9 Marbles 3 Red 3 Blue 3 Yellow Probability of Picking 2 Yellow

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A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]   Updated on: 17 Jul 2012, 01:43

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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

Originally posted by bmwhype2 on 16 Nov 2007, 06:46.
Last edited by Bunuel on 17 Jul 2012, 01:43, edited 1 time in total.

Edited the question and added the OA.

Math Expert

Joined: 02 Sep 2009

Posts: 86491

A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]   17 Jul 2012, 02:08

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:

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

Answer: D.

Hope it's clear.
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Posts: 675

[#permalink]   16 Nov 2007, 14:58

9/9*6/8*3/7 = 9/28

the answer is (D)

Intern

Joined: 26 Sep 2007

Posts: 39

[#permalink]   16 Nov 2007, 15:50

Probability of getting each color marble = (3C1*3C1*3C1)/9C3
= (3*3*3)/84
= 27/84 = 9/28
So answer is D

Senior Manager

Joined: 09 Aug 2006

Posts: 422

[#permalink]   16 Nov 2007, 23:19

KillerSquirrel wrote:

9/9*6/8*3/7 = 9/28

the answer is (D)

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!

Director

Joined: 08 Jun 2005

Posts: 675

[#permalink]   17 Nov 2007, 00:05

GK_Gmat wrote:

KillerSquirrel wrote:

9/9*6/8*3/7 = 9/28

the answer is (D)

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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Joined: 29 Aug 2007

Posts: 1339

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

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Posts: 1557

Location: New York City

[#permalink]   17 Nov 2007, 12:23

sevenplus wrote:

Probability of getting each color marble = (3C1*3C1*3C1)/9C3
= (3*3*3)/84
= 27/84 = 9/28
So answer is D

yep. most intuitive approach.

desired/total

OA is D

Director

Joined: 01 May 2007

Posts: 542

[#permalink]   17 Nov 2007, 14:50

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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Joined: 27 Oct 2008

Posts: 111

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.

6/84 is not the answer.

please help me find the mistake. thanks.

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Joined: 25 May 2011

Posts: 66

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]   23 Dec 2011, 01:36

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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Posts: 2

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 .

Intern

Joined: 29 Mar 2014

Posts: 9

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?

Math Expert

Joined: 02 Sep 2009

Posts: 86491

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]   30 Mar 2014, 10:42

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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Joined: 15 Aug 2013

Posts: 197

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:

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

Answer: D.

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.

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Joined: 20 Feb 2016

Posts: 6

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]   23 Jun 2016, 20:08

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?

Math Expert

Joined: 02 Aug 2009

Posts: 10471

Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]   23 Jun 2016, 21:33

salonipatil wrote:

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?

hi,
We are not doing so, because it will not effect the answer..
if you consider ORDER important, even the TOTAL ways will be multiplied by 3!..
ans = \(p=\frac{C^3_1*C^3_1*C^3_1*3!}{C^9_3*3!}=\frac{9}{28}\)
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Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]   30 Dec 2020, 20:30

russ9 wrote:

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:

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

Answer: D.

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.

Hey GMATBusters

can you please help me the query stated above

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

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)?
_________________

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Hero1kF

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

30 Dec 2020, 20:30

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