Question

In a bolt factory there are four machines A, B, C and D, manufacturing 20%, 15%, 25% and 40% of the total output respectively. Of there output 5%, 4%, 3% and 2% in the same order are defective balls. A ball is chosen at random from the factory production and it is found defective. What was the probability that the bolt was manufactured by machine A or D.​

Answer 1

Let A1, A2, A3, A4 be the events of selecting a bolt manufactured by machines A,B,C and D.

P(A1) = 0.2

P(A2) = 0.15

P(A3) = 0.25

P(A4) = 0.4

Let E be the event of selecting a defective bolt.

Given P(E/A1) = 0.05, P(E/A2) = 0.04, P(E/A3) = 0.03, P(E/A4) = 0.02

P(E) = 0.2(0.05) + 0.15(0.04) + 0.25(0.03) + 0.4(0.02) = 0.0315

We have to find P(A1 U A4/E)

P((A1∪A4)/E)=P(A1/E)+P(A4/E)

                     =0.2(0.05)0.0315+0.4(0.02)0.0315

                     =0.3175+0.254

                      =0.5715

Hope this will help you :)