jmc

We approach this problem by trying to find solutions to the equation $pq-4p-2q=2$. To do this, we can use Simon's Favorite Factoring Trick and add $8$ to both sides to get $pq-4p-2q+8=10$. This can be factored into $$(p-2)(q-4)=10$$ We now can see that there are solutions only if $p-2$ divides $10$. Thus, there are $4$ possible values of $p$ between $1$ and $10$ inclusive $(1,3,4\text{ and }7)$. It follows that the probability of picking such a $p$ is $\boxed{\frac{2}{5}}$.