A quality-control program at a plastic bottle production line involves 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 is 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. If a bottle passes inspection, what is the probability that it does not have a flaw?

(a) If a bottle fails inspection, the probability that it has a flaw is P(F | Fail) = 0.0195

(b) If a bottle passes inspection, the probability that it does not have a flaw is P(NF | Pass) = 0.99

Step-by-step explanation:

Let F denote that the bottle has a flaw, NF denote that the bottle has no flaw, Pass denote that the bottle passes the inspection test and Fail denote that the bottle fails the inspection test.

We are given with the probabilities:

P(F) = 0.0002

P(Fail | F) = 0.995

P(Pass | NF) = 0.99

We can obtain the probabilities:

P(NF) = 1 - 0.0002 = 0.9998

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

                  = 1 - 0.995

P(Pass | F) = 0.005

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

                  = 1 - 0.99

P(Fail | NF) = 0.01

a. We need to find the probability that the bottle has a flaw given that it fails the inspection i.e. P(F | Fail). We can use the Baye's theorem to find this probability.

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

               =  (0.995)*(0.0002)/[(0.01)*(0.9998) + (0.995)*(0.0002)]

               = 0.000199/(0.009998+0.000199)

               = 0.000199/0.010197

P(F | Fail) = 0.0195

b. We need to compute the probability that the bottle does not have a flaw given that it passes the inspection. i.e. P(NF | Pass)

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

                    = (0.99)*(0.9998)/[(0.005)*(0.0002) + (0.99)*(0.9998)]

                    = 0.989802/(0.000001 + 0.989802)

                    = 0.989802/0.989803

P(NF | Pass) = 0.99

the answer is y < -3 or   y > 3

-3y > 7 + 2

-3 y > 9

y < -3

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step-by-step explanation: