UPSC NDA Math Model Paper 3

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.