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

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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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.

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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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.

Went for the 4th draw as yellow. Wasted my time ;(

Posted from my mobile device
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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probability of selecting exactly one yellow marble from the bowl after three successive marbles are withdrawn from the bowl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

Given: There are three blue marbles, three red marbles, and three yellow marbles in a bowl.

Asked: What is the probability of selecting exactly one yellow marble from the bowl after three successive marbles are withdrawn from the bowl?

3 blue (B) , 3 red(R) & 3 yellow(Y) marbles in a bowl

No of favourable ways = 3C1 * 6C2 = 3 * 15 = 45

Total number of ways = 9C3 = 9*8*7/3*2*1 = 3*4*7 = 84

Probability = 45/84 = 15/28

IMO A

Hi chetan2u,

Just curious to know how would one solve this question had the question specificially asked for yellow AFTER 3 draws. (as in yellow in the 4th draw).
Thanks!
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Let Y = yellow and N = not yellow.

P(Y) on the first pick \(= \frac{3}{9}\) (Of the 9 marbles, 3 are yellow)
P(N) on the second pick \(= \frac{6}{8}\) (Of the 8 remaining marbles, 6 are not yellow)
P(N) on the third pick \(= \frac{5}{7}\) (Of the 7 remaining marbles, 5 are not yellow)
To combine these probabilities, we multiply
\(\frac{3}{9} * \frac{6}{8} * \frac{5}{7}\)
Since a good outcome will be achieved if Y is selected on the first pick, the second pick, or the third pick -- for a total of 3 options -- we multiply by 3:
\(\frac{3}{9} * \frac{6}{8} * \frac{5}{7} * 3 = \frac{15}{28}\)


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I agree with you. This question doesn't have an official source and is poorly written.

question is not worded properly. TOo much ambiguity.

Hi is the question written weirdly or am i not understanding? It says after 3 successful draws. So in that draw all 3 can be yellow in case1, 2 can be yellow in case 2 then we take 1 more and in case 3- 1 can be yellow and then we have 2 more options left.