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let A,B be the set of mistakes found by A & B respectively
n(A)=a
n(B)=b
n(A 
B)=c
n(A
B)=n(A) +n(B) - n(A
B)
=a+b-c
now total no. of mistakes is 
n(A
B)=a+b-c
and no. of mistakes found by A =a
therefore no. of mistakes A missed 
(a+b-c)-a=b-c
therefore no. of mistakes A missed 
b-c
and no. of mistakes found by B=b
therefore no. of mistakes B missed (a+b-c)-b=a-c
therefore no. of mistakes B missed a-c
Ok, I tried to avoid posting the solution as much as possible as the atmosphere when I posted the problem was a bit hostile towards me and I did not want to make it more unpleasant by posting a solution and then invite ridicule. But, now things are better so I will go ahead.
This was a problem posed by George Polya, a venerable mathematician.
What we need most here is the total number of mistakes made. How do we arrive at that with the given information?
If you look at A's performance, we notice that out of B's b mistakes, A has found c of them. Here we have to make an estimate that A is consistent at his job. So, if A has found a mistakes then the total number of mistakes is therefor ab/c.
Now, the number of mistakes found is a+b-c as many of you have already deduced.
So, the number of mistakes missed is just ab/c - a+b-c which equals (a-c)(b-c)/c.
Ok, I tried to avoid posting the solution as much as possible as the atmosphere when I posted the problem was a bit hostile towards me and I did not want to make it more unpleasant by posting a solution and then invite ridicule. But, now things are better so I will go ahead.
This was a problem posed by George Polya, a venerable mathematician.
What we need most here is the total number of mistakes made. How do we arrive at that with the given information?
If you look at A's performance, we notice that out of B's b mistakes, A has found c of them. Here we have to make an estimate that A is consistent at his job. So, if A has found a mistakes then the total number of mistakes is therefor ab/c.
Now, the number of mistakes found is a+b-c as many of you have already deduced.
So, the number of mistakes missed is just ab/c - a+b-c which equals (a-c)(b-c)/c.
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den mistakes counted by either of dem=a+b-c
mistakes missed by both=x-(a+b-c)