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A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]
 16 Nov 2007, 07:46
Question Stats:
88% (02:09) correct
11% (01:21) wrong based on 3 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? A. 2/21 B. 3/25 C. 1/6 D. 9/28 E. 11/24
Last edited by Bunuel on 17 Jul 2012, 02:43, edited 1 time in total.
Edited the question and added the OA.
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VP
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9/9*6/8*3/7 = 9/28
the answer is (D)
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Manager
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Probability of getting each color marble = (3C1*3C1*3C1)/9C3
= (3*3*3)/84
= 27/84 = 9/28
So answer is D
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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!
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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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Re: Probability - Marbles of each color [#permalink]
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?
2/21
3/25
1/6
9/28
11/24
alternatively:
= (3c1 x 3c1 x 3c1)/9c3
= 9/28
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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
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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, 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?
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, 14:56
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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Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]
23 Dec 2011, 02:36
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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Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]
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 .
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Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 [#permalink]
17 Jul 2012, 03:08
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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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Re: A basket contains 3 blue, 3 red and 3 yellow marbles. If 3
[#permalink]
17 Jul 2012, 03:08
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