A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted?
(A) 2/21
(B) 3/25
(C) 1/6
(D) 9/28
(E) 11/24
P(exactly n times) = P(one way) * total possible ways.
Let R = red, B = blue, Y = yellow.
P(one way):
P(1st marble is R) = 3/9. (9 marbles, 3 of them R).
P(2nd marble is B) = 3/8. (8 marbles left, 3 of them B).
P(3rd marble is Y) = 3/7. (7 marbles left, 3 of them Y).
Since we want all of these events to happen together, we multiply the fractions:
P(RBY) = 3/9 * 3/8 * 3/7 = 3/56.
Total possible ways:
Any arrangement of RBY will yield exactly 1 R, 1 B and 1 Y.
Thus, the result above must be multiplied by the number of ways to arrange the 3 elements RBY.
Number ways to arrange RBY = 3! = 6.
Thus, P(1 of each color) = 3/56 * 6 = 9/28.
The correct answer is
D.
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