GATE CS 2010 Q27 — Probability: Random Variables and Distributions

To find the probability that a randomly chosen divisor of $10^{99}$ is a multiple of $10^{96}$, we can analyze the prime factorization of these numbers.

Step 1: Find the total number of divisors of $10^{99}$

The prime factorization of $10^{99}$ is: $$10^{99} = 2^{99} \times 5^{99}$$

Any divisor $d$ of $10^{99}$ is of the form: $$d = 2^a \times 5^b$$ where $0 \le a \le 99$ and $0 \le b \le 99$.

The total number of choices for $a$ is $100$ (from $0$ to $99$), and for $b$ is $100$ (from $0$ to $99$). Thus, the total number of divisors is: $$T = 100 \times 100 = 10000$$

Step 2: Find the number of favorable divisors

For a divisor $d = 2^a \times 5^b$ to be a multiple of $10^{96} = 2^{96} \times 5^{96}$, the exponents $a$ and $b$ must satisfy: $$96 \le a \le 99 \quad \text{and} \quad 96 \le b \le 99$$

The possible values for $a$ are $\{96, 97, 98, 99\}$ (a total of $4$ choices).
The possible values for $b$ are $\{96, 97, 98, 99\}$ (a total of $4$ choices).

Thus, the number of favorable divisors is: $$F = 4 \times 4 = 16$$

Step 3: Calculate the probability

The probability $P$ is the ratio of favorable divisors to the total number of divisors: $$P = \frac{F}{T} = \frac{16}{10000} = \frac{1}{625}$$

Therefore, the correct option is A.