Question

A computer manufacturer uses chips from three sources. Chips from sources A, B, and C are defective with probabilities .005, .001, and .010, respectively. If a randomly selected chip is found to be defective, find the probability that the manufacturer was A; that the manufacturer was C. Assume that the proportions of chips from A, B, and C are 0.5, 0.1, and 0.4, respectively

Answer 1

Part 1:

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

Solution:

Let $X$ denote the event that the randomly selected chip is defective.

Also, define $A$ to be the event that the chip was manufactured by company A. Define the events $B , C$ similarly.

So, in this case, we want to find,

$$P(\text{ the chip is manufactured by A | the chip is defective}),\\$$

which is same as, $P(A|X)$ .

Now we apply Bayes' Theorem to compute this probability.

$\text{{ } { } }\ P(A|X)\\ =\dfrac{P(X|A)P(A)}{P(X|A)P(A)+P(X|B)P(B)+P(X|C)P(C)} \\\text{{ }{ }[as the events A, B, C are mutually exclusive and exhaustive]}\\$

Substituting the values given in the question, we obtain,

$\text{ { }{ }{ }{ }}P(A|X)\\ =\dfrac{0.005\times 0.5}{(0.005\times 0.5)+(0.001\times 0.1)+(0.010\times 0.4)}\\ =\dfrac{0.0025}{0.0025+0.0001+0.004}\\ =\dfrac{0.0025}{0.0066}\\ =\dfrac{25}{66}\approx 0.3788$

Hence the required probability is $\dfrac{25}{66}$ , or approximately $0.3788$

Part 2:

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

Solution:

For this part, we should compute $P(C|X)$ .

Similar to above, we apply Bayes' theorem and obtain,

$\text{{ } { } }\ P(C|X)\\ =\dfrac{P(X|C)P(C)}{P(X|A)P(A)+P(X|B)P(B)+P(X|C)P(C)}$

$=\dfrac{0.010\times 0.4}{(0.005\times 0.5)+(0.001\times 0.1)+(0.010\times 0.4)}\\ =\dfrac{0.004}{0.0025+0.0001+0.004}\\ =\dfrac{0.004}{0.0066}\\ =\dfrac{20}{33}\approx 0.6061$

Hence the required probability is $\dfrac{20}{33}$ , or approximately $0.6061$