Answer to Question #208069 in Statistics and Probability for Ishanka Amarasingh

Question

A bag contains 6 blue balls, 5 green balls and 4 red balls. Three are selected at random without replacement. Find the probability that

(a) they are all blue

(b)two are blue and one is green

(c) there is one of each colour

Answer 1

$$6+5+4=15$$

(a)

$$P(BBB)=\dfrac{6}{15}(\dfrac{6-1}{15-1})(\dfrac{6-2}{15-2})=\dfrac{4}{91}$$

Or

$$P(BBB)=\dfrac{\dbinom{6}{3}\dbinom{5}{0}\dbinom{4}{0}}{\dbinom{15}{3}}=$$

$$=\dfrac{20(1)(1)}{455}=\dfrac{4}{91}$$

(b)

$$P(2B\ and\ G)=\dfrac{\dbinom{6}{2}\dbinom{5}{1}\dbinom{4}{0}}{\dbinom{15}{3}}=$$

$$=\dfrac{15(5)(1)}{455}=\dfrac{15}{91}$$

(c)

$$P(B\ and\ G\ and\ R)=\dfrac{\dbinom{6}{1}\dbinom{5}{1}\dbinom{4}{1}}{\dbinom{15}{3}}=$$

$$=\dfrac{6(5)(4)}{455}=\dfrac{24}{91}$$