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Probability Problem 2 Please Help solve this one!!!

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Probability Problem 2 Please Help solve this one!!! [#permalink]

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New post 27 Mar 2010, 00:19
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Hi I am unable to solve this question please help!!!

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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Re: Probability Problem 2 Please Help solve this one!!! [#permalink]

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New post 27 Mar 2010, 03:11
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

Adding both cases we get

29/42 which is the answer
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Re: Probability Problem 2 Please Help solve this one!!! [#permalink]

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New post 27 Mar 2010, 04:10
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utin wrote:
Hi I am unable to solve this question please help!!!

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

Kudos [?]: 62 [0], given: 2

Manager
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Joined: 27 Mar 2010
Posts: 117

Kudos [?]: 15 [0], given: 17

Re: Probability Problem 2 Please Help solve this one!!! [#permalink]

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New post 27 Mar 2010, 04:55
bangalorian2000 wrote:
utin wrote:
Hi I am unable to solve this question please help!!!

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



Many thanks!!! bangalorian2000 for your help.your answer 37/42 is correct...Thanks a lot!!!

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Re: Probability Problem 2 Please Help solve this one!!! [#permalink]

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New post 28 Mar 2010, 11:18
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

Adding both cases we get

37/42 which is the answer
_________________

- Stay Hungry, stay Foolish -

Kudos [?]: 80 [0], given: 134

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Re: Probability Problem 2 Please Help solve this one!!! [#permalink]

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Re: Probability Problem 2 Please Help solve this one!!!   [#permalink] 07 Oct 2017, 07:04
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