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# M15-18

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Math Expert
Joined: 02 Sep 2009
Posts: 47157

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16 Sep 2014, 00:55
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74% (01:25) correct 26% (02:09) wrong based on 94 sessions

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A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted?

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

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Joined: 02 Sep 2009
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16 Sep 2014, 00:55
1
Official Solution:

A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted?

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

$$P=\frac{C^1_3*C^1_3*C^1_3}{C^3_9}=\frac{9}{28}$$. Numerator represents ways to select 1 blue marble out of 3, 1 red marble out of 3 and 1 yellow marble out of 3. Denominator represents total ways to select 3 marbles out of 9.

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Joined: 21 May 2013
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28 Dec 2014, 12:27
Could you explain please why this method does not work :

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 = 9/56
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Joined: 15 Apr 2013
Posts: 189
Location: India
Concentration: General Management, Marketing
GMAT Date: 11-23-2015
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23 Oct 2015, 23:16
2
Hello,

Old thread but might be useful to someone else.

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!

Please hit kudos if you like this explanation
Intern
Joined: 13 Jun 2016
Posts: 19

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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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Joined: 02 Sep 2009
Posts: 47157

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20 Jun 2016, 07:46
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.

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:

Theory on probability problems

Data Sufficiency Questions on Probability
Problem Solving Questions on Probability

Tough Probability Questions
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Joined: 02 Aug 2009
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20 Jun 2016, 07:49
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

Hi
it will be multiplication...
$$3C1 = \frac{3!}{(3-1)!} = \frac{3!}{2!}= \frac{3*2*1}{2*1} = 3$$....
so numerator will become = 3*3*3..
denominator = $$9C3 = \frac{9!}{(9-3)!3!}= \frac{9*8*7*6!}{6!*3*2*1} = \frac{9*8*7}{3*2} = 3*4*7$$...
so fraction $$\frac{3C1*3C1*3C1}{9C3} = \frac{3*3*3}{3*4*7}= \frac{9}{28}$$
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Intern
Joined: 02 Sep 2016
Posts: 10

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27 Sep 2016, 14:20
how to know if we need to multiply or add in the numerater?
Intern
Joined: 13 Dec 2016
Posts: 43

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22 Jan 2017, 01:17
manchitkapoor wrote:
how to know if we need to multiply or add in the numerater?

I follow the below logic to decide when to multiply or add when solving Permutation / Combination problems :

If the problem states that we need to find the probability of Say A event AND B event - Use Multiplication

If the problem states that we need to find the probability of Say A event OR B event - Use Addition

Hope this helps.
Intern
Joined: 05 Sep 2017
Posts: 4

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18 Sep 2017, 02:32
why do we use combinations and not permutations here? order does matter, I think
Intern
Joined: 12 Feb 2018
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21 May 2018, 03:00
Bunuel wrote:
Official Solution:

A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted?

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

$$P=\frac{C^1_3*C^1_3*C^1_3}{C^3_9}=\frac{9}{28}$$. Numerator represents ways to select 1 blue marble out of 3, 1 red marble out of 3 and 1 yellow marble out of 3. Denominator represents total ways to select 3 marbles out of 9.

hi bunuel,
could you please explain the solution using probability approach?
according to probability approach,
choosing blue marble = 3/9
choosing red marble = 3/6 ( as the second marble choosen shouldnt be a blue one,the total marbles left will be 6)
choosing yellow marble = 3/3
total probability = 3/9 * 3/6 * 3/3 = 1/6 (answer)
Re: M15-18 &nbs [#permalink] 21 May 2018, 03:00
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# M15-18

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