MENU
Question -

A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?



Answer -

Let E1┬аbethe event of time consumed by machine A, E2┬аbe the event oftime consumed by machine B and E3┬аbe the event of time consumedby machine C. Let X be the event of producing defective items.

Then P (E1)= 50% = 50/100 = ┬╜

P (E2) =30% = 30/100 = 3/10

P (E3) =20% = 20/100 = 1/5

As we a headed coinhas head on both sides so it will shows head.

Also P (X|E1)= P (defective item produced by A) = 1% = 1/100

And P (X|E2)= P (defective item produced by B) = 5% = 5/100

And P (X|E3)= P (defective item produced by C) = 7% = 7/100

Now the probabilitythat item produced by machine A, being given that defective item is produced,is P (E1|A).

By using BayesтАЩtheorem, we have

P (E1|A) =5/34

Comment(S)

Show all Coment

Leave a Comment

Free - Previous Years Question Papers
Any questions? Ask us!
NodeJS Interview Questions Answers
ASP.Net Interview Questions Answers
Interview Questions Answers
×