Five people are in a team. Two of them are selected at random to attend a competition.
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?
5C2 = 5*4 / 1*2 = ?
1/5 if chosen 1st
1/4 if chosen 2nd
1/5 + 1/4 = 9/20
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
Thanks guys, 5C2 for i) was correct, and 2/5 for ii) was also correct :)
i) To calculate the number of different groups of two that can be selected from a team of five people, we can use the combination formula.
The combination formula is given by:
C(n, r) = n! / (r!(n-r)!)
In this case, we want to select 2 people from a team of 5, so we can calculate it as follows:
C(5, 2) = 5! / (2!(5-2)!)
= 5! / (2!3!)
= (5 * 4 * 3!) / (2 * 1 * 3!)
= (5 * 4) / (2 * 1)
= 10
Therefore, there are 10 different groups of two that can be selected from the team of five.
ii) If Mary is one of the five people in the team, we need to calculate the probability that she is selected to attend the competition.
The probability of an event is defined as the number of favorable outcomes divided by the total number of possible outcomes.
In this case, if Mary is one of the five people in the team, the total number of possible outcomes is still the same as before, which is 10 (as calculated in part i).
The number of favorable outcomes is now 1, as Mary needs to be selected in all of these cases.
Therefore, the probability that Mary is selected to attend the competition is:
1 / 10 = 0.1 or 10%