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# A basket contains 3 blue, 3 red and 3 yellow marbles. If 3

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CEO
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A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]  16 Nov 2007, 06:46
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A basket contains 3 blue, 3 red 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. 2/21
B. 3/25
C. 1/6
D. 9/28
E. 11/24
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 Jul 2012, 01:43, edited 1 time in total.
Edited the question and added the OA.
VP
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9/9*6/8*3/7 = 9/28

Manager
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Probability of getting each color marble = (3C1*3C1*3C1)/9C3
= (3*3*3)/84
= 27/84 = 9/28
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KillerSquirrel wrote:
9/9*6/8*3/7 = 9/28

KS, can you please explain how you get 3/7?

I'm assuming you did the following:
The 1st chosen can be any color and it doesn't really matter so let's say the first chosen is Blue.

The probability that the 2nd chosen is not blue = 1-2/8 = 6/8

Now what?? How do I proceed? Thanks!
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GK_Gmat wrote:
KillerSquirrel wrote:
9/9*6/8*3/7 = 9/28

KS, can you please explain how you get 3/7?

I'm assuming you did the following:
The 1st chosen can be any color and it doesn't really matter so let's say the first chosen is Blue.

The probability that the 2nd chosen is not blue = 1-2/8 = 6/8

Now what?? How do I proceed? Thanks!

You are right about the first marble , you can choose any marble out of nine = 9/9 = 1 (nine out of nine).

For the second marble you are also correct - since you removed one marble from the basket you now have eight left but you can't choose the color already been choosen so 6/8 (six out of eight - three for one color and three for the other color).

For the third marble the logic is the same - you can choose one of the three marbles from the color not already choosen and you have seven mables to choose from meaning 3/7 (three out seven).

All of this is true only when the marbles taken out are not returned to the basket - if they were the probability then was ?? - can you solve this ?

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Re: Probability - Marbles of each color [#permalink]  17 Nov 2007, 00:15
bmwhype2 wrote:
A basket contains 3 blue, 3 red 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?

2/21
3/25
1/6
9/28
11/24

alternatively:

= (3c1 x 3c1 x 3c1)/9c3
= 9/28
CEO
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sevenplus wrote:
Probability of getting each color marble = (3C1*3C1*3C1)/9C3
= (3*3*3)/84
= 27/84 = 9/28

yep. most intuitive approach.

desired/total

OA is D
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For me it was easier to just write it out.

I need 3 different colors so, we'll say I wanted a red, blue, yellow.

So chance of picking a red = 3/9

On 2nd pick chance of picking a blue = 3/8

On 3rd pick chance of picking a yellow = 3/7.

Then I listed out the different 3 ball combos I could have.

BRY
BYR
RBY
RYB
YRB
YBR

So 6 different combos. So the prob of pick r,b,y is 3/9 * 3/8 * 3/7 = 3/56

Multiply that by 6 and you have 9/28 prob of picking three different colored balls.
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Re: Probability - Marbles of each color [#permalink]  28 Sep 2009, 03:16
A basket contains 3 blue, 3 red 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?

2/21
3/25
1/6
9/28
11/24

Soln:
= 3C1 * 3C1 * 3C1/9C3
= 27/(9 * 8 * 7/6)
= 27/84
= 9/28
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Re: Probability - Marbles of each color [#permalink]  24 Aug 2011, 13:56
Hey guys,

I understand the solution, but i need someone to explain me why the following is not working:

total groups of 3 - 9C3

total options of 3 different color:
abc
acb
bac
bca
cab
cba

so - 6 options to get 3 different colors.

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Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]  23 Dec 2011, 01:36
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We can choose 3 balls from 9 balls. Total possibilities: $$C^9_3$$
We should choose 1 ball from each color. Favorite possibilities: $$C^3_1*C^3_1*C^3_1$$

$$p=\frac{C^3_1*C^3_1*C^3_1}{C^9_3}=\frac{9}{28}$$
D
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Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]  16 Jul 2012, 22:56
Is ' picking of 3 marbles at random' same as 'picking 3 marbles one by on without replacement' ? Although looking at the solution these two scenarios appear to be similar. However, when 3 marbles are picked up together (in one shot) then should the problem not be treated differently. Please clarify .
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Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]  17 Jul 2012, 02:08
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Expert's post
malikshilpa wrote:
Is ' picking of 3 marbles at random' same as 'picking 3 marbles one by on without replacement' ? Although looking at the solution these two scenarios appear to be similar. However, when 3 marbles are picked up together (in one shot) then should the problem not be treated differently. Please clarify .

Mathematically the probability of picking 3 marbles simultaneously, or picking them one at a time (without replacement) is the same.

A basket contains 3 blue, 3 red 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. 2/21
B. 3/25
C. 1/6
D. 9/28
E. 11/24

We need to find the probability of BRY (a marble of each color).

Probability approach:

$$P(BRY)=\frac{3}{9}*\frac{3}{8}*\frac{3}{7}*3!=\frac{9}{28}$$, we are multiplying by 3! since BRY scenario can occur in several different ways: BRY, RYB, RBY (# of permutations of 3 distinct letters BRY is 3!).

Combination approach:

$$P(BRY)=\frac{C^1_3*C^1_3*C^1_3}{C^3_9}=\frac{9}{28}$$.

Hope it's clear.
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Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]  14 Aug 2013, 01:11
Expert's post
Bumping for review and further discussion.
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Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]  29 Mar 2014, 16:19
Just wanted to clarify here, as I feel like I'm looking at this wrong
n should be 3 (because this is the total set)
r should be 1 (this is the number of objects being extracted)

so shouldn't the combination formula be C 3 over 1? Instead of C 1 over 3?
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Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]  30 Mar 2014, 10:42
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Expert's post
TheBookie wrote:
Just wanted to clarify here, as I feel like I'm looking at this wrong
n should be 3 (because this is the total set)
r should be 1 (this is the number of objects being extracted)

so shouldn't the combination formula be C 3 over 1? Instead of C 1 over 3?

$$C^1_3$$, $$C^3_1$$, 3C1, mean the same thing choosing 1 out of 3, just different ways to write. Can it be choosing 3 out of 1???
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Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]  30 Mar 2014, 12:15
Bunuel wrote:
TheBookie wrote:
Just wanted to clarify here, as I feel like I'm looking at this wrong
n should be 3 (because this is the total set)
r should be 1 (this is the number of objects being extracted)

so shouldn't the combination formula be C 3 over 1? Instead of C 1 over 3?

$$C^1_3$$, $$C^3_1$$, 3C1, mean the same thing choosing 1 out of 3, just different ways to write. Can it be choosing 3 out of 1???

Agreed, but for the sake of consistency, I see the combination formula written as C n over r
http://www.mathwords.com/c/combination_formula.htm

Thanks
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Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]  20 Apr 2014, 11:47
Bunuel wrote:
malikshilpa wrote:
Is ' picking of 3 marbles at random' same as 'picking 3 marbles one by on without replacement' ? Although looking at the solution these two scenarios appear to be similar. However, when 3 marbles are picked up together (in one shot) then should the problem not be treated differently. Please clarify .

Mathematically the probability of picking 3 marbles simultaneously, or picking them one at a time (without replacement) is the same.

A basket contains 3 blue, 3 red 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. 2/21
B. 3/25
C. 1/6
D. 9/28
E. 11/24

We need to find the probability of BRY (a marble of each color).

Probability approach:

$$P(BRY)=\frac{3}{9}*\frac{3}{8}*\frac{3}{7}*3!=\frac{9}{28}$$, we are multiplying by 3! since BRY scenario can occur in several different ways: BRY, RYB, RBY (# of permutations of 3 distinct letters BRY is 3!).

Combination approach:

$$P(BRY)=\frac{C^1_3*C^1_3*C^1_3}{C^3_9}=\frac{9}{28}$$.

Hope it's clear.

Hi Bunuel,

I have a question with the methodology here -- I can follow the work you're doing but i'm having a hard time understanding why. Can you please clarify the reasoning?

a) In your statement below, can you please elaborate WHY we multiply by 3!? -- [$$P=\frac{3}{9}*\frac{3}{8}*\frac{3}{7}*3!=\frac{9}{28}$$

If I reference the GMAT club book, I don't see it referencing the multiplication by a factorial(Page 109 of the book).

b) What is the difference between the method you used vs. doing the following? (9/9)(6/8)(3/7)

c) lastly, regarding your combinatorial approach, isn't that technically a reverse approach(pg 109 of the book)? Sorry if i'm being trivial here but i'm having a hard time connecting the dots hence the trivial questions. How can you use the 1C3 approach because don't you have 9 to begin with? Sorry - i'm going nuts here.
Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3   [#permalink] 20 Apr 2014, 11:47
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