Probability - point inside disk $X, Y \sim \mathcal{N}(0,1)$ are independent random variables. Let $D_r$ be the disk centered at the origin with radius $r$. Find $r$ such that $\mathbb{P}[(X,Y)  \in D_r ] = 0.3$.


My attempt: $(X, Y) \in D_r$ means $X^2 + Y^2 \le r^2$. What should I do next?
HINT

For any event $A$, you have
$$
\mathbb{P}[A] = \iint_{\mathbb{R}^2} \mathbb{I}_{A(x,y)}f(x,y)dxdy,
$$
and in your case, since $X$ and $Y$ are independent, $f(x,y) = \phi(x) \phi(y)$.

Convert the integration over the plane with an indicator function to integrating over a region, where the indicator is always true, and compute the probability...
The pdf of $X$ is $f_X(x)=\frac{e^{-x^2/2}}{\sqrt{2\pi}}$. Same for $Y$. The pdf for $(X,Y)$ is $f_{X,Y}(x,y)=f_X(x)f_Y(y) = \frac{e^{-(x^2+y^2)/2}}{2\pi}$. In polar coordinates,
$$
\begin{split}
\mathbb P(B_R) &amp;=
\int_{-\infty}^\infty\int_{-\infty}^\infty \mathbf 1_{x^2+y^2&lt;R}(x,y)f_{X,Y}(x,y)\,dx\,dy \\
&amp;= \int_{-\infty}^\infty\int_{-\infty}^\infty \mathbf 1_{x^2+y^2&lt;R}(x,y)\frac{e^{-(x^2+y^2)/2}}{2\pi}(x,y)\,dx\,dy \\
&amp;= \int_0^R 2\pi r \frac{e^{-r^2/2}}{2\pi}\,dr \\
&amp;= 1-e^{R^2/2}.
\end{split}
$$