Five people A,B,C,D,E are in a group that regularly play squash.
a) In how many ways can 2 people be selected from 5
b) List the possible selections of 2 people from the 5 above.
c) How many times would you expect A to have played in 35 games, if the 2 players are chosen at random each time?
+118609 Five people A,B,C,D,E are in a group that regularly play squash.
a) In how many ways can 2 people be selected from 5?
This is an unordered selection. So the number of ways will be $$\frac{5*4}{2}=\frac{20}{2}=10$$
This is because you can choose any one of the 5 first then any one of the remaining 4 second. Hence 5*4=20 BUT AB is the same as BA you cannot count it twice (this will happen for any pair you choose) so you have to divide by 2
[If you know about combinations it is 5C2]
b) List the possible selections of 2 people from the 5 above.
AB, AC, AD, AE
BC, BD, BE,
CD, CE,
DE
c) How many times would you expect A to have played in 35 games, if the 2 players are chosen at random each time?
Well of the 10 possibilities above, 4 of them include A So the probability that A is included in any individual pair is
$$\frac{4}{10} = \frac{2}{5}$$
So in 35 games A would be expected to play approx
$$\frac{2}{5}\times 35 = 14\;\;times$$.
+118609 Five people A,B,C,D,E are in a group that regularly play squash.
a) In how many ways can 2 people be selected from 5?
This is an unordered selection. So the number of ways will be $$\frac{5*4}{2}=\frac{20}{2}=10$$
This is because you can choose any one of the 5 first then any one of the remaining 4 second. Hence 5*4=20 BUT AB is the same as BA you cannot count it twice (this will happen for any pair you choose) so you have to divide by 2
[If you know about combinations it is 5C2]
b) List the possible selections of 2 people from the 5 above.
AB, AC, AD, AE
BC, BD, BE,
CD, CE,
DE
c) How many times would you expect A to have played in 35 games, if the 2 players are chosen at random each time?
Well of the 10 possibilities above, 4 of them include A So the probability that A is included in any individual pair is
$$\frac{4}{10} = \frac{2}{5}$$
So in 35 games A would be expected to play approx
$$\frac{2}{5}\times 35 = 14\;\;times$$.
