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# There are six balls in a black sack. Among those six are

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There are six balls in a black sack. Among those six are [#permalink]  27 Jun 2003, 00:35
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There are six balls in a black sack. Among those six are black ball and white one. Three balls are taken at random without repetition. What is the probability that the white ball IS taken, but the black one IS NOT?
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PS: PROBABILITY14 [#permalink]  27 Jun 2003, 01:02
Stolyar,

Is the answer 3C1*1/6*4/5*3/4 = 3/10 ?
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PS: PROBABILITY14 [#permalink]  27 Jun 2003, 02:40
I racked my brain on this one, and I have a new solution. The logic is as follows

The white ball can be drawn in the first, second or the third try.

Probability of the white ball being drawn in the first try, with the black ball NOT being drawn in the second and the third tries is 1/6*4/5*3/4=12/120

Probability of the white being drawn in second try is 5/6*1/5*3/4=15/120

Probability of the white being drawn in third is 5/6*4/5*1/4 = 20/120

The sum is 47/120

This assumes that the balls are drawn without replacement. If they are drawn with replacement, then the answer will be 3C1*1/6*5/6*5/6 = 25/72

Note: What do you mean when you say the balls are drawn without repetition?
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Employ a classical approach:

P=F/T A suck contain: WBXXXX

T=6C3

F=1C1 (the whithe ball is taken)*5C4 (the black is not)

Thus, 1C1*5C4/6C3
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PS: PROBABILITY14 [#permalink]  27 Jun 2003, 03:30
Stolyar,

I understand your approach; its quite elegant. Am I correct in assuming that there is no replacement of balls?

Still, I would like to understand the flaw in my logic, if you can help me.

Thanks!
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Re: PS: PROBABILITY14 [#permalink]  03 Jul 2003, 20:14
Prashant,

I see a flaw with your second theory....

If the white ball is picked first, I agree that we have:
1/6 * 4/5 * 3/4 = 12/120

But, if the white ball is picked second, we should have:
4/6 * 1/5 * 3/4 = (still) 12/120

And, if the white ball is picked third, we should have:
4/6 * 3/4 * 1/5 = (still) 12/120

So, I would agree with your first post....
P = 12/120 * C(3,1) = 3/10

I'm still learning with these wonderful probability exercises... if there's a mistake in my logic above, please let me know! (But pls go easy on me!)

prashant wrote:
I racked my brain on this one, and I have a new solution. The logic is as follows

The white ball can be drawn in the first, second or the third try.

Probability of the white ball being drawn in the first try, with the black ball NOT being drawn in the second and the third tries is 1/6*4/5*3/4=12/120

Probability of the white being drawn in second try is 5/6*1/5*3/4=15/120

Probability of the white being drawn in third is 5/6*4/5*1/4 = 20/120

The sum is 47/120

This assumes that the balls are drawn without replacement. If they are drawn with replacement, then the answer will be 3C1*1/6*5/6*5/6 = 25/72

Note: What do you mean when you say the balls are drawn without repetition?
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stolyar wrote:
Employ a classical approach:

P=F/T A suck contain: WBXXXX

T=6C3

F=1C1 (the whithe ball is taken)*5C4 (the black is not)

Thus, 1C1*5C4/6C3

Should be 1c1 * 4c2 / 6c3.

We have one way to choose a white ball we want to know how many ways we can pair it with two of the four other non-black balls. This divided by 6c3 gives you 3/10 or .30.
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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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that day was tough, I was tired... forgive me for being so inexact

1C1*4C2/6C3=0.3
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