Question 1161236
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It is a conditional probability problem.



The probability that the flash drive is defective is 

    P(defective) = 0.60*0.2 + 0.25*0.15 + 0.15*0.05 = 0.165.



The probability that the randomly selected flash drive, made by machine A is defective, is

    P(A and defective) = 0.60*0.2 = 0.12.


The probability that the randomly selected defective flash drive was made by machine A is

    P(A | defective) = P(A and defective)/P(defective) = {{{(0.60*0.2)/(0.60*0.2 + 0.25*0.15 + 0.15*0.05)}}} = {{{0.12/0.165}}} = 0.727273.    <U>ANSWER to (a)</U>



Similarly, the probability that the randomly selected defective flash drive was made by machine B is

    P(B | defective) = P(B and defective)/P(defective) = {{{(0.25*0.15)/(0.60*0.2 + 0.25*0.15 + 0.15*0.05)}}} = 0.227273.            <U>ANSWER to (b)</U>



           the probability that the randomly selected defective flash drive was made by machine C is

    P(C | defective) = P(C and defective)/P(defective) = {{{(0.15*0.05)/(0.60*0.2 + 0.25*0.15 + 0.15*0.05)}}} = 0.045455.            <U>ANSWER to (c)</U>



As a check,  the sum of the three conditional probabilities is equal to 1 (one).

    0.727273 + 0.227273 + 0.045455 = 1.
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Solved.