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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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81% (02:13) correct
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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]
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?
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?
Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]
23 Dec 2011, 01:36
1
This post received KUDOS
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]
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 .
Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]
17 Jul 2012, 02:08
6
This post received KUDOS
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]
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?
Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]
30 Mar 2014, 10:42
1
This post received KUDOS
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]
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???
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!).
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]
14 Oct 2015, 16:09
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