jmc

There are a total of $\left(\genfrac{}{}{0ex}{}{50}{2}\right)=1225$ ways to choose the two positive integers. Call these integers $a$ and $b$. The problem asks what the probability is that: $$ab+a+b=n-1$$where $n$ is a multiple of 5. We can add one to each side of this equation and factor: $$ab+a+b+1=(a+1)(b+1)=n$$Now, we need to count the number of values of $a$ and $b$ such that $(a+1)(b+1)$ is a multiple of 5. This will happen if at least one of the factors is a multiple of 5, which will mean $a$ or $b$ is one less than a multiple of 5.

There are 10 integers from 1 to 50 inclusive that are 1 less than a multiple of 5: $4,9,14,\dots ,49$. So, the number of ways to choose $a$ and $b$ so the product is $\text{not}$ a multiple of 5 is $\left(\genfrac{}{}{0ex}{}{40}{2}\right)=780$. Therefore, there are $1225-780=445$ ways to choose $a$ and $b$ that do satisfy the requirement, which gives a probability of: $$\frac{445}{1225}=\boxed{\frac{89}{245}}$$