John and Peter are among the nine players a basketball coach : PS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 23 Jan 2017, 21:56

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

John and Peter are among the nine players a basketball coach

Author Message
Manager
Joined: 02 Aug 2007
Posts: 231
Schools: Life
Followers: 3

Kudos [?]: 56 [0], given: 0

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

Show Tags

06 Jul 2008, 17:23
00:00

Difficulty:

(N/A)

Question Stats:

100% (02:16) correct 0% (00:00) wrong based on 5 sessions

HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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?
VP
Joined: 03 Apr 2007
Posts: 1367
Followers: 4

Kudos [?]: 613 [0], given: 10

Show Tags

06 Jul 2008, 17:53
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
Director
Joined: 12 Apr 2008
Posts: 500
Location: Eastern Europe
Schools: Oxford
Followers: 14

Kudos [?]: 221 [0], given: 4

Show Tags

06 Jul 2008, 17:57
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.
Senior Manager
Joined: 23 May 2006
Posts: 327
Followers: 2

Kudos [?]: 294 [0], given: 0

Show Tags

06 Jul 2008, 18:01
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
VP
Joined: 05 Jul 2008
Posts: 1430
Followers: 39

Kudos [?]: 360 [0], given: 1

Show Tags

06 Jul 2008, 18:06
I arrived at 35/126. But it took almost a minute to figure out the damn common multiple 7.
SVP
Joined: 30 Apr 2008
Posts: 1887
Location: Oklahoma City
Schools: Hard Knocks
Followers: 40

Kudos [?]: 571 [0], given: 32

Show Tags

06 Jul 2008, 18:40
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.

_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

GMAT Club Premium Membership - big benefits and savings

Manager
Joined: 02 Aug 2007
Posts: 231
Schools: Life
Followers: 3

Kudos [?]: 56 [0], given: 0

Show Tags

07 Jul 2008, 17:30
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
Director
Joined: 26 Mar 2008
Posts: 652
Schools: Duke 2012
Followers: 15

Kudos [?]: 126 [0], given: 16

Show Tags

07 Jul 2008, 17:36
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.
_________________

"Egotism is the anesthetic that dulls the pain of stupidity." - Frank Leahy

GMAT Club Premium Membership - big benefits and savings

Director
Joined: 12 Apr 2008
Posts: 500
Location: Eastern Europe
Schools: Oxford
Followers: 14

Kudos [?]: 221 [0], given: 4

Show Tags

07 Jul 2008, 17:56
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.
Intern
Joined: 31 Mar 2008
Posts: 14
Followers: 0

Kudos [?]: 1 [0], given: 0

Show Tags

07 Jul 2008, 17:58
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
Re: PS - Basketball team - Probability   [#permalink] 07 Jul 2008, 17:58
Display posts from previous: Sort by