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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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Difficulty:

25% (low)

Question Stats:

78% (02:17) correct
21% (02:02) wrong based on 147 sessions

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]
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

2

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

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???