Number Theory and Probability Question Compute the probability that a randomly chosen positive divisor of $10^{99}$ is an integer multiple of $10^{88}$

Choosing a random positive divisor of $10^{99}$ is equivalent to choosing two random integers $0 \leq a, b \leq 99$ (the corresponding divisor is $2^a\cdot 5^b$). Multiples of $10^{88}$ are of the form $2^x 5^y$ where $x,y \geq 88$. So we need to find the probability that $a,b \geq 88$. There are $100$ possible values for each of $a$ and $b$, and $12$ of these integers are at least 88 and at most 99. Therefore the desired probability is $(12/100)^2 = 9/625$.
