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M15-18

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Bunuel
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M15-18 [#permalink] New post   16 Sep 2014, 00:55
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72% (01:47) correct 28% (02:23) wrong based on 82 sessions
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A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are drawn at random from the basket, what is the probability that one marble of each color is among those drawn?

A. \(\frac{2}{21}\)
B. \(\frac{3}{25}\)
C. \(\frac{1}{6}\)
D. \(\frac{9}{28}\)
E. \(\frac{11}{24}\)
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D
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Bunuel
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Re M15-18 [#permalink] New post   16 Sep 2014, 00:55
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A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are drawn at random from the basket, what is the probability that one marble of each color is among those drawn?

A. \(\frac{2}{21}\)
B. \(\frac{3}{25}\)
C. \(\frac{1}{6}\)
D. \(\frac{9}{28}\)
E. \(\frac{11}{24}\)


The probability \(P\) can be calculated as \(P=\frac{C^1_3*C^1_3*C^1_3}{C^3_9}=\frac{9}{28}\). The numerator represents the number of ways to select one blue marble out of three, one red marble out of three, and one yellow marble out of three. The denominator represents the total number of ways to select three marbles out of nine.


Answer: D
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Re: M15-18 [#permalink] New post   23 Oct 2015, 23:16
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Hello,

Old thread but might be useful to someone else.

filipTGIM your approach is partial.

You are assuming a defined sequence only i.e. Blue first, red second and yellow last. But don't you think there could be other possibilities also i.e. Red-Blue-Yellow...so on...

Basically three item can be arranged in 3*2*1= 6 ways and hence in the final step you need to multiply with 06.

Probability of extracting 1 blue marble : 3/9
Probability of extracting the next one as red : 3/8
And the next one as yellow : 3/7

Which equals 3/9 * 3/8 * 3/7 = 3/56 (not 9/56)

Then we need to multiply with 06:

3/56*6 = 18/56 or 09/28

Hope it helps!
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xnthic
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Re: M15-18 [#permalink] New post   20 Jun 2016, 07:37
combinatorics is my weakness but correct my if im wrong, the numerator of the answer should be addition not multiplication that equals 9?

you have 1C3∗1C3∗1C3 / 3C9 = 9/28
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Re: M15-18 [#permalink] New post   20 Jun 2016, 07:46
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lydennis8 wrote:
combinatorics is my weakness but correct my if im wrong, the numerator of the answer should be addition not multiplication that equals 9?

you have 1C3∗1C3∗1C3 / 3C9 = 9/28


No, it's multiplication, because of Principle of Multiplication. The result is 27/84, which when reduced gives 9/28.

Principle of Multiplication
If an operation can be performed in ‘m’ ways and when it has been performed in any of these ways, a second operation that can be performed in ‘n’ ways then these two operations can be performed one after the other in ‘m*n’ ways.

Principle of Addition
If an operation can be performed in ‘m’ different ways and another operation in ‘n’ different ways then either of these two operations can be performed in ‘m+n’ ways (provided only one has to be done).

Check below post for more:
Probability Made Easy!

Theory on probability problems

Data Sufficiency Questions on Probability
Problem Solving Questions on Probability

Tough Probability Questions
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Re: M15-18 [#permalink] New post   24 Nov 2020, 21:27
I think this is a high-quality question and I agree with explanation. The formula expression is incorrect
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Re: M15-18 [#permalink] New post   27 Sep 2023, 13:43
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M15-18 [#permalink]
27 Sep 2023, 13:43
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