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The question is: how to determine $\mathbb{E}[Y]$ using the partition theorem?

The partition theorem says that if we have a partition ${\{B_1,B_2,...\}}$ of our outcomespace $\Omega$ with $\Bbb{P}[B_i]>0$, then $\Bbb{P}[A] = \sum_{i}\Bbb{P}(A\mid B_i)*\Bbb{P}(B_i)~\text{for}~A \in \mathcal{F}.$

First I noted that $\Bbb{P}(X=k)=q(1-q)^{k-1}$ and by definition we have $\Bbb{E}[Y]=\sum_{y\in Y[\Omega]} y*\Bbb{P}(Y=y).$ When I try to write out this definition with the dependence of $Y$ on $X$ I stumble into bad notation and do not see a way trough. Can you help me solve it?

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    $E(Y\mid X)=pX$ and $E(X)=1/q$ hence $E(Y)=$ $____$. – 2017-01-16
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    kNiEsSoKk: Care to explain why you keep deleting your questions immediately after getting answers? A) that's a bit rude. B) some may get the impression that you don't want your teacher to see you post those questions. – 2017-10-01
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    This is my only other question I posted so far, right? And I did not delete it? – 2017-10-01

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Using the tower property of conditional expectation, \begin{align*} E(Y) &= E(E(Y \mid X)) \\ &= E(\text{Binom}(X,p) \mid X) \\ &= E(Xp) \\ &= pE(\text{Geom}(q)) = p\cdot \frac{1}{q} \end{align*}