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# probability

Author Message
TAGS:
Manager
Status: GMAT Preperation
Joined: 04 Feb 2010
Posts: 105
Concentration: Social Entrepreneurship, Social Entrepreneurship
GPA: 3
WE: Consulting (Insurance)
Followers: 1

Kudos [?]: 16 [0], given: 15

probability [#permalink]  16 Oct 2010, 04:28
A jar contains only red, yellow, and orange marbles. If there are 3 red, 5 yellow, and 4 orange
marbles, and 3 marbles are chosen from the jar at random without replacing any of them, what
is the probability that 2 yellow, 1 red, and no orange marbles will be chosen?
A. 1/60
B. 1/45
C. 2/45
D. 3/22
E. 5/22
Manager
Joined: 04 Jun 2010
Posts: 113
Concentration: General Management, Technology
Schools: Chicago (Booth) - Class of 2013
GMAT 1: 670 Q47 V35
GMAT 2: 730 Q49 V41
Followers: 11

Kudos [?]: 98 [1] , given: 43

Re: probability [#permalink]  16 Oct 2010, 05:22
1
KUDOS
vanidhar wrote:
A jar contains only red, yellow, and orange marbles. If there are 3 red, 5 yellow, and 4 orange
marbles, and 3 marbles are chosen from the jar at random without replacing any of them, what
is the probability that 2 yellow, 1 red, and no orange marbles will be chosen?
A. 1/60
B. 1/45
C. 2/45
D. 3/22
E. 5/22

In total you have 3 different scenarios: (Y-Yellow , R-Red)
YYR, RYY,YRY after you will calculate the probability for each of these scenarios you'll have the answer.
YYR =\frac{5}{12}*\frac{4}{11}*\frac{3}{10} = \frac{5*4*3}{12*11*10}
RYY =\frac{3}{12}*\frac{5}{11}*\frac{4}{10} = \frac{3*5*4}{12*11*10}
YRY = \frac{5}{12}*\frac{3}{11}*\frac{4}{10} = \frac{5*3*4}{12*11*10}

I intentionally don't calculate the fractions through because it is easier to reduce the fraction this way.
Sum them up => \frac{3*(5*4*3)}{12*11*10} = \frac{3}{22} => answer D
_________________

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Re: probability   [#permalink] 16 Oct 2010, 05:22
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