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# A bag contains 3 red, 2 white, and 6 blue marbles. What is

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Manager
Joined: 19 Jun 2003
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A bag contains 3 red, 2 white, and 6 blue marbles. What is [#permalink]

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09 Aug 2004, 13:33
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A bag contains 3 red, 2 white, and 6 blue marbles. What is the probability of drawing, in order, 2 red, 1 blue, and 2 white marbles?

I tried to solve this first using dependent events (only probablity techniques) and got the correct answer.

Than, I tried using hypergeometric distribution, couldn't get the same result..

Could somebody solve this using hypergeometric distribution? Or may be advise whether this type of question is suitable for hypergeometric distribution?

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Intern
Joined: 27 Jul 2004
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10 Aug 2004, 00:33
Is this the answer using simple probability

{(3C2)*(6C1)*(2C2)} / (11C5)

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Senior Manager
Joined: 22 Jun 2004
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10 Aug 2004, 05:42

the required probability = (3/11) (2/10) (6/9) (2/8) (1/7) = 1/770

Do let me know if I am correct.

BTW what is hypergeometric distribution? Is it different from binomial, poisson's and exponential distributions? If not, to the best of my understanding, this problem does not fit in any of binomial, poisson's and exponential distributions.

Awaiting OA and explanation on hypergeometric distribution.

afife76 wrote:
A bag contains 3 red, 2 white, and 6 blue marbles. What is the probability of drawing, in order, 2 red, 1 blue, and 2 white marbles?

I tried to solve this first using dependent events (only probablity techniques) and got the correct answer.

Than, I tried using hypergeometric distribution, couldn't get the same result..

Could somebody solve this using hypergeometric distribution? Or may be advise whether this type of question is suitable for hypergeometric distribution?

_________________

Awaiting response,

Thnx & Rgds,
Chandra

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Manager
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10 Aug 2004, 08:42
http://www.gmatclub.com/content/courses ... bility.php

So it is :
p = aCa' * bCb' * cCc' * ... * zCz' / (a+b+c+..+z)C(a'+b'+c'+...+z')

So for this question:
r=3
r'= 2
w=2
w'=2
b=6
b'=1

p= 3C2 * 2C2 * 6C1 / (3+2+6) C (2+2+1)
p= 18 / 462

But when you calculate thru the regular probability techniques:

A bag contains 3 red, 2 white, and 6 blue marbles. What is the probability of drawing, in order, 2 red, 1 blue, and 2 white marbles?

3/11 * 2/10 * 6/9 * 2/8 * 1/7 = 1/770

What I am missing ??

1/770 is the OA.

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CIO
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10 Aug 2004, 17:37
afife76 wrote:
What I am missing ??

I think you are missing sleep over something trivial and beyond the scope of the test. I would never ever worry about something like this on the gmat.

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Intern
Joined: 22 Feb 2011
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22 Feb 2011, 20:06
Acctually hypergeometric distribution doesn't consider "sequence"(order) factor, so you need to multiply the answer derived from hypergeometric distribution by the sequencial possibilities:

P = (18/462) * 2!1!2!/5! = 1/770

where the 1st 2! means the possibility of drawing first 2 as red and the 2nd 2! means the same for last 2 white.
and 5! is simply the all possible orders of 5 balls.

I know this is very old post, but I do really hope someone can confirm me whether the above is correct?

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SVP
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23 Feb 2011, 01:30
3/11 * 2/10 * 6/9 * 2/8 * 1/7

= 3/11 * 1/5 * 2/3 * 1/4 * 1/7

= 1/55 * 1/14

= 1/770
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Manager
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23 Feb 2011, 22:02
Possible with slot method.

___ ___ ___ ___ ___

First choice...to continue on you need to draw one of the 3 reds out of a bag of 11...(3/11)

(3/11) ___ ____ ____ ____

Next choice, to continue on you need one of the two reds out of a bag of 10 ...(2/10) = (1/5)

(3/11) (1/5) ____ ____ ____

Next choice. to continue you need one of the 6 blues, out of a bag of 9. (6/9) = (2/3)

(3/11) (1/5) (2/3) ____ ____

Next choice. To continue on you need one of the two white marbles from a bag of 8. (2/8) = (1/4)

(3/11) (1/5) (2/3) (1/4) ____

Last choice..you need the white, out of 7. (1/7)

(3/11) (1/5) (2/3) (1/4) (1/7)

Simplify 1/770.

Doable...

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Re: PS: hypergeometric distribution   [#permalink] 23 Feb 2011, 22:02
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