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# John and Peter are among the nine players a basketball coach

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Manager
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John and Peter are among the nine players a basketball coach [#permalink]

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06 Jul 2008, 18:23
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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

Can someone solve using combinations?

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VP
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06 Jul 2008, 18:53
x-ALI-x wrote:
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

Can someone solve using combinations?

Number of ways the team of 5 can be formed from a group of 9 students = 9C5

Assume P and J are chosen. Nos of ways other 3 players are chosen from the remaining 7 = 7C3

Probability of P and J on a team = 7C3/9C5 = 5/18

D

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Senior Manager
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06 Jul 2008, 18:57
Quote:
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?

My approach with combinations:

Total number of outcomes = C(5, 9)

Favourable outcomes = C(2,2)*C(3,7)

C(2,2) – number of ways to select two desired players from two… Really doesn’t needed, since it’s 1; I just put it here for clarity sake.
C(3,7) – number of ways to select three other players (NOT John and Peter) from remaining 7

Overall, P= C(3,7)/ C(5, 9) = 5/18.

This is D.

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06 Jul 2008, 19:01
IMO D.

Jon and Peter are fixed so you are left with 7 players.

The total no. of ways to pick five players out of nine = 9C5 = 126
The total no. of ways to pick 3 players out of 7 = 7C3 = 35

Hence the proability = 35/126 = 5/18

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06 Jul 2008, 19:06
I arrived at 35/126. But it took almost a minute to figure out the damn common multiple 7.

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06 Jul 2008, 19:40
Something that helped me reduce things quickly is to break them down into their prime factors

35 = 5 * 7
126 = 2 * 63 = 2 * 7 * 9 = 2 * 7 * 3 * 3 = Right there you know the 7's cancel out and you have

$$\frac{35}{126} = \frac{5*7}{2*63}=\frac{5*7}{2*7*9}=\frac{5}{18}$$

icandy wrote:
I arrived at 35/126. But it took almost a minute to figure out the damn common multiple 7.

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Manager
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07 Jul 2008, 18:30
greenoak wrote:

My approach with combinations:

Total number of outcomes = C(5, 9)

Favourable outcomes = C(2,2)*C(3,7)

C(2,2) – number of ways to select two desired players from two… Really doesn’t needed, since it’s 1; I just put it here for clarity sake.
C(3,7) – number of ways to select three other players (NOT John and Peter) from remaining 7

Overall, P= C(3,7)/ C(5, 9) = 5/18.

This is D.

OA is D

Why can't we just account for Jan and Peter only? Why do we have to account for the other 3 players as well?
Ali

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07 Jul 2008, 18:36
icandy wrote:
I arrived at 35/126. But it took almost a minute to figure out the damn common multiple 7.

Me too, the first time I did this problem I ended up guessing because I *couldn't* get the right answer.
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Senior Manager
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07 Jul 2008, 18:56
Quote:
Why can't we just account for Jan and Peter only? Why do we have to account for the other 3 players as well?
Ali

Because all favourable teams hold John and Peter AND three other players. J and P are the same every time (obviously ), but the other three players change. That’s why we need C(3,7) – it gives us the number of combinations of those other players, and thus, the number of different teams with J and P.

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07 Jul 2008, 18:58
1
KUDOS
i took this route, although not using combinations...

probability Peter is chosen = 5/9
prob Jan is chosen, given Peter has already been chosen = 4/8

5/9 * 4/8 = 5/18

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Re: John and Peter are among the nine players a basketball coach [#permalink]

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24 Mar 2017, 13:55
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Re: John and Peter are among the nine players a basketball coach   [#permalink] 24 Mar 2017, 13:55
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