There are six balls in a black sack. Among those six are : PS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 20 Jan 2017, 04:06

### 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

# There are six balls in a black sack. Among those six are

Author Message
SVP
Joined: 03 Feb 2003
Posts: 1603
Followers: 8

Kudos [?]: 245 [1] , given: 0

There are six balls in a black sack. Among those six are [#permalink]

### Show Tags

27 Jun 2003, 00:35
1
KUDOS
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 six balls in a black sack. Among those six are black ball and white one. Three balls are taken at random without repetition. What is the probability that the white ball IS taken, but the black one IS NOT?
Manager
Joined: 24 Jun 2003
Posts: 146
Location: India
Followers: 2

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

### Show Tags

27 Jun 2003, 01:02
Stolyar,

Is the answer 3C1*1/6*4/5*3/4 = 3/10 ?
Manager
Joined: 24 Jun 2003
Posts: 146
Location: India
Followers: 2

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

### Show Tags

27 Jun 2003, 02:40
I racked my brain on this one, and I have a new solution. The logic is as follows

The white ball can be drawn in the first, second or the third try.

Probability of the white ball being drawn in the first try, with the black ball NOT being drawn in the second and the third tries is 1/6*4/5*3/4=12/120

Probability of the white being drawn in second try is 5/6*1/5*3/4=15/120

Probability of the white being drawn in third is 5/6*4/5*1/4 = 20/120

The sum is 47/120

This assumes that the balls are drawn without replacement. If they are drawn with replacement, then the answer will be 3C1*1/6*5/6*5/6 = 25/72

Note: What do you mean when you say the balls are drawn without repetition?
SVP
Joined: 03 Feb 2003
Posts: 1603
Followers: 8

Kudos [?]: 245 [1] , given: 0

### Show Tags

27 Jun 2003, 03:22
1
KUDOS
Employ a classical approach:

P=F/T A suck contain: WBXXXX

T=6C3

F=1C1 (the whithe ball is taken)*5C4 (the black is not)

Thus, 1C1*5C4/6C3
Manager
Joined: 24 Jun 2003
Posts: 146
Location: India
Followers: 2

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

### Show Tags

27 Jun 2003, 03:30
Stolyar,

I understand your approach; its quite elegant. Am I correct in assuming that there is no replacement of balls?

Still, I would like to understand the flaw in my logic, if you can help me.

Thanks!
Intern
Joined: 23 Jun 2003
Posts: 16
Followers: 0

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

### Show Tags

03 Jul 2003, 20:14
Prashant,

I see a flaw with your second theory....

If the white ball is picked first, I agree that we have:
1/6 * 4/5 * 3/4 = 12/120

But, if the white ball is picked second, we should have:
4/6 * 1/5 * 3/4 = (still) 12/120

And, if the white ball is picked third, we should have:
4/6 * 3/4 * 1/5 = (still) 12/120

So, I would agree with your first post....
P = 12/120 * C(3,1) = 3/10

I'm still learning with these wonderful probability exercises... if there's a mistake in my logic above, please let me know! (But pls go easy on me!)

prashant wrote:
I racked my brain on this one, and I have a new solution. The logic is as follows

The white ball can be drawn in the first, second or the third try.

Probability of the white ball being drawn in the first try, with the black ball NOT being drawn in the second and the third tries is 1/6*4/5*3/4=12/120

Probability of the white being drawn in second try is 5/6*1/5*3/4=15/120

Probability of the white being drawn in third is 5/6*4/5*1/4 = 20/120

The sum is 47/120

This assumes that the balls are drawn without replacement. If they are drawn with replacement, then the answer will be 3C1*1/6*5/6*5/6 = 25/72

Note: What do you mean when you say the balls are drawn without repetition?
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 25

Kudos [?]: 205 [0], given: 0

### Show Tags

19 Jul 2003, 15:26
stolyar wrote:
Employ a classical approach:

P=F/T A suck contain: WBXXXX

T=6C3

F=1C1 (the whithe ball is taken)*5C4 (the black is not)

Thus, 1C1*5C4/6C3

Should be 1c1 * 4c2 / 6c3.

We have one way to choose a white ball we want to know how many ways we can pair it with two of the four other non-black balls. This divided by 6c3 gives you 3/10 or .30.
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

SVP
Joined: 03 Feb 2003
Posts: 1603
Followers: 8

Kudos [?]: 245 [0], given: 0

### Show Tags

19 Jul 2003, 23:22
that day was tough, I was tired... forgive me for being so inexact

1C1*4C2/6C3=0.3
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13459
Followers: 575

Kudos [?]: 163 [0], given: 0

Re: There are six balls in a black sack. Among those six are [#permalink]

### Show Tags

27 Jul 2015, 07:18
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: There are six balls in a black sack. Among those six are   [#permalink] 27 Jul 2015, 07:18
Display posts from previous: Sort by