Answer to Question #219719 in Statistics and Probability for Bless

Question

A bag contains two red, three green and four black balls of identical size except for colour. Three balls are drawn at random without replacement.

a) Show that the probability that two balls have the same colour and the third is different is 55 84 .

b) What is the probability that at least one is red.

Answer 1

$$2+3+4=9$$

a)

$$P(2R, 1G, 0B)=\dfrac{\dbinom{2}{2}\dbinom{3}{1}\dbinom{4}{0}}{\dbinom{9}{3}}$$

$$=\dfrac{1(3)(1)}{84}=\dfrac{3}{84}$$

$$P(2R, 0G, 1B)=\dfrac{\dbinom{2}{2}\dbinom{3}{0}\dbinom{4}{1}}{\dbinom{9}{3}}$$

$$=\dfrac{1(1)(4)}{84}=\dfrac{4}{84}$$

$$P(1R, 2G, 0B)=\dfrac{\dbinom{2}{1}\dbinom{3}{2}\dbinom{4}{0}}{\dbinom{9}{3}}$$

$$=\dfrac{2(3)(1)}{84}=\dfrac{6}{84}$$

$$P(0R, 2G, 1B)=\dfrac{\dbinom{2}{0}\dbinom{3}{2}\dbinom{4}{1}}{\dbinom{9}{3}}$$

$$=\dfrac{1(3)(4)}{84}=\dfrac{12}{84}$$

$$P(1R, 0G, 2B)=\dfrac{\dbinom{2}{1}\dbinom{3}{0}\dbinom{4}{2}}{\dbinom{9}{3}}$$

$$=\dfrac{2(1)(6)}{84}=\dfrac{12}{84}$$

$$P(0R, 1G, 2B)=\dfrac{\dbinom{2}{0}\dbinom{3}{1}\dbinom{4}{2}}{\dbinom{9}{3}}$$

$$=\dfrac{1(3)(6)}{84}=\dfrac{18}{84}$$

$$P(2\ same\ \&\ 1\ different)=\dfrac{3}{84}+\dfrac{4}{84}+\dfrac{6}{84}+\dfrac{12}{84}$$

$$+\dfrac{12}{84}+\dfrac{18}{84}=\dfrac{55}{84}$$

b)

$$P(0R, 2G, 1B)=\dfrac{12}{84}$$

$$P(0R, 1G, 2B)=\dfrac{18}{84}$$

$$P(0R, 3G, 0B)=\dfrac{\dbinom{2}{0}\dbinom{3}{3}\dbinom{4}{0}}{\dbinom{9}{3}}$$

$$=\dfrac{1(1)(1)}{84}=\dfrac{1}{84}$$

$$P(0R, 0G, 3B)=\dfrac{\dbinom{2}{0}\dbinom{3}{0}\dbinom{4}{3}}{\dbinom{9}{3}}$$

$$=\dfrac{1(1)(4)}{84}=\dfrac{4}{84}$$

$$P(at\ least\ 1R)=1-\dfrac{12}{84}-\dfrac{18}{84}-\dfrac{1}{84}-\dfrac{4}{84}=\dfrac{49}{84}$$