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

Oldest

Best Reply

Author

Message

TAGS:

SVP

Joined: 03 Feb 2003

Posts: 1683

Followers: 4

Kudos [?]: 16 [0], given: 0

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

27 Jun 2003, 01:35

00:00

Question Stats:

0% (00:00) correct

0% (00:00) wrong

based on 0 sessions

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: 150

Location: India

Followers: 1

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

PS: PROBABILITY14 [#permalink]

27 Jun 2003, 02:02

Stolyar,

Is the answer 3C1*1/6*4/5*3/4 = 3/10 ?

Manager

Joined: 24 Jun 2003

Posts: 150

Location: India

Followers: 1

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

PS: PROBABILITY14 [#permalink]

27 Jun 2003, 03:40

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

Comments Stolyar?

Note: What do you mean when you say the balls are drawn without repetition?

SVP

Joined: 03 Feb 2003

Posts: 1683

Followers: 4

Kudos [?]: 16 [0], given: 0

[#permalink]

27 Jun 2003, 04:22

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: 150

Location: India

Followers: 1

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

PS: PROBABILITY14 [#permalink]

27 Jun 2003, 04:30

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

Re: PS: PROBABILITY14 [#permalink]

03 Jul 2003, 21:14

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!) :wink:

prashant wrote:

GMAT Instructor

Joined: 07 Jul 2003

Posts: 773

Location: New York NY 10024

Schools: Haas, MFE; Anderson, MBA; USC, MSEE

Followers: 5

Kudos [?]: 9 [0], given: 0

[#permalink]

19 Jul 2003, 16:26

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: 1683

Followers: 4

Kudos [?]: 16 [0], given: 0

[#permalink]

20 Jul 2003, 00:22

agree guys, my bad
that day was tough, I was tired... forgive me for being so inexact

1C1*4C2/6C3=0.3

[#permalink] 20 Jul 2003, 00:22

	Similar topics	Author	Replies	Last post
Similar Topics:	There are six balls in a black sack. Among those six are	stolyar	10	27 Jun 2003, 02:56
	A black sack contains three green balls, five yellow balls,	stolyar	4	03 Jul 2003, 13:30
	A bag contains six black, seven white and eight green balls.	Konstantin Lynov	4	08 Aug 2003, 01:14
	There are six balls in a black sack. Among those six are	Riuscita	4	06 Apr 2006, 19:48
	There are six balls in a black sack. Among those six are	old_dream_1976	4	16 Apr 2006, 16:49

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

Main navigation

GMAT Resources

Partners

MBA Resources

GMAT ® is a registered trademark of the Graduate Management Admission Council ® (GMAC ®). GMAT Club's website has not been reviewed or endorsed by GMAC.

GMAT Courses

Admissions Consulting

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

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

Main navigation

GMAT Resources

Partners

MBA Resources