Answer: 0.9923

## Explanation The problem involves calculating the probability that fewer than 3 machines break down during a particular day. This is a binomial probability problem with: - **n = 100** (total number of machines) - **θ = 0.004** (probability of a single machine breaking down) - **q = 1 - θ = 0.996** (probability of a machine not breaking down) "Fewer than 3" means 0, 1, or 2 machines break down. ### Binomial Probability Formula: \[ P(X = k) = \binom{n}{k} \theta^k (1-\theta)^{n-k} \] ### Calculations: 1. **P(0 breakdowns):** \[ P(X = 0) = \binom{100}{0} (0.004)^0 (0.996)^{100} = 1 \times 1 \times (0.996)^{100} \approx 0.6698 \] 2. **P(1 breakdown):** \[ P(X = 1) = \binom{100}{1} (0.004)^1 (0.996)^{99} = 100 \times 0.004 \times (0.996)^{99} \approx 0.2690 \] 3. **P(2 breakdowns):** \[ P(X = 2) = \binom{100}{2} (0.004)^2 (0.996)^{98} = 4950 \times 0.000016 \times (0.996)^{98} \approx 0.05347 \] ### Total Probability: \[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \] \[ P(X < 3) = 0.6698 + 0.2690 + 0.05347 = 0.9923 \] ### Why Other Options Are Incorrect: - **A (0.007726):** This is too small and might represent the probability of exactly 2 breakdowns or some other incorrect calculation. - **B (0.6698):** This is only the probability of 0 breakdowns, not "fewer than 3." - **D (0.269):** This is only the probability of exactly 1 breakdown. ### Key Concept: This problem demonstrates the **binomial distribution** application for rare events with a large number of trials. Since θ is small (0.004), the Poisson approximation could also be used, but the exact binomial calculation yields the correct result of **0.9923**.

Author: Nikitesh Somanthe