Kurt, a painter, has 9 jars of paint 4 of which are yellow, : PS Archive
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# Kurt, a painter, has 9 jars of paint 4 of which are yellow,

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Director
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Kurt, a painter, has 9 jars of paint 4 of which are yellow, [#permalink]

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04 Oct 2004, 16:59
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Kurt, a painter, has 9 jars of paint 4 of which are yellow, 2 are red and the remaining jars are brown. Kurt will combine 3 jars of paint into a new container to make a new color which he will name according to the following conditions:
Brun Y if the paint contains 2 jars of brown paint and no yellow.
Brun X if the paint contains 3 jars of brown paint.
Jaune X if the paint contains at least 2 jars of yellow.
Jaune Y if the paint contains exactly 1 jar of yellow.

What is the probability that the new color will be Jaune?
(A) 5/42
(B) 37/42
(C) 1/21
(D) 4/9
(E) 5/9
Director
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04 Oct 2004, 17:25
1-5/42 = 37/42. Ans. B

5/42 = prob of not getting a yellow.
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04 Oct 2004, 21:48
venksune - you are right, the prob of not getting a yellow is 5/42, which means that 37/42 is the probability of getting AT LEAST one yellow.

It's not what the question asks, so it can't be right.

I obviously messed up somewhere in my calculations.... because I get an answer that's not listed as an answer choice. ... but I would appriciate if someone could post a solution.

-Irene
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05 Oct 2004, 00:06
Irene,
The other approach is - slightly tedious though...

JX has atleast 2 yellows and JY has exactly 1 yellow. So, we have..
4C15C2/9C3 (for JY) + 4C2*5C1/9C3 + 4C3*5C0/9C3 (for JX; the new color have only 3 jars of paint). Going by this logic also, we will get 37/42.
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05 Oct 2004, 01:35
37/42
My way:

p(at least 2 yellow) + p(exactly 1 yellow) = 1-p(0 yellow).
p(0 yellow) = 5/9 * 4/8 * 3/7 = 5/42.
1-5/42 = 37/42.
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05 Oct 2004, 01:46
Actually, i don't understand - why is the total 9C3 ?

Some of the items of the 9 are identicle...
Director
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05 Oct 2004, 04:54
I have the same questions Dookie !

Numerator and denominator using the same nCp concept there is no need to multiply or divide by n! or p!

I think the 2 approachs are equivalent (i mean yours and Venksune's one) as it is a probability and not a comb prob.

Anyway a guru's advice is welcome !
05 Oct 2004, 04:54
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