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A box contains five red and six white balls. If four balls

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Konstantin Lynov
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A box contains five red and six white balls. If four balls [#permalink] New post   08 Aug 2003, 03:04
A box contains five red and six white balls. If four balls are taken out of the box, find the probability that:

1) two are read and two are white;
2)all are the same color;
3) at least three are white;



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Re: A box contains five red and six white balls. If four balls [#permalink] New post   08 Aug 2003, 10:05
A) (5C2 * 6C2)/11C4 = 5/22
B) (5C4 + 6C4)/11C4 = 1/33
C) ((6C3 * 5C1) + 6C4)/11C4 = 23/132
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Re: A box contains five red and six white balls. If four balls [#permalink] New post   08 Aug 2003, 15:13
For the first one, I know this is the correct answer, but
I don't know how it should be.

I calculated P = 5/11 * 4/10 * 6/9 * 5/8

What am I missing?
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Re: A box contains five red and six white balls. If four balls [#permalink] New post   08 Aug 2003, 16:17
kpadma wrote:
For the first one, I know this is the correct answer, but
I don't know how it should be.

I calculated P = 5/11 * 4/10 * 6/9 * 5/8

What am I missing?


First of all 5c2 * 6c2 / 11c4 = 10 * 15 / 330 = 5/11 not 5/22.

You are calculating the probability of just one way of combining 2 whites and 2 reds. If you multiply your solution by the number of ways 2W and 2R can combine (or 4c2) you will get the correct answer.
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kpadma
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Re: A box contains five red and six white balls. If four balls [#permalink] New post   08 Aug 2003, 20:04
AkamaiBrah wrote:
kpadma wrote:
For the first one, I know this is the correct answer, but
I don't know how it should be.

I calculated P = 5/11 * 4/10 * 6/9 * 5/8

What am I missing?


First of all 5c2 * 6c2 / 11c4 = 10 * 15 / 330 = 5/11 not 5/22.

You are calculating the probability of just one way of combining 2 whites and 2 reds. If you multiply your solution by the number of ways 2W and 2R can combine (or 4c2) you will get the correct answer.


Thank you for your response. The trouble with me is either I get it
completely or not. In this case, I've not got it.

Let us assume that we are taking 4 white balls out of 5 white and 6 red
balls then we say P = 5/11 * 4/10 * 3/9 * 2/8.
Why we need to multiply in with 4C2 if we are choosing more than one
color. Why does the order of picking the balls matter?
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AkamaiBrah
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Re: A box contains five red and six white balls. If four balls [#permalink] New post   08 Aug 2003, 20:39
kpadma wrote:
AkamaiBrah wrote:
kpadma wrote:
For the first one, I know this is the correct answer, but
I don't know how it should be.

I calculated P = 5/11 * 4/10 * 6/9 * 5/8

What am I missing?


First of all 5c2 * 6c2 / 11c4 = 10 * 15 / 330 = 5/11 not 5/22.

You are calculating the probability of just one way of combining 2 whites and 2 reds. If you multiply your solution by the number of ways 2W and 2R can combine (or 4c2) you will get the correct answer.


Thank you for your response. The trouble with me is either I get it
completely or not. In this case, I've not got it.

Let us assume that we are taking 4 white balls out of 5 white and 6 red
balls then we say P = 5/11 * 4/10 * 3/9 * 2/8.
Why we need to multiply in with 4C2 if we are choosing more than one
color. Why does the order of picking the balls matter?


It doesn't matter. But you are calculating the SPECIFIC probability that the ball will come up in a specific sequence (2R then 2W). Hence, you are missing the other five ways.



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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Re: A box contains five red and six white balls. If four balls [#permalink]
08 Aug 2003, 20:39
Moderator:
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