Question

A computer manufacturer uses chips from three sources. Chips from sources A, B, and C are defective with probabilities 0.001, 0.005, and 0.01, respectively. If a randomly selected chip is found to be defective, what is the probability that the manufacturer was A; that the manufacturer was C. My clarification: Assume that the computer manufacturer uses equal numbers of chips from each source.

Answer 1

Answer:

a) 1/16

b) 5/8

Step-by-step explanation:

Let's abuse a little of the notation and also call the events A, B and C as

A= the chip comes from source A

B= the chip comes from source B

C= the chip comes from source C

Since the computer manufacturer uses equal numbers of chips from each source

P(A)=P(B)=P(C)=1/3

Let D be the event

D = the chip is defective

If a randomly selected chip is found to be defective, what is the probability that the manufacturer was A

Here we want to find P(A | D) the conditional probability that the chip comes from source A given that is defective.

By the Bayes' Theorem

$\large\bf P(A | D)=\frac{P(D|A)P(A)}{P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)}=\\\\=\frac{0.001*(1/3)}{0.001*(1/3)+0.005*(1/3)+0.01*(1/3)}=\\\\=\frac{0.001}{0.001+0.005+0.01}=\frac{0.001}{0.016}=\frac{1}{16}$

If a randomly selected chip is found to be defective, what is the probability that the manufacturer was C

Applying the same theorem

$\large\bf P(C| D)=\frac{P(D|C)P(C)}{P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)}=\\\\=\frac{0.01*(1/3)}{0.001*(1/3)+0.005*(1/3)+0.01*(1/3)}=\\\\=\frac{0.01}{0.001+0.005+0.01}=\frac{0.01}{0.016}=\frac{10}{16}=\frac{5}{8}$