A factory has 3 machines A, B, and C. Machine A

Solution

We are given:

• Machine A produces 40% of the total items, with 5% defective
• Machine B produces 35% of the total items, with 4% defective
• Machine C produces the remaining 25% , with 2% defective

Define events Let:

• P(A) = 0.40, P(B) = 0.35, P(C) = 0.25
• P(D|A) = 0.05 (probability defective given produced by A)
• P(D|B) = 0.04
• P(D|C) = 0.02

We want to find P(B|D) : probability the item came from B , given that it is defective This is a Bayes’ Theorem problem: Apply Bayes' Theorem P(B|D) = [P(B) × P(D|B)] / [P(A)×P(D|A) + P(B)×P(D|B) + P(C)×P(D|C)] Substitute values: P(B|D) = [0.35 × 0.04] / [0.40×0.05 + 0.35×0.04 + 0.25×0.02] Compute numerator:
= 0.35 × 0.04 = 0.014 Compute denominator: = (0.40 × 0.05) + (0.35 × 0.04) + (0.25 × 0.02) = 0.02 + 0.014 + 0.005 = 0.039 Now: P(B|D) = 0.014 / 0.039 = 14 / 39