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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?

Re: PS - Basketball team - Probability [#permalink]

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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

Re: PS - Basketball team - Probability [#permalink]

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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

Re: PS - Basketball team - Probability [#permalink]

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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

Re: PS - Basketball team - Probability [#permalink]

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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.

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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