How big is the chance that a arbitrary man is taller than a arbitrary woman? I'm a first year mathematics student, and I'm having trouble with computing the following:
Assume that in a country the height $X$ of men is normally distributed, with $\mu_X = 180$ (the expected value), $\sigma_X = 6$ (the standard deviation) and the same for the height of women, $Y$, with $\mu_Y = 173$ and $\sigma_Y = 5$.
How big is the chance that a arbitrary man is taller than a arbitrary woman? 
I don't know how to solve this question. I'm able to compute chances like $\mathbb{P}(X \geqslant k)$ or $\mathbb{P}(X \leqslant k)$ for some $k \in \mathbb{R}$ using the formula $\mathbb{P}(X \leqslant k) = \Phi(\frac{k + \frac{1}{2} - \mu}{\sigma})$. I think I have to compute in some way $\mathbb{P}(X > Y)$, but I don't see how to do that. Could you please help me with this?
Thanks in advance! 

The probability density function of a $N(\mu,\sigma)$ random variable $X$ is given by:
$$ f_X(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$
hence our probability is given by the integral:
$$ \mathbb{P}[X\geq Y]=\frac{1}{60\pi}\iint_{x>y}\exp\left(-\frac{(x-180)^2}{72}-\frac{(y-173)^2}{50}\right)\,dx\,dy\approx\color{red}{81.49\%}.$$
