Question
A factory has 3 machines A, B, and C. Machine A
produces 40% of the total output, B produces 35%, and C the remaining. It is known that 5% of items from A, 4% from B, and 2% from C are defective. If an item is found defective, what is the probability it was produced by machine B?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
- 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
= 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
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