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

 It is currently 26 Oct 2016, 21:02

### 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

# PS-Probability

Author Message
Senior Manager
Joined: 05 Jun 2008
Posts: 307
Followers: 2

Kudos [?]: 120 [0], given: 0

### Show Tags

02 Dec 2008, 05:20
1
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

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

### HideShow timer Statistics

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

There are 8 members; among them are Kelly and Ben. A committee of 4 is to be chosen out of the 8. What is the probability that Ben is chosen to be in the committee and Kelly is not?

How many possible ways are there to solve this question ?
Manager
Joined: 02 Nov 2008
Posts: 60
Followers: 1

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

### Show Tags

02 Dec 2008, 09:51
1. 6C3

Ben is already chosen need 3 more committe members

Kelly not chosen . Assume she is not there. Hence total member left is 6
SVP
Joined: 17 Jun 2008
Posts: 1570
Followers: 11

Kudos [?]: 241 [0], given: 0

### Show Tags

02 Dec 2008, 12:45
walker wrote:
Why not (2C1 * 6C3) / (8C4) ?

walker, unless I am missing something, Ben is to be selected and Kelly is not to be selected. Hence, for Ben, 1C1.
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 520

Kudos [?]: 3328 [0], given: 360

### Show Tags

02 Dec 2008, 12:57
I think we should take into account two possible cases: (Ben,x,x,x)(Kelly,x,x,x) and (Kelly,x,x,x)(Ben,x,x,x)
Therefore, we need chose Ben or Kelly for first committee (2C1) and three others not Ben and not Kelly (3C6)
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Director
Joined: 01 Aug 2008
Posts: 769
Followers: 4

Kudos [?]: 582 [0], given: 99

### Show Tags

02 Dec 2008, 13:18
Walker,

Great explanation. I also got the same result as scthakur got. But I see now why I should use 2c1 instead of 1c1...

Thanks.
Senior Manager
Joined: 05 Jun 2008
Posts: 307
Followers: 2

Kudos [?]: 120 [0], given: 0

### Show Tags

03 Dec 2008, 05:45
walker wrote:
Why not (2C1 * 6C3) / (8C4) ?

Because Ben has already been chosen. Then I have to choose 3 more people from the remaining, excluding Kelly, that is, three from six people, that’s 6c3.
so the total is (1c1.6c3)/8c4 which is 2/7
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 520

Kudos [?]: 3328 [0], given: 360

### Show Tags

03 Dec 2008, 06:48
Yeah, you are right. 2C1 would be correct if we had 2 committees but here we have only one committee.
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

SVP
Joined: 29 Aug 2007
Posts: 2492
Followers: 67

Kudos [?]: 719 [0], given: 19

### Show Tags

03 Dec 2008, 09:35
walker wrote:
I think we should take into account two possible cases: (Ben,x,x,x)(Kelly,x,x,x) and (Kelly,x,x,x)(Ben,x,x,x)
Therefore, we need chose Ben or Kelly for first committee (2C1) and three others not Ben and not Kelly (3C6)

Why are you considering (Kelly, x, x, x)? Kelley should be excluded.
(Ben,x,x,x) is fine.
so the prob = 6c3 / 8c4

Quote:
There are 8 members; among them are Kelly and Ben. A committee of 4 is to be chosen out of the 8. What is the probability that Ben is chosen to be in the committee and Kelly is not?

How many possible ways are there to solve this question ?

_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Re: PS-Probability   [#permalink] 03 Dec 2008, 09:35
Display posts from previous: Sort by