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Re: A bag contains 3 red, 2 white, and 6 blue marbles. What is [#permalink]
For hypergeometric distribution, please check
https://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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Re: A bag contains 3 red, 2 white, and 6 blue marbles. What is [#permalink]
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: A bag contains 3 red, 2 white, and 6 blue marbles. What is [#permalink]
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: A bag contains 3 red, 2 white, and 6 blue marbles. What is [#permalink]
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: A bag contains 3 red, 2 white, and 6 blue marbles. What is [#permalink]
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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