Let the mutually independent random variables $X_1, X_2$, and $X_3$ be $N(0,1)$, $N(2,4)$, and $N(-1,1)$, respectively. Compute the probability that exactly two of these three variables are less than zero.
Hint: Let $p_1$ be the probability $X_1$ is $\le 0$ (or $\lt 0$, same probability). Define $p_2$, $p_3$ analogously. The $p_i$ can be computed using properties of the normal. Of course, $p_1=1/2$. The other two are a bit more work. Then our probability is $$(1-p_1)p_2p_3+(1-p_2)p_1p_3+(1-p_3)p_1p_2.$$ The first term in our sum is the probability that $X_2$ and $X_3$ are less than $0$, but $X_1$ is not. The other terms have similar interpretations.