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
