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.