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

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A bag contains 3 red, 2 white, and 6 blue marbles. What is [#permalink] New post 09 Aug 2004, 12:33
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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. :cool

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

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

:thanks
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 [#permalink] New post 09 Aug 2004, 23:33
Is this the answer using simple probability

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

:roll:
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Re: PS: hypergeometric distribution [#permalink] New post 10 Aug 2004, 04:42
I think the answer is

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. :cool

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

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

:thanks

_________________

Awaiting response,

Thnx & Rgds,
Chandra

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 [#permalink] New post 10 Aug 2004, 07:42
For hypergeometric distribution, please check
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 ??

:thanks
1/770 is the OA.
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 [#permalink] New post 10 Aug 2004, 16:37
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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Re: PS: hypergeometric distribution [#permalink] New post 22 Feb 2011, 19:06
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?
Thanks in advance!
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Re: PS: hypergeometric distribution [#permalink] New post 23 Feb 2011, 00:30
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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Re: PS: hypergeometric distribution [#permalink] New post 23 Feb 2011, 21:02
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...
Re: PS: hypergeometric distribution   [#permalink] 23 Feb 2011, 21:02
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