Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards. What is the probability that first two cards are kings and the third card drawn is an ace?

$\frac{2}{5525}$

The correct answer is Option (1) → $\frac{2}{5525}$ ##

Let $K$ denote the event that the card drawn is king and $A$ be the event that the card drawn is an ace. Clearly, we have to find $P(KKA)$.

Now $P(K) = \frac{4}{52}$

Also, $P(K|K)$ is the probability of second king with the condition that one king has already been drawn. Now there are three kings in $(52 - 1) = 51$ cards.

Therefore $P(K|K) = \frac{3}{51}$

Lastly, $P(A|KK)$ is the probability of third drawn card to be an ace, with the condition that two kings have already been drawn. Now there are four aces in left 50 cards.

Therefore $P(A|KK) = \frac{4}{50}$

By multiplication law of probability, we have

$P(KKA) = P(K) \cdot P(K|K) \cdot P(A|KK)$

$= \frac{4}{52} \times \frac{3}{51} \times \frac{4}{50} = \frac{2}{5525}$