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M15-18
[#permalink]
16 Sep 2014, 00:55
16 Sep 2014, 00:55
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Difficulty:
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Question Stats:
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27%
(02:26)
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based on 81
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A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are drawn at random from the basket, what is the probability that one marble of each color is among those drawn?
A. \(\frac{2}{21}\)
B. \(\frac{3}{25}\)
C. \(\frac{1}{6}\)
D. \(\frac{9}{28}\)
E. \(\frac{11}{24}\)
A. \(\frac{2}{21}\)
B. \(\frac{3}{25}\)
C. \(\frac{1}{6}\)
D. \(\frac{9}{28}\)
E. \(\frac{11}{24}\)
Re M15-18
[#permalink]
16 Sep 2014, 00:55
16 Sep 2014, 00:55
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Official Solution:
A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are drawn at random from the basket, what is the probability that one marble of each color is among those drawn?
A. \(\frac{2}{21}\)
B. \(\frac{3}{25}\)
C. \(\frac{1}{6}\)
D. \(\frac{9}{28}\)
E. \(\frac{11}{24}\)
The probability \(P\) can be calculated as \(P=\frac{C^1_3*C^1_3*C^1_3}{C^3_9}=\frac{9}{28}\). The numerator represents the number of ways to select one blue marble out of three, one red marble out of three, and one yellow marble out of three. The denominator represents the total number of ways to select three marbles out of nine.
Answer: D
A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are drawn at random from the basket, what is the probability that one marble of each color is among those drawn?
A. \(\frac{2}{21}\)
B. \(\frac{3}{25}\)
C. \(\frac{1}{6}\)
D. \(\frac{9}{28}\)
E. \(\frac{11}{24}\)
The probability \(P\) can be calculated as \(P=\frac{C^1_3*C^1_3*C^1_3}{C^3_9}=\frac{9}{28}\). The numerator represents the number of ways to select one blue marble out of three, one red marble out of three, and one yellow marble out of three. The denominator represents the total number of ways to select three marbles out of nine.
Answer: D
Re: M15-18
[#permalink]
23 Oct 2015, 23:16
23 Oct 2015, 23:16
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Hello,
Old thread but might be useful to someone else.
filipTGIM your approach is partial.
You are assuming a defined sequence only i.e. Blue first, red second and yellow last. But don't you think there could be other possibilities also i.e. Red-Blue-Yellow...so on...
Basically three item can be arranged in 3*2*1= 6 ways and hence in the final step you need to multiply with 06.
Probability of extracting 1 blue marble : 3/9
Probability of extracting the next one as red : 3/8
And the next one as yellow : 3/7
Which equals 3/9 * 3/8 * 3/7 = 3/56 (not 9/56)
Then we need to multiply with 06:
3/56*6 = 18/56 or 09/28
Hope it helps!
Old thread but might be useful to someone else.
filipTGIM your approach is partial.
You are assuming a defined sequence only i.e. Blue first, red second and yellow last. But don't you think there could be other possibilities also i.e. Red-Blue-Yellow...so on...
Basically three item can be arranged in 3*2*1= 6 ways and hence in the final step you need to multiply with 06.
Probability of extracting 1 blue marble : 3/9
Probability of extracting the next one as red : 3/8
And the next one as yellow : 3/7
Which equals 3/9 * 3/8 * 3/7 = 3/56 (not 9/56)
Then we need to multiply with 06:
3/56*6 = 18/56 or 09/28
Hope it helps!
