C5) = 1/108290.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
If 5 cards are dealt at random from a standard deck of 52 cards, without replacement, determine the probability of getting three Kings and two 5's.
My answer is 1/108,290
-----
Ways to succeed:
Pick 3 kings: 4C3 = 4
Pick 2 5's: 4C2 = 6
----
Ways to succeed = 4*6 = 24
-------
Total number of 5-card hands: 52C5
--------
P(3K and 2 5's) = 24/52C5 = 0.0000092345 = 1/108,290
===========================================================
You have it.
Cheers,
Stan H.
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