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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

A basketball coach will select the members of a five-player [#permalink]

Show Tags

07 May 2012, 08:16

2

This post received KUDOS

8

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

77% (01:33) correct
23% (01:41) wrong based on 189 sessions

HideShow timer Statistics

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 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

Total # of five-player teams possible is \(C^5_9=126\); # of teams that include both John and Peter is \(C^2_2*C^3_7=35\);

Re: A basketball coach will select the members of a five-player [#permalink]

Show Tags

07 May 2012, 08:51

3

This post received KUDOS

1

This post was BOOKMARKED

There are 3 spots left after selecting John and Peter in the team. There are 7 players available to fill in the 3 positions. It can be done in 7C3 ways (favorable event) The total number of ways of selecting 5 out of 9 players is given by 9C5 ways.

Re: A basketball coach will select the members of a five-player [#permalink]

Show Tags

20 Aug 2012, 10:22

2

This post received KUDOS

1

This post was BOOKMARKED

Galiya wrote:

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

Probability of choosing John and Peter for the team plus three other players is \(\frac{2}{9}*\frac{1}{8}*\frac{7}{7}*\frac{6}{6}*\frac{5}{5}=\frac{1}{36}.\) Order choosing the two doesn't matter. There is a total number of \(5C2=\frac{5*4}{2}=10\) possibilities to choose the two (John and Peter) in the sequence of 5 players (like * * - - -, - * * - - ,... where * represents either John or Peter, no distinction between them). The requested probability is given by \(10*\frac{1}{36}=\frac{5}{18}.\)

Answer D
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: A basketball coach will select the members of a five-player [#permalink]

Show Tags

23 Aug 2012, 11:32

dyaffe55 wrote:

Would there be an issue doing it this way?

Probability of John getting chosen = 5/9, probability of Peter being subsequently chosen = 4/8. 5/9 * 4/8 = 5/18

Why is this not the recommended approach given that the use of combinatorics requires extensive multiplication?

How can you justify the probability of John getting chosen = 5/9? John is just one player of the total of 9, so what's the meaning of 5? Similar questions for Peter.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: A basketball coach will select the members of a five-player [#permalink]

Show Tags

23 Aug 2012, 13:08

My thinking was this... PA = (# of ways event can occur)/(total number of outcomes). John has 5 ways to be selected to the roster from a pool of 9 players = 5/9

Re: A basketball coach will select the members of a five-player [#permalink]

Show Tags

24 Aug 2012, 12:41

dyaffe55 wrote:

My thinking was this... PA = (# of ways event can occur)/(total number of outcomes). John has 5 ways to be selected to the roster from a pool of 9 players = 5/9

perhaps I am missing something very obvious

Mixing up things...

Number of outcomes is not 9, which is the total number of players to choose from. Total number of outcomes is the number of different teams you can form choosing 5 players out of 9. "John has 5 ways to be selected" has no meaning. You can count the number of possible teams of which John is also a member, but this is not 5.

Try to understand the posted solutions to this question.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: A basketball coach will select the members of a five-player [#permalink]

Show Tags

24 Aug 2012, 14:39

I'm not saying I don't understand the posted solutions... I do. However, I am attempting to find a valid alternative. All you keep saying is that "5 has no meaning" which really doesn't do much to refute my question.

"How can you justify the probability of John getting chosen = 5/9?"

If John is one of 9 and 5 random selections are made = 5/9.

However, I will agree that conceptually it isn't perfect after that. You could say the probability of peter getting selected next = 4/8. While I know this isn't exactly what the question asks, the result is the same. You could even add a third player and say "what is the probability that the coach chooses a team that includes John Peter and Paul." 5/9 x 4/8 x 3/7 = 6C2 / 9C5

Probability of picking random players plus picking John and Peter \(\frac{2}{9}*\frac{1}{8}*\frac{7}{7}*\frac{6}{6}*\frac{5}{5}=\frac{1}{36}.\)

Not sure where I went wrong?

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?

It should be P(JPAAA) = 1/9*1/8*7/7*6/6*5/5*5!/3! = 5/18. We multiply by 5!/3! because JPAAA can occur in several ways: JPAAA, JAPAAA, AJPAAA, ...
_________________

Re: A basketball coach will select the members of a five-player [#permalink]

Show Tags

20 Aug 2016, 04:49

well a much easier approach for me was for Peter to get selected in the team : there is 5 empty spots out of 9 contenders so his chance 5/9 for john 4 spots left out of 8 so his chance 4/8 both chance= 4/8*5/9=5/18 easier than 9 choose 5 calculation

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

Number of combinations with constraint i.e., both John and Peter included in the team =7C3=35 Total number of combinations = 9C5=126 Probability= 35/126=5/18
_________________