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Probability - John and Peter

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by saisree » Sun Oct 23, 2011 9:29 am
Please help me with this problem. I always have problem with such questions.

A basketball coach will select the members of a five-player team from among 9 players, including John and Peter. If the five players are chosen at random, what is the probability that the coach chooses a team that includes both John and Peter?

a)1/9
b)1/6
c)2/9
d)5/18
e)1/3
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by knight247 » Sun Oct 23, 2011 11:03 am
Total number of outcomes=9C5 i.e selecting five persons from 9 guys
Number of favourable = 2C2*7C3 since John and Peter have to be included hence 2C2. And we need to select 3 guys out of remaining 7. i.e 7C3
So Probability= 7C3/9C5 =[spoiler]5/18[/spoiler] hence D
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by knight247 » Sun Oct 23, 2011 11:14 am
U can try the complementary method as well,

P(Both will be in the Team)=1-P(Neither will be in the team)-P(Either one will be in the Team)


Total number of possible outcomes is 9C5= 126

Neither will be in the team
We need to select 5 ppl from among the remaining 7 hence 7C5= 21

John will be in the Team but not Peter
John can be selected in 1C1 Ways
The other 4 can ONLY be chosen from the remaining 7 ppl in 7C4=35

Peter will be in the Team but not John
Peter can be selected in 1C1 Ways
The other 4 can ONLY be chosen from the remaining 7 ppl in 7C4=35

Total=21+35+35= 91
91/126=13/18

1-13/18=5/18 Hence D

Step 2 and 3 can also be condensed into one single step as follows
From among John or Peter Select anyone in 2C1=2 Ways
The other 4 have to be chosen from among the remaining 7 ppl in 7C4=35 Ways.....
2*35=70
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