Why is $\mathbb E(X) =\mathbb E(Y)$ in this joint probability distribution?

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If we know $X|Y$ is a normal random variable with mean $Y$ and variance $2$, and $Y$ is a binomial distribution with success probability $0.3$ and the number of trials $5$, why is $\mathbb E(Y) = 0.3\times5 =\mathbb E(X)?$

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$E(X|Y)=Y$ so (by taking expectation on both sides) we get $EX=EY$. Since $Y\sim B(0,3.5)$ we have $EY=(0.3)(5)$.