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Kurt, a painter, has 9 jars of paint 4 of which are yellow,

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VP
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Kurt, a painter, has 9 jars of paint 4 of which are yellow, [#permalink] New post 11 Sep 2007, 22:50
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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Re: Probability - Paint - Kurt [#permalink] New post 11 Sep 2007, 23:04
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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 [#permalink] New post 11 Sep 2007, 23:55
So... if i have 2 jars of yellow and 3 jars of brown, is it Juane X or Brun X?
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 [#permalink] New post 12 Sep 2007, 00:15
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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 [#permalink] New post 12 Sep 2007, 00:20
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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 [#permalink] New post 12 Sep 2007, 00:25
# 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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 [#permalink] New post 12 Sep 2007, 00:35
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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 [#permalink] New post 12 Sep 2007, 01:37
Hate combinations and probability, becuase my hit rate decreases to 50%
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Re: Probability - Paint - Kurt [#permalink] New post 12 Sep 2007, 07:02
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

:)


= 1 - 5c3/9c3
= 37/42


sorry guys, did miscalculation:

= 1 - 5c3/9c3
= 1 - (10/84)
= (84 - 10)/84
= 74/84
= 37/42
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Re: [#permalink] New post 08 Nov 2011, 21:35
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
Re:   [#permalink] 08 Nov 2011, 21:35
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