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

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Manager
Joined: 24 Jun 2003
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Location: Moscow
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A box contains five red and six white balls. If four balls [#permalink]

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08 Aug 2003, 03:04
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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;
Manager
Joined: 10 Jun 2003
Posts: 210
Location: Maryland
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08 Aug 2003, 10:05
A) (5C2 * 6C2)/11C4 = 5/22
B) (5C4 + 6C4)/11C4 = 1/33
C) ((6C3 * 5C1) + 6C4)/11C4 = 23/132
Director
Joined: 03 Jul 2003
Posts: 652
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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?
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 26

Kudos [?]: 212 [0], given: 0

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08 Aug 2003, 16:17
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.
_________________

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

Director
Joined: 03 Jul 2003
Posts: 652
Followers: 3

Kudos [?]: 92 [0], given: 0

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08 Aug 2003, 20:04
AkamaiBrah 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?
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 26

Kudos [?]: 212 [0], given: 0

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08 Aug 2003, 20:39
AkamaiBrah 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.
_________________

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

08 Aug 2003, 20:39
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