Prime factorization of $10 = 2 \times 5$.

So, $10^{99} = 2^{99} \times 5^{99}$ and

No. of possible factors for $10^{99} = $ No. of ways in which prime factors can be combined

$= 100 \times 100$ (1 extra possibility for each prime number as prime factor raised to 0 is also possible for a factor)

$10^{99} = 10^{96} \times 1000$

So, no. of multiples of $10^{96}$ which divides $10^{99} =$ No. of possible factors of 1000

$= 4\times4 \left( \because 1000 = 2^3 \times 5^3\right)$ (See below)

$=16$

So, required probability $= \frac{16}{10000}$

$= \frac{1}{625}$

Correct Answer: $A$

How is number of possible factors of $1000 = 16?$

Here, we can prime factorize $1000$ as $2^3 \times 5^3$. Now, any factor of $1000$ will be some combination of these prime factors. For $2$, a factor has $4$ options - $2^0, 2^1, 2^2$ or $2^3$. Similarly $4$ options for $5$ also. This is true for any number $n$, if $n$ can be prime factorized as $a_1^{m_1} . a_2^{m_2} \dots a_n^{m_n}$, number of factors of $n$ $\\ = \left(m_1 +1 \right) \times \left(m_2 + 1\right) \times \dots \times \left(m_n +1 \right) $, the extra one in each factor term coming for power being $0.$