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,
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"
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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