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

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16 Nov 2007, 07:46

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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?

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 ?

Re: Probability - Marbles of each color [#permalink]

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17 Nov 2007, 01:15

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?

Re: Probability - Marbles of each color [#permalink]

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28 Sep 2009, 04:16

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?

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]

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23 Dec 2011, 02:36

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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

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]

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16 Jul 2012, 23:56

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 .

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]

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17 Jul 2012, 03:08

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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!).

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]

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29 Mar 2014, 17:19

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?

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]

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30 Mar 2014, 11:42

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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??? _________________

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]

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30 Mar 2014, 13:15

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???

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]

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20 Apr 2014, 12:47

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!).

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]

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14 Oct 2015, 17:09

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Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]

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23 Jun 2016, 21:08

shahideh wrote:

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

I'm trying to understand the basics, please excuse me if you find the question too silly: Why don't we multiple by 3! in combinatorics method? So first it's choosing i.e. 3C1 * 3C1* 3C1 Then why aren't the chosen balls arranged amongst themselves in 3! ways?

gmatclubot

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3
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23 Jun 2016, 21:08

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