three cards room randomly liked without replacement from an simple deck of 52 play cards. Given that the ace of spades is chosen, what is the probabilitythat all three cards room aces?

Using conditional probability$P(A|B) = P(AB)/P(B)$

$P(A) =$ cards space aces

$P(B) =$ ace the spade chosen

I ended up in ~ $P(AB) = \dfrac^3C_1^52C_3$ $3$ methods = Spade,club,heart, spade,heart,diamond and spade society diamond.

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$P(B) = (^51C_2)/(^52C_3)$ i think ace that spade chosen, chose another 2 cards

$P(A|B) = <^3C_1/^52C_3)>/<^51C_2)/(^52C3_)>$

I have looked roughly for this question however could only find with 2 cards not 3.Not certain if this is the correct method ?

probability proof-verification
edited Jul 17 "18 at 20:15

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Since you have chosen ace that spades, in fact three much more aces are continuing to be of which two more aces must be preferred the the probability is $$P=\dfrac\binom32\binom512$$no complexity or conditional probability is needed at all.

answer Jul 17 "18 in ~ 19:25

Mostafa AyazMostafa Ayaz
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