Probability of product of two numbers having a specific property Assume that $m$ and $n$ are integers picked completely at random, with $0 \leq m,n \leq 9$.
How do I calculate that
$$
\mathrm{Pr}(m \cdot n < 25) = \frac{67}{100}
$$
and not just add up all the possibilities? That is, how do I calculate that there are $67$ possibilities with $m \cdot n < 25$?
Update
By using the hint of Gautam Shenoy and the comment of HowDoIMath, I found a general solution: Assume that $m$ and $n$ are integers picked completely at random, with $0 \leq m,n \leq k$. Then
$$
\mathrm{Pr}(m \cdot n < l) = \frac{1}{(k + 1)^{2}} \cdot \sum_{n = 1}^{k}\left(\min(\lfloor l/n\rfloor,k)+2\right)
$$
 A: Hint: Condition on $n=0,1,\cdots 9$. For each value of $n \neq 0$, the valid values $m$ can take are from $0$ to $\min\{\lfloor\frac{25}{n}\rfloor,9\}$. For $n=0$, $m$ can vary of course over all $10$ values.
A: It depends on what you mean by "calculate". There will be a formula that you can plug into a computer to get the probability, but I suspect that it will be faster for the computer to cleverly check if $m\cdot n<25$ or not.
For $n=0$, check if $m=9$ works. It does, so $(0,m)$ is a solution for all $m$, since the product $m\cdot n$ is increasing in $m$.
Similarly $(1,m)$ and $(2,m)$ work for all $m$.
For $n=3$, you throw $m=9$ away, but $m=8$ works, so $(3,m)$ is a solution for all $m\leq 8$.
For $n=4$, we are down to $m=6$, so $(4,m)$ solves it for $m\leq 6$.
And finally $(5,m)$ solves it for $m\leq 4$.
Since the problem is symmetric in $m$ and $n$, there are no reason to check $n>5$.
If I count correctly, I did only $16$ multiplications, and any formula will probably involve division with remainder and more.
