GATE CS 2014 Set 2 Q1 — Probability: Random Variables and Distributions

We are given:

• Total number of computers, $N = 10$
• Number of working computers, $W = 4$
• Number of non-working computers, $NW = 10 - 4 = 6$
• Number of computers inspected at random (without replacement), $n = 4$

The system is deemed functional if at least $3$ of the $4$ inspected computers are working. Let $X$ be the number of working computers inspected. We need to find the probability $p = P(X \ge 3)$.

Since the selection is without replacement, $X$ follows a hypergeometric distribution:

First, calculate the total number of ways to select $4$ computers out of $10$:

Now, calculate the individual probabilities:

1. Probability of inspecting exactly 3 working computers ($X = 3$): $$P(X = 3) = \frac{\binom{4}{3} \times \binom{6}{1}}{\binom{10}{4}} = \frac{4 \times 6}{210} = \frac{24}{210}$$
2. Probability of inspecting exactly 4 working computers ($X = 4$): $$P(X = 4) = \frac{\binom{4}{4} \times \binom{6}{0}}{\binom{10}{4}} = \frac{1 \times 1}{210} = \frac{1}{210}$$

Adding these probabilities gives:

We need to find the value of $100p$: