Answer:
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.
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.