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broall
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In box A, there are 10 red marbles, 10 green marbles and 10 yellow Updated on: 14 Dec 2016, 08:12
Originally posted by broall on 14 Dec 2016, 06:59.
Last edited by broall on 14 Dec 2016, 08:12, edited 1 time in total.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

75% (hard)

Question Stats:

50% (03:22) correct 50% (02:00) wrong based on 16 sessions

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In box A, there are 10 red marbles, 10 green marbles and 10 yellow marbles. Jack picks randomly 4 marble from that box. What is the probability that Jack picks 4 marbles that don't have 3 different colors?

A. 323/609
B. 323/1827
C. 103/203
D. 337/1827
E. 103/609



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C
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vitaliyGMAT
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In box A, there are 10 red marbles, 10 green marbles and 10 yellow 14 Dec 2016, 08:09
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nguyendinhtuong
In box A, there are 10 red marbles, 10 green marbles and 10 yellow marbles. Jack picks randomly 4 marble from that box. What is the probability that Jack picks 4 marbles that don't have 3 different colors?

A. 323/609
B. 323/1827
C. 103/203
D. 337/1827
E. 103/609
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First we’ll find number of ways to break the restriction, that is to pick 3 different colors.

It can be done in three cases (R-red, G-green, Y-yellow):

1R1G 2Y or 1R2G1Y or 2R1G1Y

By pigeon hole principle 2 marbles will always be of the same color.

Hence our sough after probability will be:

\(\frac{10C1*10C1*10C2*3}{30C4}\)

which after simplification will give us \(\frac{100}{203}\)

The probability that the restriction won’t be broken, that is we won’t get 3 different colors is:

\(1 – \frac{100}{203} = \frac{103}{203}\)

I'm getting C. Official answer is A. Am I missing something? Bunuel please advise.
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In box A, there are 10 red marbles, 10 green marbles and 10 yellow 14 Dec 2016, 08:14
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vitaliyGMAT
nguyendinhtuong
In box A, there are 10 red marbles, 10 green marbles and 10 yellow marbles. Jack picks randomly 4 marble from that box. What is the probability that Jack picks 4 marbles that don't have 3 different colors?

A. 323/609
B. 323/1827
C. 103/203
D. 337/1827
E. 103/609

First we’ll find number of ways to break the restriction, that is to pick 3 different colors.

It can be done in three cases (R-red, G-green, Y-yellow):

1R1G 2Y or 1R2G1Y or 2R1G1Y

By pigeon hole principle 2 marbles will always be of the same color.

Hence our sough after probability will be:

\(\frac{10C1*10C1*10C2*3}{30C4}\)

which after simplification will give us \(\frac{100}{203}\)

The probability that the restriction won’t be broken, that is we won’t get 3 different colors is:

\(1 – \frac{100}{203} = \frac{103}{203}\)

I'm getting C. Official answer is A. Am I missing something? Bunuel please advise.
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Your approach is correct :-D Just some typos mistakes. Sorry about that.
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vitaliyGMAT
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In box A, there are 10 red marbles, 10 green marbles and 10 yellow 14 Dec 2016, 08:20
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You are welcome. Many thanks for kudos!
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rikson
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In box A, there are 10 red marbles, 10 green marbles and 10 yellow 10 Jan 2017, 03:02
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I also got ans C,,, how I overcome it..please response,,,,



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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Bunuel
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