A quality-control program at a plastic bottle production line involve inspecting finished bottles for flaws such as microscopic holes. The proportion of bottles that actually have such a flaw is only 0.0002. If a bottle has a flaw, the probability is 0.995 that it will fail the inspection. If a bottle does not have a flaw, the probability is 0.99 that it will pass the inspection. A. If a bottle fails inspection, what is the probability that it has a flaw?
B. Which of the following is the more correct interpretation of the answer to part (a)?
i. Most bottles that fail inspection do not have a flaw.
ii. Most bottles that pass inspection do have a flaw.
C. If a bottle passes inspection, what is the probability that it does not have a flaw?
D. Which of the following is the more correct interpretation of the answer to part (c)?
i. Most bottles that fail inspection do have a flaw.
ii. Most bottles that pass inspection do not have a flaw.
E. Explain why a small probability in part (a) is not a problem, so long as the probability in part (c) is large.

A) P(F | Fail) = 0.0195

B) i. Most bottles that fail inspection do not have a flaw.

C) P(NF | Pass) = 0.999999

D) ii: Most bottles that pass the inspection don't have a flaw.

E) probability in part C is very large and almost 100% and as such even if we fail a bottle which has no flaw, it is not a problem because we are sure that the ones that pass have no flaw.

Step-by-step explanation:

Let us first denote the terms:

F = The bottles have a flaw

NF = The bottles have no flaw

Pass = The bottles successfully passed the inspection test

Fail = The bottle failed the inspection test.

We are given:

P(F) = 0.0002

P(Fail | F) = 0.995

P(Pass | NF) = 0.99

Thus, probability of no flaw is;

P(NF) = 1 - P(F)

P(NF) = 1 - 0.0002

P(NF) = 0.9998

Also, probability that it passes the inspection when it has a flaw is;

P(Pass | F) = 1 - P(Fail | F)

P(Pass | F) = 1 - 0.995

P(Pass | F) = 0.005

Probability that it failed the test when It has no flaw is;

P(Fail | NF) = 1 - P(Pass | NF)

Thus;

P(Fail | NF) = 1 - 0.99

P(Fail | NF) = 0.01

A. Using Baye's theorem, we can find the probability that if fails inspection when it has a flaw;

P(F | Fail) = [P(Fail | F) × P(F)] / [(P(Fail | NF) × P(NF)) + (P(Fail | F) × P(F))]

P(F | Fail) = ((0.995) × (0.0002)/((0.01 × 0.9998) + (0.995 × 0.0002))

P(F | Fail) = 0.0195

B) probability in A is very small, thus;

We can say that majority of the bottles that fail inspection do not have a flaw.

C) using baye's theorem again, the probability that the bottle does not have a flaw if it passes the inspection. iis given by;

P(NF | Pass) = (P(Pass | NF) × P(NF)) / [(P(Pass | F) × P(F)) + (P(Pass | NF) × P(NF)]

P(NF | Pass) = (0.99 × 0.9998)/((0.005 × 0.0002) + (0.99 × 0.9998))

P(NF | Pass) = 0.999999

D) Probability in C above is very high, thus;

Most bottles that pass the inspection don't have a flaw.

E) probability in part C is very large and almost 100% and as such even if we fail a bottle which has no flaw, it is not a problem because we are sure that the ones that pass have no flaw.

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