Question 1124485
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Let assume that factory B produces 100 computers.

Then factory A produces 400 computers, according to the condition.


Of 100 computers produced by B, 20 are defective.

Of 400 computers produced by A, 0.3*400 = 120 are defective.


The probability under the question is  {{{20/(120+20)}}} = {{{20/140}}} = {{{1/7}}}.
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Solved.