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A black sack contains three green balls, five yellow balls,

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Director
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A black sack contains three green balls, five yellow balls, [#permalink] New post 28 Dec 2006, 14:45
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A
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C
D
E

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A black sack contains three green balls, five yellow balls, and four white ones. Three balls are taken at random without repetition. What is the probability of having all the the three balls of different colors?

A 3/22
B 1/22
C 3/11
D 1/121
E 1/33
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Last edited by Swagatalakshmi on 28 Dec 2006, 17:28, edited 1 time in total.
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 [#permalink] New post 28 Dec 2006, 16:24
Hmm, I get 1/22

(3C1*5C1*4C1)/(12C1*11C1*10C1) =1/22

I would have done 12C3 if this were with repetition.

Still not good with probability though, so not sure.
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 [#permalink] New post 28 Dec 2006, 16:34
good answer...
here is another potential explanation.
suppose the question was a bit different: what is the probability that the first is green, the second is yellow and the third is white?
oh, that's relatively easy:
it is 3/12 for the first, 5/11 for the second (we removed already one ball), and 4/10 for the third. total (3*4*5)/(12*11*10)
good... lets continue,
what is the probability to get yellow, white and green in that order?
it is 5/12 * 4/11 * 3/10 = (3*4*5)/(12*11*10)

do you see that? for any ordering you get the same probability, made of the same numbers, but in different order.

this should not surprise you too much, after all there is not reason that certain ordering would be more likely than another. this is called symmetry.

but the original question did not gave a specific order, so how we proceed?
that's easy as well... there are total of 3!=6 possible orderings of 3 colors.
so just add these...
6*(3*4*5)/(12*11*10) = 3/11

this approach (and again.... not the only and not necessarily the best one) exemplifies a wider strategy that help solving wide range of problems: a problem seem difficult or complicated with the given data, carefully throw in more arbitrary data that will help you solve it easily. then, carefully take "extra data" out to get the result of the original question.

it helps a lot in counting problems. it also helps with some geometry, number properties etc...
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 [#permalink] New post 28 Dec 2006, 17:13
Thanks hobbit. I came to my answer knowing I had missed something, but this is a great explanation.
I know you will probably frown upon doing it like this, would it be alright in all such cases to just divide by 12C3 (or equivalent ) in a problem without worrying about repetitions (or not)? I will also try to work this out shortly.



hobbit wrote:
good answer...
here is another potential explanation.
suppose the question was a bit different: what is the probability that the first is green, the second is yellow and the third is white?
oh, that's relatively easy:
it is 3/12 for the first, 5/11 for the second (we removed already one ball), and 4/10 for the third. total (3*4*5)/(12*11*10)
good... lets continue,
what is the probability to get yellow, white and green in that order?
it is 5/12 * 4/11 * 3/10 = (3*4*5)/(12*11*10)

do you see that? for any ordering you get the same probability, made of the same numbers, but in different order.

this should not surprise you too much, after all there is not reason that certain ordering would be more likely than another. this is called symmetry.

but the original question did not gave a specific order, so how we proceed?
that's easy as well... there are total of 3!=6 possible orderings of 3 colors.
so just add these...
6*(3*4*5)/(12*11*10) = 3/11

this approach (and again.... not the only and not necessarily the best one) exemplifies a wider strategy that help solving wide range of problems: a problem seem difficult or complicated with the given data, carefully throw in more arbitrary data that will help you solve it easily. then, carefully take "extra data" out to get the result of the original question.

it helps a lot in counting problems. it also helps with some geometry, number properties etc...
Director
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 [#permalink] New post 28 Dec 2006, 17:20
1/22 for me.

All scenarios seem to point to B.

(4/12)(5/11)(3/10)=1/22
(4/12)(3/11)(5/10)=1/22
(5/12)(3/11)(4/10)=1/22
(5/12)(4/11)(3/10)=1/22
(3/12)(5/11)(4/10)=1/22
(3/12)(4/11)(5/10)=1/22
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 [#permalink] New post 28 Dec 2006, 17:32
hsampath wrote:
Thanks hobbit. I came to my answer knowing I had missed something, but this is a great explanation.
I know you will probably frown upon doing it like this, would it be alright in all such cases to just divide by 12C3 (or equivalent ) in a problem without worrying about repetitions (or not)? I will also try to work this out shortly.



it is difficult to say what to do in "all such cases".... there is no rule for all cases. each question may contain a slightly different variant which may change the principle bes used to solve it....
it is important, though, to understand WHY 12C3 was used.
the denominator in probability calculations, usually represents the number of all possible cases. the numerator counts the relevant cases for the question. it is a MUST that the numerator and denominator count the same thing. it is like units of measurements in velocity. you cant just add velocities measured in MPH and mps, you need to convert.

12C3 counts the number of possibilities to choose 3 different items out of possible 12 different items, regardless of order.

12*11*10 counts the number of possibilities to choose 3 different items out of possible 12 different items with regard to order.

in this question if you count "green,yellow,white" and "yellow,white,green" as two different choices because of the order of draw - that's fine (this is how i did it). just count carefully, and divide by 12*11*10

if you count the possibility without regard to order, as ncprasad probably did (he just gave the formula, not explanation, so i'm not sure what went through his mind) - then divide by 12C3.

both ways are fine. just count carefully, and be aware what you are counting.
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 [#permalink] New post 28 Dec 2006, 17:33
ggarr wrote:
1/22 for me.

All scenarios seem to point to B.

(4/12)(5/11)(3/10)=1/22
(4/12)(3/11)(5/10)=1/22
(5/12)(3/11)(4/10)=1/22
(5/12)(4/11)(3/10)=1/22
(3/12)(5/11)(4/10)=1/22
(3/12)(4/11)(5/10)=1/22


but since there are 6 such scenarios.... and all are actually possible and independent from each other - then you must add those probabilities and get C.
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 [#permalink] New post 28 Dec 2006, 19:57
C.

1. All possibilities. order does not matter. it is a combination problem.:

12C3.


2. desired possibilities. you just get each ball out of each color. so
it is 3*5*4.

so 3*5*4/12C3 = 3/11.
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Re: PS:Probability [#permalink] New post 29 Dec 2006, 14:14
Swagatalakshmi wrote:
A black sack contains three green balls, five yellow balls, and four white ones. Three balls are taken at random without repetition. What is the probability of having all the the three balls of different colors?

A 3/22
B 1/22
C 3/11
D 1/121
E 1/33

-------------------------------------
C..3/11

(3C1.5C1.4C1.3!)/(12C1.11C1.10C1)
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 [#permalink] New post 31 Dec 2006, 00:30
anindyat wrote:
(3 * 5 * 4) / (12 * 11 * 10) * 3! = 3/11


Exactly. You calculate the basic scenario ((3*5*4)/(12*11*10)) and then multiply it times the number of possible arrangements for the 3 balls. It´s C.
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 [#permalink] New post 31 Dec 2006, 00:55
[(3C1/12C1) * (5C1/11C1) * (4C1/10C1) ]3P3

=3/11
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 [#permalink] New post 03 Sep 2007, 16:34
Another old problem....

I think I'm completely in the dark here. Can anyone explain exactly what tipped you off to make this a combinatorics problem? I was just multiplying straight probability for each scenario to get the answer. Thanks.
  [#permalink] 03 Sep 2007, 16:34
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