The probability that a randomly chosen positive divisor of $$10^{2023}$$ is an integer multiple of $$10^{2001}$$ is

All the factors of  $$10^{2001}$$ are factors of  $$10^{2023}$$. So, the factor of  $$10^{2023}$$ that are multiples of $$10^{2001}$$ will be the factors that are factors of  $$10^{2023}$$ but not for $$10^{2001}$$.

Number of factors of $$10^{2001}$$ =  $$(2\cdot5)^{2001}$$ is $$(2002)^2$$

Number of factors of  $$10^{2023}$$ = $$(2024)^2$$

Probability = $$\frac{\left(2024^2-2002^2\right)}{2024^2}=\frac{\left(4026\cdot22\right)}{2024^2}$$