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John and Peter are among the nine players a basketball coach

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Intern
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John and Peter are among the nine players a basketball coach [#permalink] New post 17 Aug 2006, 19:41
John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?


a) 1/9
b) 1/6
c) 2/9
d) 5/18
e) 1/3
Intern
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first try [#permalink] New post 17 Aug 2006, 19:48
this is what i tried:

find out the probability of not choosing John and Peter

7/9 * 6/8 * 5/7 * 4/6 * 3/5 = 1/6

1 - 1/6 = 5/6

where did i go wrong?
VP
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 [#permalink] New post 17 Aug 2006, 19:57
Answer is D: 5/18.

Total possible ways of selecting a 5-member team:
a combination problem, not permutation since order does not matter:

9x8x7x6x5/(5x4x3x2x1) = 126.

The ways of selecting a 5-member team including John and Peter:
It is again a combination, not permutation. Since John and Peter are already included, you don't bother about them any more. The possible ways will be the combination of the rest of the 3 players from the rest of 7 player:
7x6x5/(3x2x1) = 35.


so the probability = 35/126 = 5/18.
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 [#permalink] New post 17 Aug 2006, 20:04
Dear minhthel,
What you have done is that you have calculated the probability choosing Jon OR Peter OR Jon + Peter (5/6). We need to find the probability only of Jon + Peter.
The way I have done it is:
Total number of combinations possible for a 5 member team from a group of 9 players = 9C5
Total number of combinations possible for a 5 member team consisting of Jon + Peter out of the group of 9 players = 7C3
I got this because out of the 5 slots in the team, 2 have now been fixed. So there are only 3 slots available for the different combinations. Also now the choice needs to be made only amongst 7 players (Jon and Peter already decided).
Hence probability of selecting Jon + Peter in the team = 7C3/9C5 = 5/18.
What is the OA please ?
Any other approaches to solving this kind of problem ?
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Prashrash.

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 [#permalink] New post 17 Aug 2006, 20:21
d) 5/18

Picking 5 of 9 = 9x8x7x6x5/5x4x3x2= 126

# of ways to pick John and Peter and 3 others from 9-2 = 7 is :
7x6x5/3x2 = 35

Prob = 35/126 = 5/18
CEO
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 [#permalink] New post 17 Aug 2006, 21:08
Total cases = 9C5
cases when these two are selected = 7C3
Prob = 7C3/9C5 = 5/18

Minthel,

If we take your method then it will take hell lot of time.
You have considered only this order:
1st selection NOT one of these two
2nd selection NOT one of these two
3rd selection NOT one of these two
4th selection NOT one of these two
5th selection NOT one of these two

We have to consider two many more cases. X is any other than J and P.
JXXXX
XJXXX
XXJXX
XXXJX
XXXXJ
then 5 more for XXXXP.
It is becoming more complex.

If you still want to solve that way then here I go:


JPXXX how may ways this can be arranged
5!/3! = 20
Take one case JPXXX - 1/9 * 1/8 * 7/7 * 6/6 * 5/5
JXPXX - 1/9 * 7/8 * 1/7 * 6/6 * 5/5

So denominators of all these cases will be 9*8*7*6*5 and numerator will be 1*1*7*6*5

So prob = 20 * 7*6*5/(9*8*7*6*5) = 20/72 = 5/18
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

  [#permalink] 17 Aug 2006, 21:08
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