We draw $5$ cards from the $52$ card deck (without returning). What is the probability that we have exactly $3$ aces if we know that the first card is a black ace?
In this task we have conditional probability. So let:
$A$ - we have exactly $3$ aces
$B$ - the first card is a black ace
Then: $$P(A|B)=\frac{P(A \cap B)}{P(B)}=\frac{|A\cap B|}{|B|}=\frac{|A\cap B|}{\binom21 \binom{51}{4}}$$ However I have a problem with $|A\cap B|$ because I know that in $\binom21 \binom32\binom{48}{2}$ I do not include the order of draws.
Since you know for sure that the first card you pick WILL be a black ace, the problem simplifies to picking only $2$ aces from $4$ cards out of the deck (which has a missing black ace). $$\text{Total Number of Combinations}=N={51\choose4}$$ $$\text{Possible Ways to Choose 2 Aces}=n={3\choose2}\times{48\choose 2}$$ $$P=\frac nN\approx0.013541$$