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Probability

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gmatblast
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Probability [#permalink] New post   11 Feb 2004, 13:57
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?



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Re: Probability [#permalink] New post   11 Feb 2004, 14:17
P(2W&2R) = 3/8
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Re: Probability [#permalink] New post   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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Re: Probability [#permalink] New post   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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Re: Probability [#permalink] New post   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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Re: Probability [#permalink] New post   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....................
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Re: Probability [#permalink] New post   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.
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Re: Probability [#permalink] New post   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.



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Re: Probability [#permalink]
13 Feb 2004, 02:10
Moderator:
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