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

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stolyar
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A certain box holds 3 green balls, 2 white balls, and 1 blue [#permalink] New post   15 Jul 2003, 09:44
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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?



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Re: A certain box holds 3 green balls, 2 white balls, and 1 blue [#permalink] New post   15 Jul 2003, 10:01
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?
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Re: A certain box holds 3 green balls, 2 white balls, and 1 blue [#permalink] New post   Updated on: 16 Jul 2003, 02:24
Select all three which are not blue..

So, 5C3/6C3.

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

Originally posted by evensflow on 15 Jul 2003, 10:09.
Last edited by evensflow on 16 Jul 2003, 02:24, edited 1 time in total.
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Re: A certain box holds 3 green balls, 2 white balls, and 1 blue [#permalink] New post   15 Jul 2003, 19:48
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?
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Re: A certain box holds 3 green balls, 2 white balls, and 1 blue [#permalink] New post   16 Jul 2003, 01:18
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.



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Re: A certain box holds 3 green balls, 2 white balls, and 1 blue [#permalink]
16 Jul 2003, 01:18
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