Comparing two normal distributions

Question

Given a normal distribution $X$~$N(60,9^2)$ with a random variable $A$ and a normal distribution $Y$~$N(50,7^2)$ with a random variable $B$, how do I go about finding the probability $P(B>A)$?

(Given that A and B are independent events).

Answer 1

If ther are dependent you cannot do this. If they are independent then $C=B-A$ has normal distribution with mean $50-60$ and variance $9^{2}+7^{2}$. You can compute $P(C>0)$ by integrating the density function from $0$ to $\infty$.

Answer 2

Hint: If $A$ and $B$ are independent, you have that $A-B \sim \mathcal{N}(10,7^2+9^2)$ and $P(B>A) = P(A-B<0)$.

If they are dependent and are jointly normally distributed, you need to use the joint distribution.