Use a tree diagram

Question

Question 3 Two factories manufacture computer memory chips. Suppose that 20% of the chips from Factory I are defective and that 5% of the chips from Factory II are defective. Each week Factory I produces four times as many chips as Factory II. Use a tree diagram. (a) What is the probability that a randomly selected chip is satisfactory? (b) What is the probability that a randomly selected chip is defective? (c) Given that the chip is defective, what is the probability that it came from Factory I?

Answer 1

USING BAYESIAN STATISTICS INSTEAD OF A TREE DIAGRAM

JOINT PROBABILITY OF A AND B:

P(A^B) = P(A|B) P(B) = P(B|A) P(A)

P(A|B) = probability of A given that B occurs

Let

A = Factory I

A' = Factory II

B = - is defective

P(A|-) P(-) = P(-|A) P(A)

Either comes from Factory 1 or Factory 2

P(A) + P(A') = 1

P(A) = 4 P(A') four times as many factories

5 P(A') = 1

P(A') = 0.20 , Factory II

P(A) = 1 - P(A') = 0.80, Factory I

Defect probablity

P(-) = P(-|A) P(A) + P(-|B) P(B)

P(-|A) = 0.20

P(-|B) = 0.05

P(-) = (0.20)(0.80) + (0.05)(0.20) = 0.17 or 17% probability of getting a defective

P(+) = 1 - P(-) = 0.83 or 83% probability of getting an ok item

What is the probability that a defective part came from A = Factory 1 ?

P(A|-) = P(-|A) P(A)/P(-)

P(A|-) = P(-|A) P(A)/P(-)

P(A|-) = (0.20) (0.80)/(0.17) = (20/17)(.80) = 0.94 or 94%

So knowing that a part is defective (-) raises the probability it came from Factory 1 (A) produces 0% of the parts to a probability of 94% that it came from that factory.

A Bayesian would say that the information was used to transform the prior probability of having come from Factory 1 from an initial estimate of 80% to a final (a posteriori) probability of 94%.

Presumably, the decision tree approach would reach the same answer.