Question

In a bolt manufacturing factory, machines A,B, and C produce 25%, 30% and 45% of the total output, respectively. Of their outputs, 7%, 6% and 4% are defective bolts, respectively. If a bolt drawn at random from the production is found to be defective, what is the probability that it was manufactured by machine C?

Answer 1

Let, A₁ be the event that the drawn bolt is manufactured by machine A

A₂ be the event that the drawn bolt is manufactured by machine B

A₃ be the event that the drawn bolt is manufactured by machine C

B be the event that the drawn bolt is defective

Now, we are given,

P(A₁) = 25% = 0.25, P(A₂) = 30% = 0.3, P(A₃) = 45% = 0.45, P(B | A₁) = 7% = 0.07, P(B | A₂) = 6% = 0.06, P(B | A₃) = 4% = 0.04

The required probability

= The probability that the defective bolt was manufactured by machine C

= The probability that the bolt was manufactured by machine C given it was defective

= P(A₃ | B)

Here the events A₁, A₂ and A₃ are mutually exclusive and exhaustive.

By using Bayes' theorem,

P(A₃ | B) = $\frac{P(A_3).P(B |A_3)}{P(A_1).P(B |A_1)+P(A_2).P(B |A_2)+P(A_3).P(B |A_3)}$

= $\frac{0.04 \times 0.45}{0.07 \times 0.25+0.06 \times 0.3+0.04 \times 0.45}$

= 0.336 (rounded to 3 decimal places)

Answer: If a bolt drawn at random from the production is found to be defective, the probability that it was manufactured by machine C is 0.336.