bay's theorem based quesion

Question

in a university office two clerks are working to checking application form.on an average the first clerk 60percent while clerk second check remaining .the error rates are by and first and second clerk respectively 0.4 and 0.5.an application form is selected at random from the total forms found to be having an error.find the probability form was checked.

1-by the first clerk
2-by the second clerk

Answer 1

Hi Kashish,

                 These types of inference problems are done using a tree unfortunately this site has no drawing tools. I will just talk you through the calculations.

 

If i understand the problem correctly student #1 processes 60% and student #2 does 40% of the applications. We are also told that #1's error rate is 40% meaning for every 100 applications he processes 40 have errors and 60 are correct.

Student #2's error rate is 50% meaning 50 wrong and 50 right. I don't know about you but I would fire both of these students :)

 

We now have to calculate the probabilities of the 4 possible outcomes the first is p(#1/E)*p(E)=.6*.4=.24

 

The second is p(#1/C)*P(C)=.6*.6=.36

The third is    p(#2/E)*p(E)=.4*.5=.20

The fourth is  p(#2/C)*p(C)=.4*.5=.20

The meaning of these results are; out of 100 applications 24 were processed in error by student #1, 36 were processed by student #1 correctly and likewise for #2 20 in error and 20 correct.

 

If we now randomly choose an application and it has an error the probability that it was processed by student #1is p(E/#1)=p(#1/E)/p(E) (Bayes theorem)=.24/(.24+.2)=.24/.44=.545. So the probability is slightly over 50%

and p(E/#2)=.2/.44=.455. 

So even though student #1 has a lower error rate #1 processes more applications and if you pick one at random and it has an error chances are it was processed by student #1, Cool eh?

 

Hope this helps

Jim