Marbles

Total number of ways in which 3 balls can be randomly selected out of 9 = 9C3 = 84. n(T) = 84.

One blue ball can be selected from 3 blue balls in 3C1 = 3 ways. n(B) = 3. Similarly, n(R) = 3 and n(Y) = 3.

P = (n(B) x n(R) x n(Y)) / n(T) = 27/84 = 9/28.

Probability for selecting 3 marbles with 3 different color from total of 9 marbles

= 3c1x3c1x3c1/ 9c3 = 9/28

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.

Hey Mitch
I approached the problem in a slightly different manner. Don't know whether I am correct.

1. Ways of selecting the first marble = 3 * (3/9) = 1 { as there are 3 marbles of one color, and there are 3 different colors}

2. Ways of selecting the first marble = 2 * (3/8) = 3/4 { as there are 3 marbles of one color, and 2 remaining colors to choose from}

3. Ways of selecting the third marble = 1 * (3/7) = 3/7 { as there are 3 marbles of one color, and 1 remaining color to choose from}

4. Total no. of ways = 1 * 3/4 * 3/7 = 9/28

Thanks
Suruchi

GMATGuruNY wrote:

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.