Three cards are randomly chosen without replacement from an ordinary deck of 52 playing cards. Given that the ace of spades is chosen, what is the probability that all three cards are aces?

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

$P(A) =$ cards are aces

$P(B) =$ ace of spade chosen

I ended up at $P(AB) = \dfrac{^3C_1}{^{52}C_3}$
$3$ ways = Spade,club,heart, spade,heart,diamond and spade club diamond.

$P(B) = (^{51}C_2)/(^{52}C_3)$ assume ace of spade chosen, chose another 2 cards

$P(A|B) = [^3C_1/^{52}C_3)]/[^{51}C_2)/(^{52}C3_)]$

I have looked around for this question but could only find with 2 cards not 3. Not sure if this is the correct way ?

  • 2
    $\begingroup$ I think that looks fine as it is. $\endgroup$ – Allen O'Hara Jul 17 '18 at 19:09

Since you have chosen ace of spades, in fact three more aces are remaining of which two more aces should be chosen the the probability is $$P=\dfrac{\binom{3}{2}}{\binom{51}{2}}$$no complexity or conditional probability is needed at all.

  • $\begingroup$ Thanks, yeah i was forced to use conditional in this problem, but your answer helped me check my answer is correct ! thanks $\endgroup$ – MasterYoshi Jul 17 '18 at 19:36
  • 1
    $\begingroup$ You're welcome!! $\endgroup$ – Mostafa Ayaz Jul 17 '18 at 19:36

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