The excesise is next:
An urn contains four white and six black balls. A random sample of size 4 is chosen. Let $X$ denote the number of white balls in the sample. An additional ball is now selected from the remaining six balls in the urn. Let $Y$ equal 1 if this ball is white and 0 if it is black.
Find (a) $E [Y |X = 2]$ Analogous to the first part of the resolution of this question we have
$$\begin{aligned} &\bullet\,\, E[Y|X=2]=\sum_yyP(Y=y|X=2)=P(Y=1|X=2)=\frac{2}{6}=\frac13.\\ \end{aligned}$$ Where $P(Y=1|X=2)=P(Z=1)=\frac26$ with $Z$ equal $1$ if this ball is white and $0$ if it is black about the event of having an urn with $2$ white balls and $4$ black balls.
If i want calculate $P(Y=1|X=2)$ without a auxiliar Random Variable $Z$ then $P(Y=1|X=2)=\frac{P(Y=1,X=2)}{P(X=2)}$ but i dont know how to calculate $P(Y=1,X=2)$.