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hard probability problem [#permalink]
26 Nov 2009, 12:45

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

83% (05:59) correct
17% (38:14) wrong based on 13 sessions

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

Re: hard probability problem [#permalink]
31 Mar 2010, 08:04

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

gmatclubot

Re: hard probability problem
[#permalink]
31 Mar 2010, 08:04

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