I don't understand what the random variables here mean exactly. $B$ is a $5$ element subset of the set $A$.
$X$ over here is defined by $B$ intersection ${a,b,c,d}$. Does this mean that ${a,b,c,d}$ is part of the $5$ element subset of $B$? I'm assuming I have to use the indicator variables to solve this.
If I define $X = 1$ as $B$ intersection ${a,b,c,d}$,
so $E(X) = P(X=1) = $${5}\choose{2}$ * ${4}\choose{2}$ / ${8}\choose{2}$ ?
I know for independence, $P(X=x)*P(Y=y) = P(X=x $ $ and $ $ Y=y)$
But again, I'm not sure what the defined random variables mean, how should I solve this problem, it's been a struggle for the last couple of hours.
As is pointed out, $|A|$ denotes the cardinality (the number of elements) of $A$. Thus we may write $$ X = \sum_{x=a,b,c,d} 1_{x \in B},$$ and $$Y = \sum_{x=e,f,g,h}1_{x\in B}$$ where $1_{z\in E}$ denotes the indicator function (taking values in $\{0,1\}$ whose value is $1$ if and only if the $z\in E$.) We can immediately observe that $$ X+Y =|B| = 5. $$ This readily implies that $X$ and $Y$ are dependent. And since expectation operation is linear, we have $$ E[X] = \sum_{x=a,b,c,d} E[1_{x \in B}] = \sum_{x=a,b,c,d} P(x \in B). $$ For arbitrary $x$, it holds that $$ P(x\in B) = \frac{\binom{7}{4}}{\binom{5}{8}} = \frac{5}{8}. $$Hence, this gives $$ E[X] = 4\cdot\frac{5}{8} = \frac{5}{2}, $$ and $$ E[Y] =\frac{5}{2}$$ also follows.