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Math Expert V
Joined: 02 Sep 2009
Posts: 58434

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Difficulty:   35% (medium)

Question Stats: 75% (01:27) correct 25% (02:02) wrong based on 110 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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Math Expert V
Joined: 02 Sep 2009
Posts: 58434

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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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Intern  Joined: 21 May 2013
Posts: 7

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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
Verbal Forum Moderator Joined: 15 Apr 2013
Posts: 178
Location: India
Concentration: General Management, Marketing
GMAT Date: 11-23-2015
GPA: 3.6
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6
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!

Please hit kudos if you like this explanation
Intern  Joined: 13 Jun 2016
Posts: 18

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 58434

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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.

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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Math Expert V
Joined: 02 Aug 2009
Posts: 7991

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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

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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how to know if we need to multiply or add in the numerater?
Intern  S
Joined: 13 Dec 2016
Posts: 41

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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  B
Joined: 05 Sep 2017
Posts: 4

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why do we use combinations and not permutations here? order does matter, I think
Intern  B
Joined: 12 Feb 2018
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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)
Manager  S
Joined: 08 Jan 2013
Posts: 103

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This can be solved either by combinatorics or probability.

With combinatorics, ordering is not important as we are dealing with selection only, so the answer will be 3C1(for blue)*3C1(for red)*3C1(for yellow)/9C3(total 3 from 9) => 9/28.

With probability chaining or trees, ordering is important. So, it can be calculated as 3/9(for first color)*3/8(for second color)*3/7(for third color)*6(i.e. 3! for arrangement of 3 numbers) => 9/28.
Intern  B
Joined: 18 Aug 2018
Posts: 4

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WillGetIt wrote:
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!

Please hit kudos if you like this explanation

Why are you multiplying by 6 after getting 3/56?

Also, should'nt this question have said "without replacement" so we'd know for sure?
Intern  B
Joined: 29 Jan 2018
Posts: 5

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WillGetIt wrote:
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!

Please hit kudos if you like this explanation

I followed this approach but in a different way: given that order does not matter:
1. We can extract any color on the first attempt, so prob=9/9
2. We can extract only 2 colors on the second attempt (the one that were not extract previously), so prob =6/8
3. We need to extract the missing color from the remain marbles, so prob =3/7

The product of probs is (9/9)*(6/8)*(3/7)= 18/56 = 9/28 Re: M15-18   [#permalink] 15 Nov 2018, 22:36
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