Answer to Question #91985 in Statistics and Probability for maham

Question #91985
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
1
Expert's answer
2019-08-01T13:22:50-0400

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


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



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