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# hard probability problem

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Manager
Joined: 18 Jul 2009
Posts: 53
Followers: 3

Kudos [?]: 109 [0], given: 7

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26 Nov 2009, 13:45
00:00

Difficulty:

(N/A)

Question Stats:

73% (04:33) correct 27% (15:41) wrong based on 31 sessions

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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

I am getting

jaune x+jaune y

4/9*5/8*4/7 +( 4/9*3/8*5/7 + 4/9*3/8*2/7)

= 41/126

I know if i use other method i.e (1-probability of no yellow) then answer is

1-5/9*4/8*3/7 = 1-5/42=37/42

Can someone tell me where I am going wrong in 1st method
CEO
Joined: 17 Nov 2007
Posts: 3588
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 566

Kudos [?]: 3797 [1] , given: 360

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26 Nov 2009, 14:22
1
KUDOS
Expert's post
3 * 4/9*5/8*4/7 +( 3 * 4/9*3/8*5/7 + 4/9*3/8*2/7) = (3*4*5*4 + 3*4*3*5 + 4*3*2)/(9*8*7) = (4*5+3*5+2)/(3*2*7) = 37/42

In other words, you forgot that besides y?? there are two other ways: ?y? and ??y. The same in the case of yy?.
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Manager
Joined: 18 Jul 2009
Posts: 53
Followers: 3

Kudos [?]: 109 [0], given: 7

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26 Nov 2009, 14:37
Thanks walker.

+1 for you
Intern
Joined: 31 Mar 2010
Posts: 17
WE 1: Top-Tier MC
Followers: 0

Kudos [?]: 3 [0], given: 12

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31 Mar 2010, 09:04
Any yellow paint in the mix results in a jaune paint once combined.

So, probability of jaune paint is (1 - probability of all combinations that do not contain any yellow)

# of combinations = 9C3 = 84 {9 pots, choose 3}
# of combinations that do not contain any yellow = 5C3 = 10 {5 non-yellow pots, choose 3}
So, probability of combinations that DO NOT contain any yellow 10/84

So, probability of combinations that DO contain yellow = 1 - 10/84 = 74/84 = 37/42

Nick

Edit: Spelling and clarity
Re: hard probability problem   [#permalink] 31 Mar 2010, 09:04
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