I am given: $Y$ equals $1$ if a success was achieved in the first try (of something, doesn't matter what), and $0$ otherwise.
I need to prove the expectation of the Geometric distribution using $Y$ and using the law of total expectation.
I'm not really sure where to start and I need some directions. I know that $E(Y) = E(E(Y | X))$ but what should be my $X$? And how do I even start the proof?
but what should be my $X$? And how do I even start the proof?
Let $X$ be the count of tries until success in an indefinite sequence of independent Bernoulli trials with identical success rate $p$. That is, $~X\sim\mathcal{Geo_1}(p)$
$$\mathsf P(X=x) ~=~ p\,(1-p)^{x-1} \cdot\mathsf 1_{x\in\Bbb N^+}$$
Specify that $Y$ is the indicator of success on the first try of this geometric distribution.$$Y =\begin{cases} 1 &:& X=1\\0 &:& X>1\end{cases}$$
By the tower principle, $\mathsf E(Y) ~=~ \mathsf E\bigl(\mathsf E(Y\mid X)\bigr)$, aka the Law of Total Expectation:
$$\begin{align}\mathsf E(Y) ~&=~ \sum_{x=1}^\infty \mathsf P(X=x)~\mathsf E(Y\mid X=x) \\[0ex] &~~\vdots\end{align}$$
