1. **Problem Statement:** A bag contains 3 white, 4 black, and 5 blue balls (total 12 balls). Four balls are drawn. Find the probability that the drawn balls are 2 white, 1 black, and 1 blue. 2. **Total number of ways to draw 4 balls from 12:** $$n(S) = \binom{12}{4} = 495$$ This is the total sample space size. 3. **Number of favorable outcomes (event A):** We want exactly 2 white, 1 black, and 1 blue ball. Number of ways to choose 2 white balls from 3: $$\binom{3}{2} = 3$$ Number of ways to choose 1 black ball from 4: $$\binom{4}{1} = 4$$ Number of ways to choose 1 blue ball from 5: $$\binom{5}{1} = 5$$ Total favorable outcomes: $$n(A) = \binom{3}{2} \times \binom{4}{1} \times \binom{5}{1} = 3 \times 4 \times 5 = 60$$ 4. **Calculate the probability:** $$P(A) = \frac{n(A)}{n(S)} = \frac{60}{495}$$ Simplify the fraction: $$\frac{60}{495} = \frac{\cancel{15} \times 4}{\cancel{15} \times 33} = \frac{4}{33}$$ 5. **Final answer:** The probability of drawing 2 white, 1 black, and 1 blue ball is: $$\boxed{\frac{4}{33}}$$ **Explanation:** - We use combinations because the order of drawing does not matter. - The total ways to draw 4 balls is the combination of 12 balls taken 4 at a time. - The favorable ways multiply the combinations of each color chosen. - Probability is favorable outcomes divided by total outcomes. - Simplify the fraction to get the final probability. This step-by-step approach clarifies the original solution and corrects the calculation of favorable outcomes and probability.