Five people are in a team. Two of them are selected at random to attend a competition.

Question

i) How many different groups two can be selected?
ii) If Mary is one of the five people in the team, what is the probability that she is selected to attend the competition?

Answer 1

5C2 = 5*4 / 1*2 = ?

1/5 if chosen 1st
1/4 if chosen 2nd
1/5 + 1/4 = 9/20

Answer 2

I don't agree with the answer to the second part.

If Mary is to be in the pair to be selected, then you just need one more of the remaining 4, which is C(4,1) or 4
prob (Mary in the group of 2 selected) = 4/10 = 2/5

or
number of pairs without Mary = C(4,2) = 6
so number of pairs with Mary = C(5,2) - C(4,2) = 10-6 = 4

or , list them, suppose we have A,B,C,D,M
sets of 2 at a time:
AB,AC,AD,AM, BC, BD, BM, CD, CM, DM <---- 4 containing M

Answer 3

Thanks guys, 5C2 for i) was correct, and 2/5 for ii) was also correct :)