Question

1. A box containing 6 black balls, 4 green balls, and 3 red balls. 3 balls are drawn in succession, wherein each ball being replaced in the box before the next draw is made. What is the probability that

(a) all 3 are the same color?
(b) each color is represented?

Answer 1

a). The probability that 3 black balls are drown is: $p_1=(\frac{6}{13})^3$.

The probabiltiy that green balls are drawn is: $p_2=(\frac{4}{13})^3$ .

For red balls we have: $p_3=(\frac{3}{13})^3$ . We used the multiplication rule to get these probabilities (see https://stattrek.com/probability/probability-rules.aspx) The probability that all three balls have the same color is: $p=p_1+p_2+p_3=\frac{27+216+64}{2197}\approx0.1397.$

b). The probability that all balls have different colors is: $3!\frac{3\cdot4\cdot6}{13^3}\approx0.1966$ .We used the factor 3!, since drawn balls can be situated in different order.