 Hello, and welcome to the session. I am Deepika here. Let's discuss a question which says A factory has two machines A and B past record shows that machine A produced 60% of the items are for output and machine B produced 40% of the items further 2% of the items produced by machine A and 1% produced by machine B. What defective all the items are put into One stock by and then one item is chosen at random from this and is found to be defective What is the probability that it was produced by machine B? Now here in this question, we will use Bayes theorem Now according to Bayes theorem if even e2 so on ENR events Which constitute a partition of sample space S That is even e2 so on ENR pairwise disjoint and Even union e2 union so on union EN is equal to S and ABNE event with non-zero probability then probability of EI Upon a is equal to probability of EI into probability of a upon EI over Sigma probability of EJ into probability of a upon EJ J varying from 1 to N So this is a key idea behind that question We will take the help of this key idea to solve the above question So let's start the solution Now according to the given question a factory has two machines a and b past record shows that machine a Produced 60% of the items of output and machine B produced 40% of the items so let even and e2 be the events that The machine a and machine B produces the items respectively therefore probability of E1 is equal to 60% which is equal to 60 over 100 and this is again equal to 3 over 5 and probability of e2 is equal to 40 percent which is equal to 40 over 100 and this is again equal to 2 over 5 Again, we are given 2% of the items produced by machine a and 1% produced by machine B What defective all the items are put into one stop by and then one item is Choosing at random from this and is found to be defective We have to find the probability that it was produced by machine B so let a be the event of getting our defective item then probability of a upon e1 That is probability that the defective item was produced by machine a is equal to 2% which is equal to 2 over 100 and this is again equal to 0.02 again probability of a upon e2 that is the probability that defective item was produced by machine B and This is equal to 1% Which is again equal to 1 over 100 and this is equal to 0.01 now We have to find the probability that the machine B Produced that defective item. It is given by probability of e2 upon a So by using this theorem we have probability e2 upon a is equal to probability of e2 into probability of a upon e2 over probability of e1 into probability of a upon e1 plus probability of e2 into probability of a upon e2 now we have probability of even is equal to 3 over 5 probability of e2 is equal to 2 over 5 probability of a upon e1 is equal to zero point zero two and And probability of A upon E2 is equal to 0.01. So on substituting these values, we have probability of E2 upon A is equal to 2 over 5 into 0.01 over 3 over 5 into 0.02 plus 2 over 5 into 0.01 and this is again equal to 0.02 over 5 upon 0.06 over 5 plus 0.02 over 5 and this is equal to 0.02 over 0.08 and this is again equal to 1 over 4. Hence the probability that the machine B produced the effective item is 1 over 4. So this completes our session. I hope the solution is clear to you. Bye and have a nice day.