# Calculating P(A>B), where A and B are normal distribution

In the problem we have that A ~ N(7, 11/60) and B ~ N(7.3, 7/20) and the question is what is the probability that A gives a higher value that B. Since the textbook we have for the course doesn't include information about this type of question, I figured that it might just required to do manipulation with the normal law formula. So I came up with this :

$\int_{-\infty}^{\infty}(\frac{1}{\frac{11}{60}*\sqrt{2\pi}} * e^{-0.5* \frac{t-7}{\frac{11}{60}}^2 })* (\int_{-\infty}^{t}\frac{1}{\frac{7}{20}*\sqrt{2\pi}}*e^{ -0.5*\frac{x-7.3}{\frac{7}{20}}^2 }dx) dt$

What I tough could be a way to solve this was by multiplicating the probability to get a lower value of B (right part of the integral) by the probability of A (left part of the integral) for each value of A. In theory this might work, but I'm unable to actually calculate this with either my calculator or Wolfram Alpha. Is there something I'm overlooking in this problem ?

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An easy way of solving this problem for the case when $A$ and $B$ are jointly normal, is to use the fact that $A-B$ is also a normal random variable with mean equal to $E[A] -E[B]$ and variance equal to $\text{var}(A) + \text{var}(B) - 2\cdot\text{cov}(A,B)$. So then the question simplifies to
What is the probability than a normal random variable with given mean and variance is greater than $0$?
the answer to which can be obtained from tables of the standard normal distribution function. You don't say anything about whether $A$ and $B$ are jointly normal or not, though your approach seems to indicate that you are treating them as independent normal random variables (which happen to be jointly normal). Note that this approach cannot be used when $A$ and $B$ are normal but not jointly normal random variables because in this case $A-B$ is not a normal random variable.
@Alex Excerpt from an answer I wrote on stats.SE. Consider two standard normal random variables. If they are jointly normal, then $p(x,y)$ is the bivariate normal density. As a special case, if they are independent, then $p(x,y) = p(x)p(y)$. But they could be marginally normal random variables that are not jointly normal with joint density $$p(x,y)=\begin{cases}2p(x)p(y),&\text{if}~x\geq 0, y\geq 0,\text{or}~x<0,y<0,\\0,&\text{otherwise.}\end{cases}$$ For even more possibilities, see here. –  Dilip Sarwate Oct 23 '12 at 1:49