Bunuel wrote:

Official Solution:

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. \(\frac{2}{21}\)

B. \(\frac{3}{25}\)

C. \(\frac{1}{6}\)

D. \(\frac{9}{28}\)

E. \(\frac{11}{24}\)

\(P=\frac{C^1_3*C^1_3*C^1_3}{C^3_9}=\frac{9}{28}\). Numerator represents ways to select 1 blue marble out of 3, 1 red marble out of 3 and 1 yellow marble out of 3. Denominator represents total ways to select 3 marbles out of 9.

Answer: D

hi bunuel,

could you please explain the solution using probability approach?

according to probability approach,

choosing blue marble = 3/9

choosing red marble = 3/6 ( as the second marble choosen shouldnt be a blue one,the total marbles left will be 6)

choosing yellow marble = 3/3

total probability = 3/9 * 3/6 * 3/3 = 1/6 (answer)