Question

Factories A, B and C produce computers. Factory A produces 4 times as manycomputers as factory C, and factory B produces 7 times as many computers asfactory C. The probability that a computer produced by factory A is defective is0.04, the probability that a computer produced by factory B is defective is 0.02,and the probability that a computer produced by factory C is defective is 0.03. Acomputer is selected at random and found to be defective. What is the probabilityit came from factory A?

Answer 1

Answer:

The  probability is   $P(A') = 0.485$

Step-by-step explanation:

Let assume that the number of computer produced by factory C is  k = 1  

 So  From the  question we are told that

       The number produced by  factory A is  4k =  4

        The  number produced by factory B is  7k  = 7

        The  probability of defective computers from A is  $P(A) = 0.04$

        The  probability of defective computers from B is  $P(B) = 0.02$

        The  probability of defective computers from C is $P(C) = 0.03$

Now the probability of factory A producing a defective computer out of the 4 computers produced is  

       $P(a) = 4 * P(A)$

substituting values

        $P(a) = 4 * 0.04$

        $P(a) = 0.16$

The probability of factory B producing a defective computer out of the 7 computers produced is  

       $P(b) = 7 * P(B)$

substituting values

        $P(b) = 7 * 0.02$

        $P(b) = 0.14$

The probability of factory C producing a defective computer out of the 1 computer produced is  

       $P(c) = 1 * P(C)$

substituting values

        $P(c) = 1 * 0.03$

        $P(b) = 0.03$

So the probability that the a computer produced from the three factory will be defective is  

     $P(t) = P(a) + P(b) + P(c)$

substituting values

     $P(t) = 0.16 + 0.14 + 0.03$

     $P(t) = 0.33$

Now the probability that the defective computer is produced from factory A is

      $P(A') = \frac{P(a)}{P(t)}$

       $P(A') = \frac{ 0.16}{0.33}$

        $P(A') = 0.485$