Crossing Roads

We can think of the problem this way:

He walks 10 blocks without waiting at all for the traffic lights to turn (so he may not end at his house). This means that the student has a $50\%$ chance of moving North and a $50\%$ chance of moving East at any intersection. As his house is at $(5,5)$ , if he moves more than $5$ blocks in one of the directions, he is "penalized" one minute for every "wrong" block he walks on top of the $10$ minutes he needs to walk $10$ blocks (e.g. if he turns out to walk $6$ blocks North and $4$ blocks East, he is "penalized" for one minute, as in the actual situation, instead of walking the last block North, he will instead wait for the traffic light to turn so that he can walk a block East).

The probability of him walking $a$ blocks in one direction and $(10-a)$ blocks in the other direction is $\frac{10C_{a}}{2^{10}} \times 2=\frac{10C_{a}}{2^{9}}$ (except the case where he walks $5$ blocks in each direction, where the $\times 2$ is not needed), as the total number of ways he can walk $10$ blocks is $2^{10}$ . Therefore, the total time he is "penalized" for is

$5\frac{10C_{0}}{2^{9}}+4\frac{10C_{1}}{2^{9}}+3\frac{10C_{2}}{2^{9}}+2\frac{10C_{3}}{2^{9}}+1\frac{10C_{4}}{2^{9}}+0\frac{10C_{5}}{2^{10}}=\frac{315}{256}$

As $m=315$ and $n=256$ , the answer is $\boxed{571}$ .