Question

As items come to the end of a production line, an inspector chooses which items are to go through a complete inspection. Twenty percent of all items produced are defective. Seventy percent of all defective items go through a complete inspection, and 25% of all good items go through a complete inspection. Given that an item is completely inspected, what is the probability it is defective? (Round your answer to four decimal places.) 14. [-18 Points) DETAILS WACKERLYSTAT7 2.E.175. MY NOTES ASK YOUR TEACHER Three events, A, B, and C, are said to be mutually independent if P(ANB) = P(A) P(B), P(BNC) = P(B) x P(C), P(ANC) = P(A) X P(C), PAN BNC) = P(A) x P(B) x PC). Suppose that a balanced coin is independently tossed two times. Define the following events. A: Head appears on the first toss. B: Head appears on the second toss. C: Both tosses yield the same outcome. Are A, B, and C mutually independent? P(A) = P(B) = P(C) = , so P(A) x P(B) = P(B) x PCC) = P(A) P(C) = Computing the probabilities of the intersections we find P(ANB) = ,P(BNC) = PIAN BOC) = . Thus, A, B, and C --Select--- mutually independent. and the product of all three probabilities is P(A) x P(B) x PCC) = and P(ANC) = . Further,

Answer 1

We would be looking at the first question here as:

Q1) We are given here that:
P(defective) = 0.2, therefore P(not defective) = 1 - 0.2 = 0.8

Also, we are given here that:
P( complete inspection | defective ) = 0.7 and P( complete inspection | not defective) = 0.25

Using law of total probability, we have here:
P( complete inspection) = P( complete inspection | defective ) P(defective) + P( complete inspection | not defective) P(not defective) = 0.7*0.2 + 0.25*0.8 = 0.34

Given that the item is completely inspected, the probability that it is defective is computed using Bayes theorem here as:

P( defective | complete inspection) = P( complete inspection | defective ) P(defective) / P( complete inspection)

= 0.7*0.2 / 0.34

= 0.4118

Therefore 0.4118 is the required probability here.