Question

a manufacturer has three machine operators a, b and c. the first operator a produces 1% defective items, where as the other two operators b and c produce 5% and 7% defective items respectively. a is on the job for 50% of the time, b is on the job for 30% of the time and c is on the job for 20% of the time. a defective item is produced, what is the probability that it was produced by a?4

Answer 1

To find the probability that the defective item was produced by operator A, we need to use Bayes' theorem. Therefore, the probability that the defective item was produced by operator A is 0.04, or 4%.

Step-by-step explanation:

To find the probability that the defective item was produced by operator A, we need to use Bayes' theorem. Let event A be the event that operator A produced the defective item. Let event B be the event that a defective item was produced. We want to find P(A|B), the probability that A occurred given that B occurred. Bayes' theorem states:

P(A|B) = (P(B|A) * P(A)) / P(B)

We know that P(A) = 0.01 (1% of the time), P(B|A) = 1 (100% of the time operator A produces a defective item), P(B) = (P(B|A) * P(A)) + (P(B|B) * P(B)) + (P(B|C) * P(C)) = (1 * 0.01 * 0.5) + (0.05 * 0.3) + (0.07 * 0.2) = 0.025.

Substituting these values into Bayes' theorem, we get:

P(A|B) = (1 * 0.01) / 0.025 = 0.04

Therefore, the probability that the defective item was produced by operator A is 0.04, or 4%.