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February 23rd, 2012, 05:00 AM   #1
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Probability of selecting things in order Part 2

Say there are 15 red balls, 10 green balls, and 3 yellow balls within a container.

What is the probability that I will select 1 red ball first, then 1 green ball, then 1 more green ball, then 1 red ball, and then 1 yellow ball, in that order?

My best guess would be (15 / 2 * (10 / 27) * (9 / 26) * (14 / 25) * (3 / 24) = 1/208 = about 0.48%

What would be the permutation way to calculate this? Is 30Permutation5 = 17,100,720 going to be 17,100,720 different ways of picking 5 balls in certain order from the total? Which includes choosing all the same color, but isn't this including choosing 5 yellow balls all in a row, even though there are only 3 of them?

The balls are not put back after taking them out.
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February 23rd, 2012, 05:13 AM   #2
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Re: Probability of selecting things in order Part 2

Quote:
Originally Posted by daigo
Say there are 15 red balls, 10 green balls, and 3 yellow balls within a container.

What is the probability that I will select 1 red ball first, then 1 green ball, then 1 more green ball, then 1 red ball, and then 1 yellow ball, in that order?
Does the question say if you put the ball back in the container after you have picked one?
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February 23rd, 2012, 07:23 AM   #3
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Re: Probability of selecting things in order Part 2

No, you don't put it back
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February 23rd, 2012, 08:44 AM   #4
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Re: Probability of selecting things in order Part 2

should be

(15 / 28 * (10 / 27) * (9 / 26) * (14 / 25) * (3 / 24)
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February 23rd, 2012, 08:49 AM   #5
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Re: Probability of selecting things in order Part 2

Oh yea, sorry, I fixed the calculation.

Do you know how to set up the permutation/combination formula for this solution?
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February 23rd, 2012, 09:23 AM   #6
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Re: Probability of selecting things in order Part 2

Quote:
Originally Posted by daigo
My best guess would be (15 / 2 * (10 / 27) * (9 / 26) * (14 / 25) * (3 / 24) = 1/208 = about 0.48%
For your benefit: your result usually (standard, I think) shown this way: = .0048 or 0.48%
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