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# Probability

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Senior Manager
Joined: 11 Nov 2003
Posts: 355
Location: Illinois
Followers: 1

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

Probability [#permalink]  11 Feb 2004, 12:57
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?
Director
Joined: 03 Jul 2003
Posts: 656
Followers: 2

Kudos [?]: 21 [0], given: 0

P(2W&2R) = 3/8
SVP
Joined: 30 Oct 2003
Posts: 1797
Location: NewJersey USA
Followers: 5

Kudos [?]: 41 [0], given: 0

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
Senior Manager
Joined: 11 Nov 2003
Posts: 355
Location: Illinois
Followers: 1

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

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
SVP
Joined: 30 Oct 2003
Posts: 1797
Location: NewJersey USA
Followers: 5

Kudos [?]: 41 [0], given: 0

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
Manager
Joined: 28 Jan 2004
Posts: 203
Location: India
Followers: 2

Kudos [?]: 12 [0], given: 4

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....................
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 13

Kudos [?]: 51 [0], given: 0

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

Manager
Joined: 28 Jan 2004
Posts: 203
Location: India
Followers: 2

Kudos [?]: 12 [0], given: 4

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

Thanks.
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