Question

a bag 10 white, 15 red and 8 green balls. A single draw of 3 balls is made. what is the probability that a white, a red and green balls are drawn

Answer 1

$$10+15+8=33$$

$$P(1W\&1B\&1G)=\dfrac{\dbinom{10}{1}\dbinom{15}{1}\dbinom{8}{1}}{\dbinom{33}{3}}=\dfrac{10(15)(8)}{5456}$$

$$=\dfrac{75}{341}\approx0.22$$