This scenario models a binomial distribution with parameters $n = 100$ and $p = 0.004$. We want the probability that fewer than 3 machines break down, which equates to $P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$.

Using the binomial probability formula $P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}$: $P(X = 0) = \binom{100}{0} (0.004)^0 (0.996)^{100} \approx 0.6698$ $P(X = 1) = \binom{100}{1} (0.004)^1 (0.996)^{99} \approx 100 \times 0.004 \times 0.6725 \approx 0.2690$ $P(X = 2) = \binom{100}{2} (0.004)^2 (0.996)^{98} \approx 4950 \times 0.000016 \times 0.6752 \approx 0.0535$

Adding these probabilities: $P(X < 3) = 0.6698 + 0.2690 + 0.0535 = 0.9923$