Let P(A), P(B), P(C) be the probability of machine parts A, B, and C defective respectively. Given: P(A) = 0.02, P(B) = 0.10, P(C) = 0.05 Let P(A’), P(B’), P(C’) be the probability of machine parts A, B, and C not defective respectively. P(A’) = 1 - P(A) = 1 – 0.02 = 0.98 P(B’) = 1 - P(B) = 1 – 0.1 = 0.9 P(C’) = 1 - P(C) = 1 – 0.05 = 0.95 Probability of machine not stops working = P (A’ ∩ B’ ∩ C’) = P(A’) × P(B’) × P(C’) ⇒ P (A’ ∩ B’ ∩ C’) = (0.98) × (0.9) × (0.95) = 0.837 = 0.84. Hence, option (3) is correct.