Problem 1 10 points A manufacturing process produces micropr

Problem 1 (10 points) A manufacturing process produces microprocessors using an entirely new technology. Historical results suggest that 25% of the final product is damaged and therefore not usable. In the past, the manufacturer has used an extremely expensive (but extremely reliable) diagnostic procedure to test if a microprocessor is damaged. In order to cut costs, management is considering a very inexpensive test that is not as reliable. Preliminary data suggest that if a microprocessor is damaged, then the probability that the new test would be negative (i.e., the test declares that the microprocessor is damaged), is 85%. If the microprocessor is working properly, the probability that the test is negative is 10%. Clearly there are two types of errors that can occur with the new test. The first type of error is that a damaged microprocessor might test positive, and so be shipped to a customer. The second type of error is that a perfectly good microprocessor might test negative and so be unnecessarily discarded. Please figure out the probabilities that each of these two errors might occur, that is, a. (5 points) The probability that a microprocessor is damaged given that it tests positive; b. (5 points) The probability that a microprocessor is working properly given that it tests negative.

Solution

First, let us define some sets or events in order to deal this mathematically.

G = The microprocessor works fine
B = The microprocessor is faulty
T = The test indicates that the microprocessor is fine
F = The test indicates that the microprocessor is faulty

Now note that what are all probabilities that are given.

P(B) =25%, So, P(G) = 75%

P(F|B) = 85%, So, P(T|B) = 15%

P(F|G) = 10%, So, P(T|G) = 90%

We need to find the following two probabilities.

P(B|T) and P(G|F)

We will apply Baye\'s theorem to find these probabilities. The formulations are -

P(B|T) = P(B) x P(T|B) / P(T) and P(G|F) = P(G) x P(F|G) / P(F)

The calculations are shown in the following tabular form.

Answer:

(a) P(B|T) = 0.0526

(b) P(G|F) = 0.260

T
Prior	Conditional	Joint	Posterior
P(G)	0.75	P(T|G)	0.90	P(G) x P(T|G)	0.675	P(G|T)	0.947368
P(B)	0.25	P(T|B)	0.15	P(B) x P(T|B)	0.038	P(B|T)	0.052632
Total	1			P(T) =	0.713		1.000
F
Prior	Conditional	Joint	Posterior
P(G)	0.75	P(F|G)	0.10	P(G) x P(F|G)	0.075	P(G|F)	0.26087
P(B)	0.25	P(F|B)	0.85	P(B) x P(F|B)	0.213	P(B|F)	0.73913
Total	1			P(F) =	0.287		1.000