On Pr(X>Y) when X and Y are independent normal Let X∼N(6,1) and Y∼N(7,1) be two independent normal variables. Find Pr(X>Y). the answer is 0.2389 but I do not know how to do it.I have tried adding them and subtracting but i am still clueless.

We have $Pr(X>Y)=Pr(X-Y>0)$.  Because $X$ and $-Y$ are independent Gaussians, their sum is normal (mean the sum of the means and variance the sum of the variances).  Therefore $X+(-Y)\sim N(6-7,1+1)=N(-1,2)$.
Depending on what calculational tools you have available to you, you may be able to evaluate $P(N(-1,2)>0)$ directly.  However, it is instructional to convert this quantity into something you can look up in a table for the standard normal.  Therefore, we can compute
$$P(N(-1,2)>0)=P(N(0,2)>1) = P\left(\frac{N(0,2)}{\sqrt{2}}>\frac{\sqrt{2}}{2}\right)=P\left(N(0,1)>\frac{\sqrt{2}}{2}\right). $$
