It is C.
Now the possible committees that include M :
Totally we need 3 people out which M is already there. So the remaining two could be any 2 from the remaining 5 ie 5C2 =10
So totally there are 10 committees that include M.
Now coming to our favourable committees i.e. committees including M and A.
Since there are already two people out of the three needed the remaining person could be anyone from the remaining 4 people.
So there are 4 committees that include M and N.
So it is 4/10 = 40%
Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)