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# Five people A,B,C,D,E are in a group that regularly play squash.

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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?

Feb 25, 2015

#1
+95356
+5

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.

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

$$\frac{2}{5}\times 35 = 14\;\;times$$.

Feb 25, 2015

#1
+95356
+5

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.

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

$$\frac{2}{5}\times 35 = 14\;\;times$$.

Melody Feb 25, 2015