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A jar contains only red, yellow, and orange marbles. If there are 3 re

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A jar contains only red, yellow, and orange marbles. If there are 3 re [#permalink] New post 16 Oct 2010, 04:28
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 27 Sep 2014, 15:50, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: A jar contains only red, yellow, and orange marbles. If there are 3 re [#permalink] New post 16 Oct 2010, 05:22
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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: A jar contains only red, yellow, and orange marbles. If there are 3 re [#permalink] New post 27 Sep 2014, 15:39
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Re: A jar contains only red, yellow, and orange marbles. If there are 3 re [#permalink] New post 28 Sep 2014, 00:44
5C2*3C1=10*3=30
12C3=220

3/22

Answer: D
Re: A jar contains only red, yellow, and orange marbles. If there are 3 re   [#permalink] 28 Sep 2014, 00:44
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