Probability with dependent Bernoulli and normal distribution Question. Let $X\sim\text{Bernoulli}\left(\frac12\right)$ and $Y\sim\text{Norm}\left(2(X-\frac12),1\right)$. Calculate $\mathbb{P}(X=1\mid Y\geqslant 1).$
As you can see, the random variable $X$ is nested into the expectation of the normal distribution. I have never seen a situation like this before.
My try:
$$\mathbb{P}(X=1\mid Y\geqslant 1)=\mathbb{P}(Y\geqslant 1 \mid X=1)\frac{\mathbb{P}(X=1)}{\mathbb{P}(Y\geqslant 1)}$$
and
$$\mathbb{P}(Y\geqslant 1\mid X=x)=\int_1^\infty \frac{1}{\sqrt{2\pi}}e^{-(s-2(x-1/2))^2/2}\,ds=\int_{2-2x}^\infty \frac{1}{\sqrt{2\pi}}e^{-u^2/2}\,du$$
I honestly have no idea how to proceed.
Could someone provide help?
 A: Doing it your way, we get
$$\Bbb P (Y\geq 1 | X = 1) = \int_{0}^\infty \frac{1}{\sqrt{2\pi}}e^{-u^2/2}\;du = \frac{1}{2}$$
$$\Bbb P(X = 1) = \frac{1}{2}$$
\begin{align}\Bbb P(Y\geq 1) &= \Bbb P(Y\geq 1 | X = 1)\Bbb P (X = 1) + \Bbb P(Y\geq 1 | X = 0)\Bbb P(X = 0)\\
&= \frac{1}{2}\cdot \frac{1}{2} + (1 - \Phi(2))\cdot \frac{1}{2} \\
&\approx 26.1375\%
\end{align}
So you end up with
$$\Bbb P(X = 1 | Y\geq 1) \approx \frac{25\%}{26.1375\%} \approx 95.65\%$$
which is exactly the same as in zoli's answer!
EDIT: Just in case, by $\Phi$, I mean
$$\Phi(x) = \int_{-\infty}^x \frac{1}{\sqrt{2\pi}}e^{-z^2/2}\;dz$$
A: The distribution of $Y$ is $N(-1,1)$ with probability $\frac12$ and $N(1,1)$ with the same probability. So the pdf of $Y$,
$$f_Y(y)=\frac12\frac1{\sqrt{2\pi}}\left(e^{\frac{(y+1)^2}2}+e^{\frac{(y-1)^2}2}\right).$$
So 
$$P(Y\geq 1)=\frac12\frac1{\sqrt{2\pi}}\int_1^{\infty}\left(e^{\frac{(y+1)^2}2}+e^{\frac{(y-1)^2}2}\right)\ dy.$$
And 
$$P(X=i\cap Y\geq 1)=P(Y\geq 1\mid X=1)\frac12=\frac12\frac1{\sqrt{2\pi}}\int_1^{\infty}e^{\frac{(y-1)^2}2}\ dy.$$
So
$$P(X=1\mid Y\geqslant 1)=\frac{P(X=1\cap Y\geq 1)}{P(Y\geq 1)}=\frac{\int_1^{\infty}e^{\frac{(y-1)^2}2}\ dy}{\int_1^{\infty}\left(e^{\frac{(y+1)^2}2}+e^{\frac{(y-1)^2}2}\right)\ dy}.$$
