Probability of normal distribution calculation.

Question

I am studying normal distribution now and encounter one problem. We have two normal distribution $X_1$ and $X_2$, $X_1 \sim N(0,1)$ and $X_2 = 2 + 3x_1$ what is the $P(max(X_1,X_2)<0)$?

Thanks for your time.

Answer 1

$\max \{x,2+3x\} <0$ iff $x <-2/3$. Hence the anwser is $P\{ X_1 <-2.3\} =\int_{-\infty}^ {2/3} \phi (x) \, dx =\Phi(-2/3)$.

Answer 2

Since $X_2=2+3X_1$, we have $P(\max(X_1,2+3X_1)<0)$. Now, you have that $$\max(X_1,2+3X_1)<0 \iff \begin{cases} X_1 < 0 \\ 2+3X_1 < 0 \end{cases} \iff X_1 < -2/3$$ The only thing you need to compute is $P(X_1 < -2/3)$. Because you know that $X_1$ is $N(0,1)$, this is easy to find.

Hope that help!