In a bolt factory, machines A, B and C manufacture respectively 25%, 35% and 40% of the total bolts. Of their output 5, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product. Then the probability that the bolt drawn is defective is

0.0345

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

$E_1 = $ the bolts is manufactured by machine A;

$E_2 = $ the bolts is manufactured by machine B;

$E_3 = $the bolts is manufactured by machine C, and A = the bolt is defective. 

Then $ P(E_1)=\frac{25}{100}=\frac{1}{4}, P(E_2)=\frac{35}{100}, P(E_3)=\frac{40}{100}.$

$P(\frac{A}{E_1})+ $ Probability that the bolt drawn is defective given the condition that it is manufactured by machine A $= \frac{5}{100}$.

Similarly $P(\frac{A}{E_2})=\frac{4}{100}$ and $P(\frac{A}{E_3})=\frac{2}{100}.$

Using the law of total probability, we have $P(A) = P(E_1)P(\frac{A}{E_1})+P(E_2)P(\frac{A}{E_2})+P(E_3)P(\frac{A}{E_3})$

$=\frac{25}{100}×\frac{5}{100}+\frac{35}{100}×\frac{4}{100}+\frac{4}{100}×\frac{2}{100}=0.0345.$