<s>[INST] Given that exactly 2 Jacks appear, what is the expected number of Aces that appear? Pick 5 cards from standard 52 cards without replacement. Given that exactly 2 jack cards appear, find the expected number of ace cards that appear.
ATTEMPT
Let $X$ be number of aces chosen and $Y$ number of jacks chosen so we want find $E(X|Y=2)$. In the definition we have
$$ E(X|Y=2) = \sum_{x=0}^5 x \frac{ p_{XY}(x,2) }{p_Y(2)} $$
First we find 
$$ p_{XY}(1,2) = P(X=1 \cap Y=2) = \frac{ {4 \choose 1 } {4 \choose 2} {13 \choose 2} }{ {52 \choose 5} } $$
$$ p_{XY}(2,2) = P(X=2 \cap Y=2) = \frac{ {4 \choose 2} {4 \choose 2} {13 \choose 2} }{ {52 \choose 5} } $$
$$ p_{XY}(3,2) = P(X=3 \cap Y=2) = \frac{ {4 \choose 3 } {4 \choose 2} {13 \choose 2} }{ {52 \choose 5} } $$
after $x=3$ we have $0$ since we cant have more that 5 cards. finally we find 
$$ p_Y(2) = P(Y=2) = \frac{ {4 \choose 2}{13 \choose 3} }{ {52 \choose 5} } $$
Now, pluggin in into the first equation should give the answer. Is this a correct approach to tackle this problem?
 [/INST] 
Now, pluggin in into the first equation should give the answer. Is this a correct approach to tackle this problem?

Yes, that would work, although your evaluations are off.
$p_{\small Y}(2) =\mathsf P(Y{=}2) = \left.\binom 4 2\binom {48}{3}\middle/\binom{52}{5}\right.$ is the probability for selecting $2$ from $4$ Jacks and $3$ from $48$ non-Jacks when selecting $5$ from $52$ cards.
$p_{\small X,Y}(x,2) =\mathsf P(X{=}x, Y{=}2) =\left.\binom 42\binom 4x\binom {44}{3-x}\middle/\binom{52}{5}\right. $ is the probability for selecting $2$ from $4$ Jacks, $x$ from $4$ aces, and $3-x$ from $44$ non-Jack-or-Aces when selecting $5$ from $52$ cards.
Thus $p_{\small X\mid Y}(x\mid 2) =\mathsf P(X{=}x\mid Y{=}2) =\left.\binom 4x\binom {44}{3-x}\middle/\binom{48}{3}\right. $ is the probability for selecting $x$ from $4$ aces, and $3-x$ from $44$ non-Jack-or-Aces when selecting $3$ from $48$ non-Jack cards.
Alternatively, I place the hand with the non-Jack cards face down on the table, and the two Jacks face up, asserting they are the only Jacks in the hand.   I point to one of the face down cards and ask: What is the expectation that this card is an Ace?     I do this for each from the three face-down cards in the hand, and the answer is the same.
So, by using the Linearity of Expectation, the expected count for Aces in the hand when given exactly two Jacks are in the hand, will be...
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