Question

5. 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

$\binom{6+5+4}{3}=\binom{15}{3}=\frac{15!}{3!12!}$ is the general number of possibilities to get 3 balls out of 15 ones

a)$\binom{6}{3}$ is the number of possibilities to get 3 blue balls out of 6 ones, so probability that all balls are blue is $\frac{\binom{6}{3}}{\binom{15}{3}}=\frac{\frac{6!}{3!3!}}{\frac{15!}{3!12!}}=\frac{6!}{3!}\frac{12!}{15!}=\frac{4\cdot 5\cdot 6}{13\cdot 14\cdot 15}=\frac{4}{13\cdot 7}=\frac{4}{91}$

b)$\binom{6}{2}$ and $\binom{5}{1}$ are the number of possibilities to get 2 blue balls out of 6 ones and the number of possibilities to get 1 green ball out of 5 ones, so $\binom{6}{2}\binom{5}{1}$ is number of possibiliies to get 2 blue balls and 1 green ball. Probability that two balls are blue and one ball is green is $\frac{\binom{6}{2}\binom{5}{1}}{\binom{15}{3}}=\frac{\frac{6!}{2!4!}\frac{5!}{1!4!}}{\frac{15!}{3!12!}}=\frac{15\cdot 5}{\frac{13\cdot 14\cdot 15}{1\cdot 2\cdot 3}}=\frac{15\cdot 5}{13\cdot 7\cdot 5}=\frac{15}{91}$

c)$\binom{6}{1}$, $\binom{5}{1}$ and $\binom{4}{1}$ are the number of possibilities to get 1 blue ball out of 6 ones,

the number of possibilities to get 1 green ball out of 5 ones and the number of possibilities to get 1 red ball out of 5 ones, so $\binom{6}{1}\binom{5}{1}\binom{4}{1}$ is number of possibilities to get 1 blue, 1 green and 1 red balls. Probability that there are balls of each colour is $\frac{\binom{6}{1}\binom{5}{1}\binom{4}{1}}{\binom{15}{3}}=\frac{\frac{6!}{1!5!}\frac{5!}{1!4!}\frac{4!}{1!3!}}{\frac{15!}{3!12!}}=\frac{6\cdot 5\cdot 4}{\frac{15\cdot 14\cdot 13}{1\cdot 2\cdot 3}}=\frac{6\cdot 5\cdot 4}{13\cdot 7\cdot 5}=\frac{24}{91}$

Answer: a)$\frac{4}{91}$ , b)$\frac{15}{91}$ , c)$\frac{24}{91}$