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

 It is currently 20 Oct 2016, 14:40

### 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

# Jury

Author Message
TAGS:

### Hide Tags

Senior Manager
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 327
Followers: 14

Kudos [?]: 672 [0], given: 20

### Show Tags

26 May 2010, 01:04
00:00

Difficulty:

(N/A)

Question Stats:

100% (10:10) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

Can someone try to solve this?
Attachments

del1.jpg [ 11.15 KiB | Viewed 1582 times ]

_________________

press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

Intern
Joined: 28 Feb 2010
Posts: 9
Followers: 0

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

### Show Tags

26 May 2010, 03:17

2/3 of 15 is 10Men
1/3 of 15 is 5Women

Jury consist of 10 Men and 5Women

We need to choose 12 from 15 - 15C12.

However of the 12, [highlight]"atleast "[/highlight] 2/3 should be men which means we should have atleast 8 or more men. This leads to the following selections

8m 4w
9m 3w
10m 2w

so the prob is (10c8*5c4 + 10c9*5c3 + 10c10*5c2)/15c12 ===> 67/91
Intern
Joined: 29 Dec 2009
Posts: 33
Followers: 0

Kudos [?]: 3 [0], given: 2

### Show Tags

07 Jun 2010, 12:55
Why is (10C8 * 7C4)/15C12 incorrect?

10C8 should give us the no: of groups of 8 men and 7C4 should give us the groups of the remaining 4?
Senior Manager
Joined: 25 Jun 2009
Posts: 306
Followers: 2

Kudos [?]: 119 [0], given: 6

### Show Tags

08 Jun 2010, 07:39
dimitri92 wrote:
Can someone try to solve this?

2/3 men out of 15 = 10 men
1/3 women = 5

now if we need to have have at least 2/3 men in 12 member jury, that means we need to have at least 8 men,

Now the $$P (of having at least 2/3 men in 12 member jury)= 1 - P ( not having 2/3 men in the jury)$$

now, there is only one possiblity when we cant have 8 men in the 12 member jury i.e when all the women are selected. ( 5 women and 7 men)

So the $$probablity of having 5 women and 7 men = \frac{5C_5 * 10C_7}{15C_12}$$
$$15C_12$$ = total ways of choosing 12 members out of 15
$$P (of having at least 2/3 men in 12 member jury) = 1 - \frac{10*9*8*3*2}{3*2*15*14*13} = 1- \frac{24}{91}$$ $$=\frac{67}{91}$$
Re: Jury   [#permalink] 08 Jun 2010, 07:39
Similar topics Replies Last post
Similar
Topics:
49 If a jury of 12 people is to be selected randomly from a 22 04 Feb 2008, 16:40
Display posts from previous: Sort by

# Jury

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.