SOLUTION: Factories A, B and C produce computers. Factory A produces 3 times as many computers as factory C, and factory B produces 5 times as many computers as factory C. The probability th

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Question 1168486: Factories A, B and C produce computers. Factory A produces 3 times as many computers as factory C, and factory B produces 5 times as many computers as factory C. The probability that a computer produced by factory A is defective is 0.033, the probability that a computer produced by factory B is defective is 0.03, and the probability that a computer produced by factory C is defective is 0.035.
A computer is selected at random and it is found to be defective. What is the probability it came from factory A?

Answer by ikleyn(52781) About Me  (Show Source):
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Factories A, B and C produce computers. Factory A produces 3 times as many computers as factory C,
and factory B produces 5 times as many computers as factory C. The probability that a computer produced by factory A
is defective is 0.033, the probability that a computer produced by factory B is defective is 0.03,
and the probability that a computer produced by factory C is defective is 0.035.
A computer is selected at random and it is found to be defective. What is the probability it came from factory A?
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Let x = # of computers produced by factory C.

Then factory A produces 3x computers, while factory B produces 5x computers.


In total, the amount of all produced computers is  Total = x + 3x + 5x = 9x.


The amount of defective computers produced by A is 0.033(3x);  by factory B is 0.03*(5x), and by factory C is 0.035x.


The total amount of defective computers produced by A, B and C together is  Total_defective = 0.033*(3x) + 0.03(5x) + 0.035x.


The problem's question is about the CONDITIONAL probability  P(comp is from A | comp is defective).



By the definition of the conditional probability,


    P(comp is from A | comp is defective) = P%28comp_is_from_A_AND_is_defective%29%2FP%28comp_is_defective%29.     (!)



The probability that the computer if from A and is defective is  %280.033%2A3x%29%2FTotal = 0.099x%2F%289x%29 = 0.011.


The probability that the computer is defective is  

    Total_defective%2FTotal = %280.033%2A%283x%29+%2B+0.03%285x%29+%2B+0.035x%29%2F%289x%29 = %280.033%2A3+%2B+0.03%2A5+%2B+0.035%29%2F9 = 0.031556.



THEREFORE, the conditional probability (1) under the problem's question is

    P(comp is from A | comp is defective) = P%28comp_is_from_A_AND_is_defective%29%2FP%28comp_is_defective%29 = 0.011%2F0.031556 = 0.348587.    ANSWER

Solved.

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I presented the solution for you in the most detailed way.

When a person gains some experience, he (or she) can do such calculations in a couple of lines.