Let $X1, \ X2, \ X3, \ X4$ denote the number of hearts, diamonds, clubs, and spades drawn from 10 draws with replacement from a standard 52 card deck of playing cards.
What is $E\ [X2\ |\ X1 + X4 = 5]$ ?
Is the answer $1/4 * 5 = 5/4$ ? My reasoning is that we know that 5 of the 10 cards are not $X2$, so we are only concerned with the other 5. So we multiply the other 5 by the probability of getting a heart, i.e. $1/4$. Thanks in advance
On
$\mathsf E(X_\diamondsuit\mid X_\heartsuit+X_\spadesuit=5)$ is the expected count for diamonds in 10 draws (with replacement) given that we know exactly five of the cards are either hearts or spades.
That is the expected count for diamonds among five cards that are each either diamonds or clubs, with equal likelyhood.
$$\frac{5}{2}$$
Your answer $\dfrac14*5$ would be correct for $E[X2 | X1+X4 \geq 5]$. As would your reasoning. But $X1 + X4$ is exactly 5, no more.
Your reasoning starts out correctly: we know that 5 cards are not X2, so we can disregard them. This leaves us with 5 cards.
This is where you cut a crucial corner. We know that the remaining 5 cards are not X1 or X4, because X1 + X4 is already 5 from the first 5 cards. Another card in that camp would push the sum over 5. We know that the remaining 5 cards are not X1 or X4, so they must be X2 or X3 with equal probability: $\dfrac12$ for both.
Ultimately, the correct answer is $5*0 + 5*\dfrac12 = \dfrac52$
That's 5 cards that are X1 or X4 and 5 cards that are X2 or X3.