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

00:00

A

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Difficulty:

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Question Stats:

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

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?

Re: PS: hypergeometric distribution [#permalink]
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.

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?

Re: PS: hypergeometric distribution [#permalink]
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!

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...