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

Probability - Part 2

Solution:

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.
$\therefore P(K|K) = \frac{3}{51}$
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.
$\therefore P(A|KK) = \frac{4}{50}$
By multiplication law of probability, we have
$P(KKA) = P(K) . P(K|K). P(A|KK)$
$=\frac{4}{52} \times \frac{3}{51} \times \frac{4}{50}$
$= \frac{2}{5525}$