Answer :

Let E_{1} be the event that machine A is defective. P(E_{1}) = 1%

Let E_{2} be the event that machine B is defective. P(E_{2}) = 5%

Let E_{3} be the event that machine C is defective. P(E_{3}) = 7%

Let A be the event that the item chosen is defective.

P(the defective item is manufactured by A) = P(A|E_{1}) = 50%

P(the defective item is manufactured by B) = P(A|E_{2}) = 30%

P(the defective item is manufactured by C) = P(A|E_{3}) = 20%

Now, According to the Bayes’ theorem,

Given E_{1}, E_{2}, E_{3}…….E_{n} are mutually exclusive and exhaustive events, we can find the conditional probability P(E_{i}|A) for any event A associated with E_{i} is given by:

Therefore,

The probability that defective item is produced by A,

Hence, the probability that the defective item is produced by machine A is .

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