Divisor of ten to the 99 that is a multiple of ten to the 80

Pick a positive divisor of $10^{99}$ uniformly at random. What is the probability that it is divisible by $10^{80}$?

$\tfrac1{25}$. A divisor is $2^i5^j$ with $i,j\in\{0,\dots,99\}$, giving $100^2$ choices. Divisibility by $10^{80}$ needs $i,j\ge80$, i.e. $20$ choices each, so the probability is $\tfrac{20^2}{100^2}=\tfrac1{25}$.