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# Probability

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Senior Manager
Joined: 11 Nov 2003
Posts: 356

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Location: Illinois

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11 Feb 2004, 13:57
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

There are 4 identical balls (same size). Each ball could be either RED or WHITE only. What is the probability that there are 2 RED and 2 WHITE balls?

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Director
Joined: 03 Jul 2003
Posts: 651

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11 Feb 2004, 14:17
P(2W&2R) = 3/8

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SVP
Joined: 30 Oct 2003
Posts: 1788

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Location: NewJersey USA

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11 Feb 2004, 14:34
I get 1/4 as the answer.

Each ball can either be RED or WHITE so we have 4 balls hence 2^4 = 16 combinations. Out of these 2W and 2R occur 4 times as follows

WWRR
RRWW
WRWR
RWRW
Hence 4/16 = 1/4

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Senior Manager
Joined: 11 Nov 2003
Posts: 356

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Location: Illinois

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11 Feb 2004, 14:36
anandnk wrote:
I get 1/4 as the answer.

Each ball can either be RED or WHITE so we have 4 balls hence 2^4 = 16 combinations. Out of these 2W and 2R occur 4 times as follows

WWRR
RRWW
WRWR
RWRW
Hence 4/16 = 1/4

Anand,

You missed WRRW and RWWR

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SVP
Joined: 30 Oct 2003
Posts: 1788

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Location: NewJersey USA

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11 Feb 2004, 14:37
Oops my bad. You are right
No of ways you can arrange 2 W and 2R = 4!/(2!*2!) = 6
Total combinations = 16
so P = 6/16 = 3/8

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Manager
Joined: 28 Jan 2004
Posts: 202

Kudos [?]: 28 [0], given: 4

Location: India

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12 Feb 2004, 06:14
It is not arrangement ...so there is no difference between WRRW and RWWR!!!!!!!!!!!!!!
Please let me know if my approach is incorrect.

RED BLUE
4 0
3 1
2 2
1 3
0 4

So the probability is 1/5..............
Probability is my weakest...so please pardon me if i did a stupid analysis...and if anyone is posting a reply to my reply the please just copy and send that to me as PM also because tomorrow i will not be able to track this question!!!!! i don't know how to do that if it can be done..and i am sure that is some way of doing that....................

Kudos [?]: 28 [0], given: 4

GMAT Instructor
Joined: 07 Jul 2003
Posts: 769

Kudos [?]: 235 [0], given: 0

Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE

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12 Feb 2004, 11:31
mdfrahim wrote:
It is not arrangement ...so there is no difference between WRRW and RWWR!!!!!!!!!!!!!!
Please let me know if my approach is incorrect.

RED BLUE
4 0
3 1
2 2
1 3
0 4

So the probability is 1/5..............
Probability is my weakest...so please pardon me if i did a stupid analysis...and if anyone is posting a reply to my reply the please just copy and send that to me as PM also because tomorrow i will not be able to track this question!!!!! i don't know how to do that if it can be done..and i am sure that is some way of doing that....................

You counted the "outcomes" but they're are not equally likely, hence, you cannot figure out the probability that way.
_________________

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

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Manager
Joined: 28 Jan 2004
Posts: 202

Kudos [?]: 28 [0], given: 4

Location: India

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13 Feb 2004, 02:10
Well ,Probability = favourable events/total number of events
Sorry to say , i am not yet satisfied......is somebody there to elaborate on this.

Thanks.

Kudos [?]: 28 [0], given: 4

13 Feb 2004, 02:10
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# Probability

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