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Manager
Joined: 26 Mar 2010
Posts: 94

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26 Mar 2010, 23:19
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Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

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Kurt, a painter, has 9 jars of paint:
4 are yellow
2 are red
rest are brown
Kurt will combine 3 jars of paint into a new container to make a new color, which he will name accordingly 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

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Manager
Joined: 24 Mar 2010
Posts: 69

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27 Mar 2010, 02:11
To get jaune, we have to evaluate the two latter sentences, and add them to get probability
But looking at the first and last sentences , there we realise that for the last case we cannot have 2 brown and 1 yellow as it will become brun.

So
4c2*5c1/9c3 + 4c3/9c3

is the fist jaune case
and the second is
(4c1* 5c1 *2c1)/9c3

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Manager
Joined: 01 Feb 2010
Posts: 234

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27 Mar 2010, 03:10
1
utin wrote:

Kurt, a painter, has 9 jars of paint:
4 are yellow
2 are red
rest are brown
Kurt will combine 3 jars of paint into a new container to make a new color, which he will name accordingly 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

My approach is
As in both cases of Jaune atleast 1 jar of yellow is required hence
p(Jaune) = 1-p(no yellow)
p(no yellow) = 5C3/9C3 = 5/42
p(Jaune) = 1-5/42 = 37/42
Manager
Joined: 26 Mar 2010
Posts: 94

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27 Mar 2010, 03:55
bangalorian2000 wrote:
utin wrote:

Kurt, a painter, has 9 jars of paint:
4 are yellow
2 are red
rest are brown
Kurt will combine 3 jars of paint into a new container to make a new color, which he will name accordingly 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

My approach is
As in both cases of Jaune atleast 1 jar of yellow is required hence
p(Jaune) = 1-p(no yellow)
p(no yellow) = 5C3/9C3 = 5/42
p(Jaune) = 1-5/42 = 37/42

Manager
Joined: 24 Mar 2010
Posts: 69

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28 Mar 2010, 10:18
Oops Read the question wrong. To correct it.To get jaune, we have to evaluate the two latter sentences, and add them to get probability

So
4c2*5c1/9c3 + 4c3/9c3

is the fist jaune case
and the second is
(4c1* 5c2)/9c3

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Non-Human User
Joined: 09 Sep 2013
Posts: 8847

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07 Oct 2017, 06:04
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