In a certain factory, machines A, Band C produce bolts of their production, machines A, Band C produce 2%, 1 % and 3% defective bolts, respectively. Machine A produces 35% of the total output of bolts, machine B produces 25% and machine C produces 40%. A bolt is chosen at random from the factory's production and is found to be defective. The probability it was produced on machine C, is

Let  $E_1,E_2,E_3$    and A be the events defined as follow

  $E_1:$   The bolt is manufactured by machine A.

   $E_2$ : The bolt is manufactured by machine B.

  $E_3$  :  : The bolt is manufactured by machine c

   A :  The bolt is defective.  

 Now   .

       $\begin{array}{l}P\left(E_1\right)=\frac{35}{100}=\frac{7}{20},P\left(E_2\right)=\frac{25}{100}=\frac{1}{4}\\ P\left(E_3\right)=\frac{40}{100}=\frac{4}{10}=\frac{2}{5}\end{array}$

      $P\left(\frac{A}{E_1}\right)$  :   Probability that the bolt drawn in defective given that it is 

  manufactured by machine     $P\left(\frac{A}{E_1}\right)=\frac{2}{100}=\frac{1}{50}$

Similarly,         $P\left(\frac{A}{E_2}\right)=\frac{1}{100}$

                $P\left(\frac{A}{E_3}\right)=\frac{3}{100}$

        By Baye's theorem, we have 

       $\begin{array}{l}P\left(\frac{E_3}{A}\right)=\frac{P\left(E_3\right)P\left(\frac{A}{E_3}\right)}{P\left(E_1\right)P\left(\frac{A}{E_1}\right)+P\left(E_2\right)P\left(\frac{A}{E_2}\right)+P\left(E_3\right)P\left(\frac{A}{E_3}\right)}\\ =\frac{\frac{2}{5}\times \frac{3}{100}}{\frac{7}{20}\times \frac{1}{50}+\frac{1}{4}\times \frac{1}{100}+\frac{2}{5}\times \frac{3}{100}}\end{array}$

     $\begin{array}{l}=\frac{\frac{6}{500}}{\frac{7}{1000}+\frac{1}{400}+\frac{6}{500}}=\frac{\frac{6}{500}}{\frac{14+5+24}{2000}}=\frac{\frac{6}{500}}{\frac{43}{2000}}\\ =\frac{6}{500}\times \frac{2000}{43}=\frac{24}{43}\end{array}$