I randomly pick an integer $p$ between $1$ and $10$ inclusive. What is the probability that I choose a $p$ such that there exists an integer $q$ so that $p$ and $q$ satisfy the equation $pq - 4p - 2q = 2$? Express your answer as a common fraction.
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 $\boxed{\frac{2}{5}}$.