Question

If one ball each is drawn from 3 boxes, the first containing 3 red, 2 yellow, and 1 blue, the second box contains 2 red, 2 yellow, and 2 blue, and the third box with 1 red, 4 yellow, and 3 blue. What is the probability that all 3 balls drawn are different colors?

Answer 1

$$P(3\ different)=P(RYB)+P(RBY)$$

$$+P(YRB)+P(YBR)+P(BYR)+P(BRY)$$

$$=\dfrac{3}{6}(\dfrac{2}{6})(\dfrac{3}{8})+\dfrac{3}{6}(\dfrac{2}{6})(\dfrac{4}{8})+\dfrac{2}{6}(\dfrac{2}{6})(\dfrac{3}{8})$$

$$+\dfrac{2}{6}(\dfrac{2}{6})(\dfrac{1}{8})+\dfrac{1}{6}(\dfrac{2}{6})(\dfrac{1}{8})+\dfrac{1}{6}(\dfrac{2}{6})(\dfrac{4}{8})$$

$$=\dfrac{17}{72}$$