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VP
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Kurt, a painter, has 9 jars of paint 4 of which are yellow, [#permalink]
 11 Sep 2007, 23:50
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?
(1) 5/42
(2) 37/42
(3) 1/21
(4) 4/9
(5) 5/9
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Senior Manager
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Re: Probability - Paint - Kurt [#permalink]
12 Sep 2007, 00:04
KillerSquirrel wrote: 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? (1) 5/42 (2) 37/42 (3) 1/21 (4) 4/9 (5) 5/9 My answer is C, though I am not good at probability
We can make Jaune color only using option which is marked red, because he takes 3 jars
P=Favorable/Total
Favorable=4C3=4!/3!=4
Total=9C3=9!/(6!*3!)=84
P=4/84=1/21
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GMAT Club Legend
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So... if i have 2 jars of yellow and 3 jars of brown, is it Juane X or Brun X?
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VP
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ywilfred wrote: So... if i have 2 jars of yellow and 3 jars of brown, is it Juane X or Brun X?
I think you can only choose three jars.
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GMAT Club Legend
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KillerSquirrel wrote: ywilfred wrote: So... if i have 2 jars of yellow and 3 jars of brown, is it Juane X or Brun X? I think you can only choose three jars. hahaha.. yeah, i see that now.. 3 jars it is...
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GMAT Club Legend
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# of ways to pick any three jars = 9C3 = 84
To get Jaune X --> Either YYX (where X is any color) or YYY
YYX -> 4C2 * 5C1 = 30
YYY -> 4C3 = 4
Total # of ways get Jaune X --> 34
# of ways to get Jaune Y --> YXX (where X is any two colors, apart from yellow)
YXX = 4C1 * 5C2 = 40 ways.
Total # of ways to get Jaune = 74 ways.
P = 74/84 = 37/42
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VP
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ywilfred wrote: # of ways to pick any three jars = 9C3 = 84
To get Jaune X --> Either YYX (where X is any color) or YYY
YYX -> 4C2 * 5C1 = 30
YYY -> 4C3 = 4
Total # of ways get Jaune X --> 34
# of ways to get Jaune Y --> YXX (where X is any two colors, apart from yellow)
YXX = 4C1 * 5C2 = 40 ways.
Total # of ways to get Jaune = 74 ways.
P = 74/84 = 37/42
Perfect ! that's the OA (B)
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Senior Manager
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Hate combinations and probability, becuase my hit rate decreases to 50%
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Re: Probability - Paint - Kurt [#permalink]
12 Sep 2007, 08:02
Fistail wrote: KillerSquirrel wrote: 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? (1) 5/42 (2) 37/42 (3) 1/21 (4) 4/9 (5) 5/9
= 37/42sorry guys, did miscalculation:
= 1 - 5c3/9c3
= 1 - (10/84)
= (84 - 10)/84
= 74/84
= 37/42
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Manager
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KillerSquirrel wrote: ywilfred wrote: # of ways to pick any three jars = 9C3 = 84
To get Jaune X --> Either YYX (where X is any color) or YYY
YYX -> 4C2 * 5C1 = 30
YYY -> 4C3 = 4
Total # of ways get Jaune X --> 34
# of ways to get Jaune Y --> YXX (where X is any two colors, apart from yellow)
YXX = 4C1 * 5C2 = 40 ways.
Total # of ways to get Jaune = 74 ways.
P = 74/84 = 37/42 Perfect ! that's the OA (B) Hi I just did this problem and I'm a bit confused on why my approach is wrong instead of the following YYX -> 4C2 * 5C1 = 30 YYY -> 4C3 = 4 i just did 4c2 * 7c1 , choosing 2 yellows, and the 3rd being any of the remaining (including yellow) that ends up giving me 42 instead of 34
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