Let $X_1, X_2 \sim \mathcal{N}(0, 1)$ and independent, then $\mathbb{P}(X_1>a_1, X_2 + bX_1>a_2) = ?$ Can it be expressed as CDF of $X_1$ and $X_2$?
My attempt: $Z_1 = X_1$, $Z_2 = X_2 + bX_1$, then $(Z1,Z_2)\sim\mathbb{P}(0,\Sigma)$, where the diagonal elements of $\Sigma$ are 1 and off diagonals are $b$. But I want $\mathbb{P}(X_1>a_1, X_2 + bX_1>a_2)$ in terms of cdf of $X_1,X_2$. Ofcource it will have some conditional probability terms.
We have $$ X_1>a_1\\X_2>a_2-bX_1$$ so we can write your probability as $$ \int_{a_1}^\infty\int_{a_2-bx_1}^\infty f_{X_{1},X_2}(x_1,x_2)dx_2 \; dx_1$$ where $f_{X_1,X_2}$ is the joint PDF.
Since $X_1$ and $X_2$ are independent, this factors into $f_{X_1}(x_1)f_{X_2} (x_2)$ and we can write it as $$ \int_{a_1}^\infty f_{X_1}(x_1)\int_{a_2-bx_1}^\infty f_{X_2}(x_2)dx_2\;dx_1 = \int_{a_1}^\infty f_{X_1}(x_1)(1-F_{X_2}(a_2-bx_1))dx_1$$ where $F_{X_2}$ is the CDF of $X_2.$ That's the closest I think I can get to having something in terms of the CDFs. I think you'll always have a PDF in there somewhere.
Note that also since the support of the integral is some sector of space (i.e. all points with polar angle $\theta$ in some range) and the joint density is radially symmetric, this probability will just be given by the fraction of space the support takes up, which is just a matter of drawing a picture and doing some trigonometry in terms of $a_1$ and $a_2.$
