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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
A. 323/609
B. 323/1827
C. 103/203
D. 337/1827
E. 103/609
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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
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.
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
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.
Your approach is correct
Just some typos mistakes. Sorry about that. 
