Answer to Question #119378 in Statistics and Probability for Arafeen

Question

(14). 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, A = event that a bolt has been manufactured by machine A

B = event that a bolt has been manufactured by machine B

C = event that a bolt has been manufactured by machine C

D = event that the randomly drawn bolt is defective

Then by the problem we have,

P(A) = 25% = 0.25, P(B) = 30% = 0.3, P(C) = 45% = 0.45

and P(D | A) = 7% = 0.07, P(D | B) = 6% = 0.06, P(D | C) = 4% = 0.04

Now, the probability that the defective bolt was manufactured by machine C

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

= P(C | D)

Here the events A, B and C are mutually exclusive and exhaustive.

Then by using Bayes' theorem we have,

P(C | D) = $\frac{P(C).P(D |C)}{P(A).P(D |A)+P(B).P(D |B)+P(C).P(D |C)}$

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

= 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.