If X and Y are two independent standard normal r.v.s, calculate $P(X+Y \in [0,1] \mid X \in [0,1])$ I'm not sure how to solve this. This is my attempt so far.
So if X,Y two independent standard normal r.v.s, we have:
$$\mathbb{P}(X+Y\in [0,1] \mid X \in [0,1])=\frac{\mathbb{P}(\{X+Y\in [0,1]\} \cap \{X \in [0,1]\})}{\mathbb{P}(X \in [0,1])}.$$
Moreover, we have:
\begin{split} \mathbb{P}(\{X+Y\in [0,1]\} \cap \{X \in [0,1]\}) = {} &  \int_0^{1}dx\frac{1}{\sqrt{2 \pi}}e^{-\frac{x^2}{2}}\int_{-x}^{1-x} dy\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}.
\end{split}
To calculate the integral, it looks like it might be better to switch to polar coordinates (?). Then we have:
\begin{split} \int_0^{1}dx\frac{1}{\sqrt{2 \pi}}e^{-\frac{x^2}2}\int_{-x}^{1-x} dy \frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}  = & \frac{1}{2\pi}\int_{-\frac\pi4}^{0} \, d \varphi \int_0^{\frac{1} {  \cos \varphi}}r e^{-\frac{r^2}2} \, dr \\
& +  \frac{1}{2\pi}\int_{0}^{\frac\pi2} \, d \varphi \int_0^{\frac1 { \sin \varphi + \cos \varphi}}r e^{-\frac{r^2}2} \, dr \\
= &- \frac{1}{2\pi}\int_{-\frac\pi4}^{0} \, d \varphi \int_0^{-\frac1 { 2\cos^2 \varphi}} e^t \, dt \\
&  - \frac{1}{2\pi}\int_{0}^{\frac{\pi}{2}}d \varphi \int_{0}^{-\frac{1}{2 ( \sin \varphi + \cos \varphi) ^2}} e^t\,dt, \\
\end{split}
and I don't know what to do from here. Although, I'm not even sure if my procedure doesn't have any mistakes. Thanks for any insights.

The desired probability is $$\Pr[0 \le X+Y \le 1 \mid 0 \le X \le 1] = \frac{\int_{x=0}^1 \int_{y=-x}^{1-x} f_{X,Y}(x,y) \, dy \, dx}{\int_{x=0}^1 f_X (x) \, dx}.$$  It is not necessary to separate the cases $X \le 1/2$ versus $X > 1/2$.  We find the denominator easily enough:  $$\int_{x=0}^1 f_X(x) \, dx = \Phi(1) - \tfrac{1}{2}.$$  The numerator is harder:
$$\begin{align*} \int_{x=0}^1 f_X(x) \int_{y=-x}^{1-x} f_Y(y) \, dy \, dx 
&= \int_{x=0}^1 f_X(x) \left(\Phi(1-x) - \Phi(-x)\right) \, dx \\ 
&= \int_{x=0}^1 f_X(x) (\Phi(1-x)+\Phi(x) - 1) \, dx \\ 
&=\left[\frac{\Phi(x)^2}{2} \right]_{x=0}^1 - \left(\Phi(1) - \frac{1}{2}\right) + \int_{x=0}^1 f_X(1-x) \Phi(x) \, dx \\
&= \frac{\Phi(1)^2}{2} + \frac{3}{8} - \Phi(1) + \int_{x=0}^1 f_X(1-x) \Phi(x) \, dx.
\end{align*}$$  Here, the last integral does not seem to have an elementary closed form.  The approximate value of the numerator through numeric integration is $0.125988$; the resulting probability is approximately $0.369094$.
