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# A certain box holds 3 green balls, 2 white balls, and 1 blue

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SVP
Joined: 03 Feb 2003
Posts: 1609
Followers: 6

Kudos [?]: 75 [0], given: 0

A certain box holds 3 green balls, 2 white balls, and 1 blue [#permalink]  15 Jul 2003, 08:44
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A certain box holds 3 green balls, 2 white balls, and 1 blue ball. Three balls are taken at random without repetition. What is the probability of having the blue ball among the taken ones?
SVP
Joined: 03 Feb 2003
Posts: 1609
Followers: 6

Kudos [?]: 75 [0], given: 0

The only thing that I can add is that it is 1/2.

Let us reformulate the question: what is the probability of NOT having the blue ball among the taken ones?
Manager
Joined: 08 Apr 2003
Posts: 149
Followers: 1

Kudos [?]: 8 [0], given: 0

Select all three which are not blue..

So, 5C3/6C3.

Opps!! it was a typo. I corrected the mistake.

Last edited by evensflow on 16 Jul 2003, 01:24, edited 1 time in total.
Manager
Joined: 07 Jul 2003
Posts: 56
Followers: 1

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Stop! i stuck!

let's do it slowly -
when we choose three balls, we have 6c3 possibilities.
from these choices we have 3c1 outcomes with blue one.
where I am wrong?
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 13

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RK73 wrote:
Stop! i stuck!

let's do it slowly -
when we choose three balls, we have 6c3 possibilities.
from these choices we have 3c1 outcomes with blue one.
where I am wrong?

There are 6c3 or 20 possible combinations of 3 balls.

To calculate the number of possible arrangement of 3 balls which include a blue ball assume that the arrangement has a blue ball. Now calculate the number of ways that 2 balls can be combined of the 5 left to accompany the blue one. This is 5c2 = 10. Hence, the probability is 1/2.

Another way to do this is say:
Suppose I pull the balls out 1 by 1 (w/o replacement). For the 3 balls to have the blue one, it must be the 1st, 2nd, or 3rd ball.

Pr(1st) = 1/6
pr(2nd) = 5/6*1/5 = 1/6
pr(3rd) = 5/6*4/5*1/4 = 1/6

These are mutually exclusive events, so we can add them together to get a total probability of 1/2.
_________________

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

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