Probability hand has 4 aces, given 5-card hand and ....

fred2028

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3. A 5-card poker hand is dealt and the last card is turned face up. Determine the probability that you have been dealt 4 aces, given that the card turned over is an ace.
My work:
Since this is conditional:
P(A|B) = P(A n B) / P(B) A = Dealt 4 aces B = Turned card is ace
And here is where I'm stuck. I dunno how to go about calculating the probability that of the remaining 4 cards, 3 are aces. I tried 3/4 * 13^4, and I tried 13^3 and 13^4. The answer in the back of the book is 1/20825.
Any help?
 
\(\displaystyle \L\\(\frac{3}{51})(\frac{2}{50})(\frac{1}{49})\)
 
galactus said:
\(\displaystyle \L\\(\frac{3}{51})(\frac{2}{50})(\frac{1}{49})\)
Oh I see. Aren't you supposed to take into account that it's conditional though?
 
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