Question

An aircraft emergency locator transmitter (ELT) is a device designed to transmit a signal in the case of a crash. The ACME Manufacturing Company makes 70% of the ELTs, the B. BUNNY Company makes 25% of them, and the W. E. COYOTE Company makes the other 5%. The ELTs made by ACME have a 3.5% rate of defects, the B. BUNNY ELTs have a 5% rate of defects, and the W. E. COYOTE ELTs have a 8% rate of defects (which helps to explain why W. E. COYOTE has the lowest market share)
a) If a locator is randomly selected from the general population of all locators, find the probability that it was made by the ACME Manufacturing Company.b) If a randomly selected locator is then tested and is found to be defective, find the probability that it was made by the ACME Manufacturing Company.

Answer 1

Answer:

a) P(ACME) = 0.7

b) P(ACME/D) = 0.5976

Step-by-step explanation:

Taking into account that ACME manufacturing company makes 70% of the ELTs, if a locator is randomly selected from the general population, the probability that it was made by ACME manufacturing Company is 0.7. So:

P(ACME) = 0.7

Then, the probability P(ACME/D) that a randomly selected locator was made by ACME given that the locator is defective is calculated as:

P(ACME/D) = P(ACME∩D)/P(D)

Where the probability that a locator is defective is:

P(D) = P(ACME∩D) + P(B. BUNNY∩D) + P(W. E. COYOTE∩D)

So, the probability P(ACME∩D) that a locator was made by ACME and is defective is:

P(ACME∩D) = 0.7*0.035 = 0.0245

Because 0.035 is the rate of defects in ACME

At the same way, P(B. BUNNY∩D) and P(W. E. COYOTE∩D) are equal to:

P(B. BUNNY∩D) = 0.25*0.05 = 0.0125

P(W. E. COYOTE∩D) = 0.05*0.08 = 0.004

Finally, P(D) and P(ACME/D) is equal to:

P(D) = 0.0245 + 0.0125 + 0.004 = 0.041

P(ACME/D) = 0.0245/0.041 = 0.5976