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

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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?

Re: A jar contains only red, yellow, and orange marbles. If there are 3 re [#permalink]

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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]

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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]

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29 Nov 2014, 20:29

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For me, combinations work better. Ways for selecting 2 yellow marbles out of 5: 5C2 Ways for selecting 1 red marble out of 3: 3C1 Total no. of way for selecting 3 out of 12: 12C3 Probability: 5C2*3C1/12C3=3/22

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04 Jun 2016, 08:06

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