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There are three blue marbles, three red marbles, and three yellow marb

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There are three blue marbles, three red marbles, and three yellow marb  [#permalink]

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New post 01 Nov 2019, 06:49
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Difficulty:

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Question Stats:

42% (03:04) correct 58% (02:50) wrong based on 52 sessions

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There are three blue marbles, three red marbles, and three yellow marbles in a bowl. What is the probability of selecting exactly one yellow marble from the bowl after three successive marbles are withdrawn from the bowl?

(A) 15/28
(B) 5/28
(C) 1/27
(D) 1/84
(E) 2/243

Are You Up For the Challenge: 700 Level Questions

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Re: There are three blue marbles, three red marbles, and three yellow marb  [#permalink]

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New post 01 Nov 2019, 08:23
Bunuel wrote:
There are three blue marbles, three red marbles, and three yellow marbles in a bowl. What is the probability of selecting exactly one yellow marble from the bowl after three successive marbles are withdrawn from the bowl?

(A) 15/28
(B) 5/28
(C) 1/27
(D) 1/84
(E) 2/243

Are You Up For the Challenge: 700 Level Questions


There are 3 B, 3R and 3Y....
We have to select only 1 Y in 3 selections..

Ways to select only 1 Y.... \(3C1*6C2=3*\frac{6*5}{2=}45\)....any one yellow from 3 yellows and 2 non-yellows from remaining 6
Total ways = \(9C3=\frac{9*8*7}{3*2}=3*4*7=84\)

Probability = \(\frac{45}{84}=\frac{15}{28}\)

A
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Re: There are three blue marbles, three red marbles, and three yellow marb  [#permalink]

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New post 01 Nov 2019, 08:43
Hello Chetan2u!

I have many doubts for this problem.

exactly one yellow marble from the bowl after three successive marbles
In our explanation, you consider the withdrawal of exactly one yellow within three successive trials. But question seems asking for withdrawal of exactly one yellow after three trials.

Even I approach the problem with different method, I am getting awkward answer.
First trial- no yellow- 6/9
Second trial- No yellow- 5/8
Third trial- one yellow- 1/7
I can write the above combination as YOO and this can be arranged in 3!/2!= 3 ways
so probability of getting exactly one yellow within three successive trials = 6/9*5/8*1/7*3= 5/28

Can you help me out to understand where I am wrong with this approach.



chetan2u wrote:
Bunuel wrote:
There are three blue marbles, three red marbles, and three yellow marbles in a bowl. What is the probability of selecting exactly one yellow marble from the bowl after three successive marbles are withdrawn from the bowl?

(A) 15/28
(B) 5/28
(C) 1/27
(D) 1/84
(E) 2/243

Are You Up For the Challenge: 700 Level Questions


There are 3 B, 3R and 3Y....
We have to select only 1 Y in 3 selections..

Ways to select only 1 Y.... \(3C1*6C2=3*\frac{6*5}{2=}45\)....any one yellow from 3 yellows and 2 non-yellows from remaining 6
Total ways = \(9C3=\frac{9*8*7}{3*2}=3*4*7=84\)

Probability = \(\frac{45}{84}=\frac{15}{28}\)

A
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Re: There are three blue marbles, three red marbles, and three yellow marb  [#permalink]

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New post 01 Nov 2019, 08:49
1
gvij2017 wrote:
Hello Chetan2u!

I have many doubts for this problem.

exactly one yellow marble from the bowl after three successive marbles
In our explanation, you consider the withdrawal of exactly one yellow within three successive trials. But question seems asking for withdrawal of exactly one yellow after three trials.

Even I approach the problem with different method, I am getting awkward answer.
First trial- no yellow- 6/9
Second trial- No yellow- 5/8
Third trial- one yellow- 1/7
I can write the above combination as YOO and this can be arranged in 3!/2!= 3 ways
so probability of getting exactly one yellow within three successive trials = 6/9*5/8*1/7*3= 5/28

Can you help me out to understand where I am wrong with this approach.



chetan2u wrote:
Bunuel wrote:
There are three blue marbles, three red marbles, and three yellow marbles in a bowl. What is the probability of selecting exactly one yellow marble from the bowl after three successive marbles are withdrawn from the bowl?

(A) 15/28
(B) 5/28
(C) 1/27
(D) 1/84
(E) 2/243

Are You Up For the Challenge: 700 Level Questions


There are 3 B, 3R and 3Y....
We have to select only 1 Y in 3 selections..

Ways to select only 1 Y.... \(3C1*6C2=3*\frac{6*5}{2=}45\)....any one yellow from 3 yellows and 2 non-yellows from remaining 6
Total ways = \(9C3=\frac{9*8*7}{3*2}=3*4*7=84\)

Probability = \(\frac{45}{84}=\frac{15}{28}\)

A



Hi,

It means within three successive draws, and yes someone can mistake the way you have mentioned..

Now, why you are going wrong..
Third trial - one yellow will be 3/7 as there are 3 yellows to choose from..

So answer will be (6/9)(5/8)(3/7)*3=15/28
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Re: There are three blue marbles, three red marbles, and three yellow marb  [#permalink]

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New post 01 Nov 2019, 08:58
Thanks for explanation. Actually I made mistake in third withdrawal. You are correct it is 3/7.
Moreover, Don't you think that after three successive draws is total different than within three successive draws?
Former gives sense of 4th draws.This is the main reason I did this problem wrong.
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Re: There are three blue marbles, three red marbles, and three yellow marb  [#permalink]

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New post 01 Nov 2019, 11:50
Went for the 4th draw as yellow. Wasted my time ;(

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Re: There are three blue marbles, three red marbles, and three yellow marb  [#permalink]

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New post 02 Nov 2019, 04:17
Bunuel wrote:
There are three blue marbles, three red marbles, and three yellow marbles in a bowl. What is the probability of selecting exactly one yellow marble from the bowl after three successive marbles are withdrawn from the bowl?

(A) 15/28
(B) 5/28
(C) 1/27
(D) 1/84
(E) 2/243

Are You Up For the Challenge: 700 Level Questions


total marbles = 9
probability of selecting exactly one yellow marble from the bowl after three successive marbles are withdrawn from the bowl
6/9*5/8*3/7 * 3c1 = 15/28
IMO A
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Re: There are three blue marbles, three red marbles, and three yellow marb   [#permalink] 02 Nov 2019, 04:17
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