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# M15#18

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M15#18 [#permalink]  19 Nov 2008, 18:24
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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) $$\frac{11}{24}$$

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

We can assume that the marbles are extracted from the basket one by one. After the first marble is drawn, 6 of the 8 marbles left in the basket have color which is different from the color of the first marble. Therefore, the probability that the color of the second marble will be different from the color of the first is 6/8. After two different-colored marbles are extracted from the basket, we are left with 7 marbles 3 of which have color which is different from the color of either of the first two marbles. Therefore, the probability that the color of the third marble will be different from the color of the first two is 3/7. Thus, the answer to the question is $$\frac{6}{8} * \frac{3}{7} = \frac{9}{28}$$ .

I see how to answer this question using probabilities as explained in the answer but I was hoping someone would be willing to explain how to answer this question using a combination formula.
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Re: M15#18 [#permalink]  19 Nov 2008, 22:44
snowy2009 wrote:
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?

(C) 2008 GMAT Club - m15#18

* $$\frac{2}{21}$$
* $$\frac{3}{25}$$
* $$\frac{1}{6}$$
* $$\frac{9}{28}$$
* $$\frac{11}{24}$$

We can assume that the marbles are extracted from the basket one by one. After the first marble is drawn, 6 of the 8 marbles left in the basket have color which is different from the color of the first marble. Therefore, the probability that the color of the second marble will be different from the color of the first is 6/8. After two different-colored marbles are extracted from the basket, we are left with 7 marbles 3 of which have color which is different from the color of either of the first two marbles. Therefore, the probability that the color of the third marble will be different from the color of the first two is 3/7. Thus, the answer to the question is $$\frac{6}{8} * \frac{3}{7} = \frac{9}{28}$$ .

I see how to answer this question using probabilities as explained in the answer but I was hoping someone would be willing to explain how to answer this question using a combination formula.

= (9c1/9c1) (6c1/8c1) (3c1/7c1)
= 1 (6/8) (3/7)
= 9/28
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Re: M15#18 [#permalink]  19 Nov 2008, 23:26
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(3C1 * 3C1 * 3C1) / 9C3
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Re: M15#18 [#permalink]  20 Nov 2008, 10:08
scthakur wrote:
(3C1 * 3C1 * 3C1) / 9C3

Why are the combinations in the numerator multiplied with each other?
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Re: M15#18 [#permalink]  17 Dec 2008, 07:27
snowy2009 wrote:
scthakur wrote:
(3C1 * 3C1 * 3C1) / 9C3

Why are the combinations in the numerator multiplied with each other?

This is because we have to find the probability of 3 events combined: 1 blue AND 1 red AND 1 yellow.

Hope this helps.
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Re: M15#18 [#permalink]  02 Mar 2010, 06:17
snowy2009 wrote:
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) $$\frac{11}{24}$$

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

We can assume that the marbles are extracted from the basket one by one. After the first marble is drawn, 6 of the 8 marbles left in the basket have color which is different from the color of the first marble. Therefore, the probability that the color of the second marble will be different from the color of the first is 6/8. After two different-colored marbles are extracted from the basket, we are left with 7 marbles 3 of which have color which is different from the color of either of the first two marbles. Therefore, the probability that the color of the third marble will be different from the color of the first two is 3/7. Thus, the answer to the question is $$\frac{6}{8} * \frac{3}{7} = \frac{9}{28}$$ .

I see how to answer this question using probabilities as explained in the answer but I was hoping someone would be willing to explain how to answer this question using a combination formula.

p = (3c1*3c1*3c1)/9c3 = 9/28 hence D.
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Re: M15#18 [#permalink]  02 Mar 2010, 06:58
Sorry, how is (3c1*3c1*3c1)/9c3 = 9/28. I keep getting 9/84? 3c1 cubed is 9. Why is 9c3 28? I am sure it has to do with order or something or am I just missing something?

Edit: Crap because 3 cubed is not 9. 27/84 == 9/28. I really hope I don't make stupid mistakes like this on the 17th when I take the test. (It is taking some will power not to edit my question above.)
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Re: M15#18 [#permalink]  02 Mar 2010, 07:25
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Out of 3 blue marbles, 3 red marbles, and 3 yellow marbles, probability of selecting one marble of each color is as follows:

B R Y
(3/9) * (3/8) * (3/7)

B, R, and Y can be arranged in factorial (3) = 3 * 2 * 1 ways.

Thus, resulting probability is:

(3/9) * (3/8) * (3/7) * (3 * 2 * 1) = 9/28

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Re: M15#18 [#permalink]  02 Mar 2010, 08:18
3/9 * 3/8 * 3/7 * 3! = 9/28
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Re: M15#18 [#permalink]  02 Mar 2010, 21:41
I'm still a bit confused. I got the denominator correct (9C3), which enumerates all the possible outcomes. But I only see 6 (3!) desired outcomes...not 27:

BRY
BYR
YRB
YBR
RBY
RYB

What am I missing?

EDIT: I think this is only taking into account only one of three of each color...i.e. B1 Y1 R1, etc.

Let me know if I am right!
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Re: M15#18 [#permalink]  13 Mar 2010, 08:15
(3C1 * 3C1 * 3C1)/9C3

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Re: M15#18 [#permalink]  07 Mar 2011, 20:36
dzyubam wrote:
snowy2009 wrote:
scthakur wrote:
(3C1 * 3C1 * 3C1) / 9C3

Why are the combinations in the numerator multiplied with each other?

This is because we have to find the probability of 3 events combined: 1 blue AND 1 red AND 1 yellow.

Hope this helps.

Please explain why we have written 9C3 in the denominator.
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Re: M15#18 [#permalink]  08 Mar 2011, 00:30
Denominator contains the number of total possible outcomes. In this case, the number of possible outcomes equals the number of ways 3 marbles can be drawn from the total of 9 marbles available (9C3).
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Re: M15#18 [#permalink]  08 Mar 2011, 22:44
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The faster way to calculate the probability is to follow events manually:
1. The first marble will be of any one color. Probability (P1) is 100% or 1;
2. The second marble should be of other two colors (any of 6 out of 8 remaining marbles). Probability (P2) is 6/8 or 3/4;
3. The third marble should be of the last color (any of 3 out of 7 remaining marbles). Probability (P3) is 3/7;

P=P1*P2*P3=1*3/4*3/7=9/28

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Re: M15#18 [#permalink]  10 Mar 2012, 00:35
snowy2009 wrote:
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) $$\frac{11}{24}$$

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

We can assume that the marbles are extracted from the basket one by one. After the first marble is drawn, 6 of the 8 marbles left in the basket have color which is different from the color of the first marble. Therefore, the probability that the color of the second marble will be different from the color of the first is 6/8. After two different-colored marbles are extracted from the basket, we are left with 7 marbles 3 of which have color which is different from the color of either of the first two marbles. Therefore, the probability that the color of the third marble will be different from the color of the first two is 3/7. Thus, the answer to the question is $$\frac{6}{8} * \frac{3}{7} = \frac{9}{28}$$ .

I see how to answer this question using probabilities as explained in the answer but I was hoping someone would be willing to explain how to answer this question using a combination formula.

Got D!!!

First Draw - (9 marbles/ 9 marbles) = 1
Second Draw - Since one color is selected & we dont want that same one again... (6 marbles/ 8 remaining marbles) = 3/4
Third Draw - Since two colors are selected & we dont want the same ones again... (3 marbles/ 7 remaining marbles) = 3/7

Multiply all three to get the probability...
1 * (3/4) * (3/7) = 9/28
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Re: M15#18 [#permalink]  11 Mar 2013, 01:41
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For first ball...we can pick any one out of nine..(lets say first is blue)

For second ball of different color...we have choose one ball out of 8 balls
and color of ball should not blue..it may be of red or yellow color...
so probability is 6(3 red and 3 yellow)/8...(lets say we pick a red one here)

For third ball of different color...we have choose one ball out of 7 balls
and color of ball should yellow..
so probability is 3 (yellow)/7

So probability that a marble of each color is among the extracted
$$= \frac{6}{8}*\frac{3}{7}$$
$$= \frac{9}{28}$$
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Re: M15#18 [#permalink]  11 Mar 2013, 01:49
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Expert's post
snowy2009 wrote:
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) $$\frac{11}{24}$$

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

We can assume that the marbles are extracted from the basket one by one. After the first marble is drawn, 6 of the 8 marbles left in the basket have color which is different from the color of the first marble. Therefore, the probability that the color of the second marble will be different from the color of the first is 6/8. After two different-colored marbles are extracted from the basket, we are left with 7 marbles 3 of which have color which is different from the color of either of the first two marbles. Therefore, the probability that the color of the third marble will be different from the color of the first two is 3/7. Thus, the answer to the question is $$\frac{6}{8} * \frac{3}{7} = \frac{9}{28}$$ .

I see how to answer this question using probabilities as explained in the answer but I was hoping someone would be willing to explain how to answer this question using a combination formula.

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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Re: M15#18 [#permalink]  12 Mar 2013, 22:36
snowy2009 wrote:
scthakur wrote:
(3C1 * 3C1 * 3C1) / 9C3

Why are the combinations in the numerator multiplied with each other?

So as a basic with Combinations, if we have an event 1 (3C1 - 1 blue ball extracted) followed by event 2 (3C1 - 1 red ball extracted) and event 3 (3C1 - 1 yellow ball extracted), we have to multiply all events.

Instead if this was a case of 'OR', we would have added the events.

Hope that helped.

Can someone tell me what is the difficulty level of this question?
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Re: M15#18 [#permalink]  13 Mar 2013, 01:35
1
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Expert's post
ranjitaarao wrote:
snowy2009 wrote:
scthakur wrote:
(3C1 * 3C1 * 3C1) / 9C3

Why are the combinations in the numerator multiplied with each other?

So as a basic with Combinations, if we have an event 1 (3C1 - 1 blue ball extracted) followed by event 2 (3C1 - 1 red ball extracted) and event 3 (3C1 - 1 yellow ball extracted), we have to multiply all events.

Instead if this was a case of 'OR', we would have added the events.

Hope that helped.

Can someone tell me what is the difficulty level of this question?

The difficulty level is ~650.
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Re: M15#18 [#permalink]  24 Feb 2014, 10:03
nevermind... i realize i was forgetting the parenthetical part of the selection. in the denominator... its follish lapses like these that affect my scores and keep me in the 600s
Re: M15#18   [#permalink] 24 Feb 2014, 10:03

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