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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]

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New post Updated on: 27 Sep 2014, 16:50
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Question Stats:

79% (02:16) correct 21% (02:29) wrong based on 164 sessions

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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

Originally posted by vanidhar on 16 Oct 2010, 05:28.
Last edited by Bunuel on 27 Sep 2014, 16:50, 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]

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New post 16 Oct 2010, 06:22
5
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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New post 28 Sep 2014, 01:44
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5C2*3C1=10*3=30
12C3=220

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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New post 29 Nov 2014, 21:29
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1
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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Re: A jar contains only red, yellow, and orange marbles. If there are 3 re  [#permalink]

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New post 10 Jan 2015, 04:32
1
I started by finding the 2 probabilities, without calculation, like this:
P(YYR)
P(YRY)
P(RYY)

I calculated the first one and ended in 1/22. I looked at the answer choices at this point and saw answer D: 3/22.

This helped me realise that for the 3 possible orderings the probabbility is the same. So, it should be (1/22)*(3), which indeed is 3/22.
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Re: A jar contains only red, yellow, and orange marbles. If there are 3 re  [#permalink]

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New post 02 Apr 2018, 10:05
Hi All,

I'm a fan of any type of Quant or Verbal shortcut that fits a given situation, BUT if you ever find yourself in a situation in which you can't "spot the shortcut", then you have to just get into the question and do the necessary work as efficiently as you can (re: "brute force").

In this question, we have 3 red, 5 yellow and 4 orange marbles. We're asked for the probability of grabbing 2 yellow and 1 red (in some order)?

There are 3 outcomes that satisfy this question:
YYR
YRY
RYY

Let's calculate each:
YYR = (5/12)(4/11)(3/10) = 60/1320

YRY = (5/12)(2/11)(4/10) = 60/1320

RYY = (2/12)(5/11)(4/10) = 60/1320

Total = 180/1320 = 18/132 = 6/44 = 3/22

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Re: A jar contains only red, yellow, and orange marbles. If there are 3 re &nbs [#permalink] 02 Apr 2018, 10:05
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