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%

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Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)