Probability that hand contains Ace and King of at least one suit? Compute the probability that a hand of 13 cards (drawn
randomly from a standard deck of 52) contains both the ace and the
king from at least one suit.
I think I would use the inclusion exclusion principle here, but I'm not sure how to start. Would it be something like: P(Ace+King of suit 1) + P(Ace+King of suit 2) + P(Ace+King of suit 3) + P(Ace+King of suit 4) - ... so on.?
 A: Edit:
In my original answer, I ignored the condition that the ace and king are in the same suit. That is partly due to the awkward language, which technically can support my reading.  But there is no reason for the question to mention suits at all if the problem was about just having an ace and a king,
so I'm taking the suggestion by commenter AnkitSeth that the correct reading that we are looking for the probability that the hand has, for at least one suit, the ace and king of that suit.

Let $A$ be the set of all deals.
Let $A_i$ be the deals that contain the ace and king of suit $i,$ for $i=1,2,3,4.$
Then the inclusion-exclusion gives the probability:
$$\frac{\binom{4}{1}|A_1|-\binom{4}{2}|A_1\cap A_2|+\binom{4}{3}|A_1\cap A_2\cap A_3|-\binom{4}{4}|A_1\cap A_2\cap A_3\cap A_4|}{|A|}\\=\frac{\binom{4}{1}\binom{50}{11}-\binom{4}{2}\binom{48}{9}+\binom{4}{3}\binom{46}{7}-\binom{4}{4}\binom{44}{5}}{\binom{52}{13}}$$

Original Answer
This answer finds the probability that the hand has one ace and one king.
If $p$ is the probability that you get at least one ace and one king, consider $$1-p = p_{\lnot A}+p_{\lnot K}-p_{\lnot AK}$$ where $p_{\lnot A}$ is that probability that you didn't get any ace, $p_{\lnot K}$ is the probability that you got no king, and $p_{\lnot AK}$ is the probability that you got no ace and no king.
So $$p_{\lnot A} = p_{\lnot K}=\frac{\binom{48}{13}}{\binom{52}{13}}$$ and $$p_{\lnot AK}=\frac{\binom{44}{13}}{\binom{52}{13}}$$
So $$p = \frac{\binom{52}{13} - 2\binom{48}{13} + \binom{44}{13}}{\binom{52}{13}}$$
