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 350,000 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]
17 Aug 2010, 10:27

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

81% (02:14) correct
19% (02:30) wrong based on 36 sessions

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?

Re: MGMAT Probability [#permalink]
17 Aug 2010, 13:11

you can but that will require more cases and will be time consuming.

straight forward way is to group J and P as always present on the team and as order does not matter so we just need to find the total number of ways to select rest three players = 7c3

total number of ways of selecting 5 players out of 9 = 9c5

Re: MGMAT Probability [#permalink]
17 Aug 2010, 22:22

ingoditrust wrote:

you can but that will require more cases and will be time consuming.

probability = 7c3/9c5 = 5/18

Look at my approach - it's easy, nice and not at all time consuming. But I'm somehow getting the wrong answer I decided to find the probability of not getting these two guys in the choosen group and then substruct it from 1. What's wrong with my approach?

Re: MGMAT Probability [#permalink]
11 May 2014, 16:30

I'm not sure if I'm misunderstanding the question but the way I read it, the question states that "what is the probability that jon and peter" will be selected. Correct?

If that's the case, why isn't this method working:

(5c2)/(5c9) - I need to pick J and P from a team of 5 and the number of options for the teams are 5 out of 9 players. Why is that wrong?

Additionally, I tried the direct probability approach and that seemed to work out:

(2/9)(1/8)(5!/2!3!) - This implies that the prob of picking J or P is 2 out of 9, then only J or P is left so 1 out of 9 and then I could have picked them the 1st time, 2nd, 3rd etc...

This method seems to work but I actually have a question regarding the permutation part. It obviously doesn't matter if I pick John 1st, 2nd, 3rd etc...and the same goes for peter. If order doesn't matter, then why do I need to apply the permutation part of this?

Re: MGMAT Probability [#permalink]
12 May 2014, 01:25

Expert's post

russ9 wrote:

I'm not sure if I'm misunderstanding the question but the way I read it, the question states that "what is the probability that jon and peter" will be selected. Correct?

If that's the case, why isn't this method working:

(5c2)/(5c9) - I need to pick J and P from a team of 5 and the number of options for the teams are 5 out of 9 players. Why is that wrong?

Additionally, I tried the direct probability approach and that seemed to work out:

(2/9)(1/8)(5!/2!3!) - This implies that the prob of picking J or P is 2 out of 9, then only J or P is left so 1 out of 9 and then I could have picked them the 1st time, 2nd, 3rd etc...

This method seems to work but I actually have a question regarding the permutation part. It obviously doesn't matter if I pick John 1st, 2nd, 3rd etc...and the same goes for peter. If order doesn't matter, then why do I need to apply the permutation part of this?

About 5C2 in the numerator: are we choosing 2 from 5? No, we want to choose 5 players, including John and Peter, from 9 players.

About the probability approach: though the order of the selection does not matter, John and Peter can be any from the 5 selected. We can have {John, Peter, any, any, any}, {John, any, Peter, any, any}, {John, any, any, Peter, any}, ... Now, each of these cases have equal probabilities to occur: 1/9*1/8*1*1*1 and there can be 5!/3! cases, so the overall probability is 1/9*1/8*1*1*1*5!/3! = 5/18.

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 \(C^2_2*C^3_7=35\);

This week went in reviewing all the topics that I have covered in my previous study session. I reviewed all the notes that I have made and started reviewing the Quant...

I was checking my phone all day. I wasn’t sure when I would receive the admission decision from Tepper. I received an acceptance from Goizueta in the early morning...

I started running as a cross country team member since highshcool and what’s really awesome about running is that...you never get bored of it! I participated in...