Three machines E 1 , E 2 and E 3 in a certain factory produce 50 % , 25 % and 25 % , respectively of the total daily output of electric tubes. It is known that 4 % of the tubes produced one each of machines E 1 and E 2 are defective and that 5 % of those produced on E 3 are defective. If one tube is picked up at random from a day's production, the probability that it is defective, is

Step-by-step solution

Let E → Event that the tube is defective. B 1 → Event when E 1 produces the tube. B 2 → Event when E 2   produces the tube. B 3 → Event when E 3 produces the tube. P B 1 = 50 100 = 1 2 and P E B 1 = 4 100 = 1 25 . P B 2 = 25 100 = 1 4 and P E B 2 = 4 100 = 1 25 . P B 3 = 25 100 = 1 4 and P E B 3 = 5 100 = 1 20 . Now, P E = P E B 1 P B 1 + E B 3 P B + E B 3 P B 3 . ⇒   P E = 1 2 × 1 25 + 1 4 × 1 25 + 1 4 × 1 20 ⇒   P E = 0 . 0425