Question

An aircraft emergency locator transmitter (ELT) is a device designed to transmit a signal in the case of a crash. Company A makes 50% of the ELTs, Company B makes 45% of them, and the Company C makes the other 5% The ELTs made by Company A have a 4% rate of defects, the Company B's ELTs have a 6% rate of defects, and the Company C's ELTs have a 9% rate of defects. If a randomly selected ELT is then tested and is found to be defective, what is the probability that it was made by the Company B?

Answer 1

We are given that 50% of the ELTs are made by Company A, 45% are made by Company B, and 5% are made by Company C. We also know the rate of defects for each company's ELTs: 4% for Company A, 6% for Company B, and 9% for Company C.

Let D be the event that an ELT is defective, and let Bi be the event that an ELT was made by Company i, where i = A, B, or C. We want to find the probability that an ELT is made by Company B given that it is defective, i.e., we want to find P(B|D).

Using Bayes' theorem, we can write:

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

where P(D|B) is the probability that an ELT made by Company B is defective, P(B) is the prior probability that an ELT is made by Company B, and P(D) is the overall probability of an ELT being defective.

We can calculate each of these probabilities as follows:

P(D|B) = 0.06 (given)

P(B) = 0.45 (given)

P(D) = P(D|A) * P(A) + P(D|B) * P(B) + P(D|C) * P(C)

= 0.04 * 0.5 + 0.06 * 0.45 + 0.09 * 0.05

= 0.0355

Substituting these values into Bayes' theorem, we get:

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

= 0.06 * 0.45 / 0.0355

≈ 0.762

Therefore, the probability that an ELT made by Company B is defective given that it is found to be defective is approximately 0.762, or 76.2%.