Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 14 Sep 2014, 18:21

### 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
TAGS:
Current Student
Joined: 31 Aug 2007
Posts: 370
Followers: 1

Kudos [?]: 39 [0], given: 1

John and Peter are among the nine players a basketball coach [#permalink]  21 Dec 2007, 07:26
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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?

1/9
1/6
2/9
5/18
1/3
CEO
Joined: 17 Nov 2007
Posts: 3571
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 360

Kudos [?]: 1783 [0], given: 358

Expert's post
D

P=(9-2)C(5-2)*2C2/9C5=7!*2!*5!*4!/(3!*4!*9!*2!)=5*4/(8*9)=5/18
Current Student
Joined: 31 Aug 2007
Posts: 370
Followers: 1

Kudos [?]: 39 [0], given: 1

walker wrote:
D

P=(9-2)C(5-2)*2C2/9C5=7!*2!*5!*4!/(3!*4!*9!*2!)=5*4/(8*9)=5/18

hmm walker why isn't this correct:

8*7*6*5*2 / 9*8*7*6*5 = 2/9

# of teams with the two required people divided by total number of teams
CEO
Joined: 17 Nov 2007
Posts: 3571
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 360

Kudos [?]: 1783 [0], given: 358

Expert's post
young_gun wrote:
hmm walker why isn't this correct:
8*7*6*5*2 / 9*8*7*6*5 = 2/9

Explain, please, your approach step by step. Your logic is not clear for me...How did you get 8/9, 7/8....? Maybe I missed something.
Current Student
Joined: 31 Aug 2007
Posts: 370
Followers: 1

Kudos [?]: 39 [0], given: 1

walker wrote:
young_gun wrote:
hmm walker why isn't this correct:
8*7*6*5*2 / 9*8*7*6*5 = 2/9

Explain, please, your approach step by step. Your logic is not clear for me...How did you get 8/9, 7/8....? Maybe I missed something.

sure...

total number of 5 person teams, out of 9 players, possible: 9*8*7*6*5
number of teams that include John and Peter on the same team:
J P _ _ _ J and P counted as 1 person, 8*7*6*5 * 2 (multiply by two because JP and PJ placements)...

so:
number of Peter-John teams divided by total number of teams:

8*7*6*5 * 2 / 9*8*7*6*5 = 2/9
Current Student
Joined: 31 Aug 2007
Posts: 370
Followers: 1

Kudos [?]: 39 [0], given: 1

young_gun wrote:
walker wrote:
young_gun wrote:
hmm walker why isn't this correct:
8*7*6*5*2 / 9*8*7*6*5 = 2/9

Explain, please, your approach step by step. Your logic is not clear for me...How did you get 8/9, 7/8....? Maybe I missed something.

sure...

total number of 5 person teams, out of 9 players, possible: 9*8*7*6*5
number of teams that include John and Peter on the same team:
J P _ _ _ J and P counted as 1 person, 8*7*6*5 * 2 (multiply by two because JP and PJ placements)...

so:
number of Peter-John teams divided by total number of teams:

8*7*6*5 * 2 / 9*8*7*6*5 = 2/9

hmm you know what--i just realized this should be a combination problem and not perm....
CEO
Joined: 17 Nov 2007
Posts: 3571
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 360

Kudos [?]: 1783 [0], given: 358

Expert's post
young_gun wrote:
hmm you know what--i just realized this should be a combination problem and not perm....

Exactly! I also went by wrong way using permutation at the begin, but we really choose persons, don't positions.
CEO
Joined: 29 Mar 2007
Posts: 2593
Followers: 16

Kudos [?]: 189 [0], given: 0

Re: PS probability [#permalink]  21 Dec 2007, 14:36
young_gun 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?

1/9
1/6
2/9
5/18
1/3

I set it up like this 1/9*1/8 --> 1/72 then had 1/8*1/9 --> 1/72 --> 1/36... from here I just guessed D.

Now retrying this I think I can come up with a proper explanation.

how many different scenarios do we have John and Peter in a team of 5?

Should be 5!/(5!-2!) --> 5!/3! --> 20 ways in which we could arrange john and peter on a team of 5.

20/72--> 10/36--> 5/18.

Walker can u plz confirm whether approach is correct?
Director
Joined: 03 Sep 2006
Posts: 893
Followers: 6

Kudos [?]: 124 [0], given: 33

Re: PS probability [#permalink]  21 Dec 2007, 19:27
young_gun 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?

1/9
1/6
2/9
5/18
1/3

Total number of ways of choosing 5 people out of 9 = how many combinations are possible = 9C5.

Total number of favorite combinations, any of these combinations will always include John and Peter. thus we need to know how many combinations are feasible, when we have to choose 3 people out of 7 = 7C3.

Thus probability

P= 7C3/ 9C5 = 35/126 = 5/18.

Possible Wrong approach is as follows***********************

But initially, I had taken the wrong approach, and used the permutation fot the favorable events, I did something like this:

2 ( John and Peter) are already chosen, and now we need to choose the remaining 3 out of 7. Thus first can be chosen in 7 ways, second in 6 ways and third in 5 ways.

P= 7*6*5/ 9C5.

But 7*6*5 is permutation, just the number of ways the three people can be arranged and not the total number of combinations.

Just think/remember: ORDER does not matter in selecting the 3people out of 7, therefore it is a combination problem and not permutation ( in which order matters.)
SVP
Joined: 28 Dec 2005
Posts: 1593
Followers: 2

Kudos [?]: 72 [0], given: 2

what about approach of 1-prob that both are not on the team ?

prob that they are both not on the team is what im having trouble figuring out though ...
Director
Joined: 03 Sep 2006
Posts: 893
Followers: 6

Kudos [?]: 124 [0], given: 33

pmenon wrote:
what about approach of 1-prob that both are not on the team ?

prob that they are both not on the team is what im having trouble figuring out though ...

When both are not in the team answer would be

P = 1-(7C3/9C5) = 1-(5/18)
CEO
Joined: 17 Nov 2007
Posts: 3571
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 360

Kudos [?]: 1783 [0], given: 358

Re: PS probability [#permalink]  22 Dec 2007, 00:47
Expert's post
GMATBLACKBELT wrote:
I set it up like this 1/9*1/8 --> 1/72 then had 1/8*1/9 --> 1/72 --> 1/36... from here I just guessed D.

Looks like permutation...

GMATBLACKBELT wrote:
how many different scenarios do we have John and Peter in a team of 5?

Should be 5!/(5!-2!) --> 5!/3! --> 20 ways in which we could arrange john and peter on a team of 5.

20/72--> 10/36--> 5/18.

I guess your direction is not correct but you got right answer !!!
when you use 5!/3! it seems you should use the fact that there is the nine players but you did not use that....
CEO
Joined: 21 Jan 2007
Posts: 2770
Location: New York City
Followers: 7

Kudos [?]: 230 [0], given: 4

Re: PS probability [#permalink]  14 Feb 2008, 08:13
total = 9c5
teams with peter and tom = 7c3

desired/total = 7c3/9c5
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Senior Manager
Joined: 18 Jun 2007
Posts: 297
Followers: 2

Kudos [?]: 19 [0], given: 0

Re: PS probability [#permalink]  17 Feb 2008, 12:00
I got this correct in 1 min.
but interestingly when i tried different approach ans. was different..please tell me what I am doing wrong...
total combinations 9c5 = 126

total combinations w/o those 2 players in the team (9 - 2)C5 => 7C5 = 21
probability that those 2 would not be selected 21/126 = 1/6
probability that those 2 would be selected 1-1/6 = 5/6
and guess what, it is not even an option <blushes>

what am i doing wrong?
Re: PS probability   [#permalink] 17 Feb 2008, 12:00
Similar topics Replies Last post
Similar
Topics:
John and Peter are among the nine players a basketball coach 2 30 Aug 2008, 15:11
11 John and Peter are among the nine players a basketball coach 25 31 Jul 2008, 15:56
John and Peter are among the nine players a basketball coach 9 06 Jul 2008, 17:23
Mike and Ted are among the nine players a basketball coach 7 08 May 2007, 01:18
John and Peter are among the nine players a basketball coach 5 17 Aug 2006, 19:41
Display posts from previous: Sort by