A poker hand of $5$ cards is dealt from a normal deck of $52$ cards. What is the probability that the hand contains more than one ace, given that it contains an ace of spades?
I'm assuming that I need to use conditional probability to solve this question. Specifically, let X be the number of aces in a five-card hand. Then, I need to solve
$$Pr( X \gt1 |\text{Ace of Spades}) = \frac{Pr(X\gt1 ~\text{and Ace of Spades})}{ Pr(\text{Ace of Spades})} $$
Here is how I tried to find the Pr(Ace of Spades)
$$Pr(\text{Ace of Spades}) = \frac{\binom{4}{1}\times \binom{48}{4}}{\binom{52}{5}}$$
I'm not sure how to go about solving for $Pr(X \gt 1~\text{and Ace of Spades})$.
Your formula for $\mathbb P(A\spadesuit)$ looks a little off -- it should be $\frac{\binom{51}{4}}{\binom{52}{5}}=\frac{5}{52}$. This is because if we want the probability of drawing the ace of spades, then we need to allow the other four cards to be anything.
Your expression looks like the probability of drawing exactly one ace (not necessarily the ace of spades).
Anyhow, it's probably easier to think of the probability as $$\mathbb P(X>1\mid A\spadesuit)=1-\mathbb P(X=1\mid A\spadesuit),$$ But the probability of drawing exactly one ace, conditional on drawing the ace of spades, is $$\frac{\mathbb P(X=1\cap A\spadesuit)}{\mathbb P(A\spadesuit)}=\frac{\binom{48}{4}/\binom{52}{5}}{\binom{51}{4}/\binom{52}{5}}=\frac{3243}{4165}\implies\mathbb P(X>1\mid A\spadesuit)=\frac{922}{4165}.$$