Answer to Question #145610 in Statistics and Probability for SHIVA n

Question

A box contains 22 balls of which 7 are red, 9 are white, and 6 are black. Four balls are drawn
at random from the box without replacement. Find the probability that
(a) three red balls and one white ball are drawn,
(b) the four balls are all red,
(c) the balls are all of the same colour,
(d) there is at least one ball of each colour.

Answer 1

(a) $P(4R, 0W,0B)=\dfrac{\dbinom{7}{4}\dbinom{9}{0}\dbinom{6}{0}}{\dbinom{22}{4}}$

$\dbinom{22}{4}=7315$

$\dbinom{7}{3}=\dfrac{7!}{3!(7-3)!}=\dfrac{7(6)(5)}{1(2)(3)}=35$

$\dbinom{9}{1}=9$

$\dbinom{6}{0}=1$

$$P(3R, 1W,0B)=\dfrac{35(9)}{7315}=\dfrac{9}{209}$$

(b) $P(3R, 1W,0B)=\dfrac{\dbinom{7}{3}\dbinom{9}{1}\dbinom{6}{0}}{\dbinom{22}{4}}$

$\dbinom{22}{4}=\dfrac{22!}{4!(22-4)!}=\dfrac{22(21)(20)(19)}{1(2)(3)(4)}=7315$

$\dbinom{7}{4}=\dfrac{7!}{4!(7-4)!}=\dfrac{7(6)(5)}{1(2)(3)}=35$

$\dbinom{9}{0}=1$

$\dbinom{6}{0}=1$

$$P(4R, 0W,0B)=\dfrac{35}{7315}=\dfrac{1}{209}$$

(c) $P(0R, 4W,0B)=\dfrac{\dbinom{7}{0}\dbinom{9}{4}\dbinom{6}{0}}{\dbinom{22}{4}}$

$\dbinom{22}{4}=7315$

$\dbinom{7}{0}=1$

$\dbinom{9}{4}=\dfrac{9!}{4!(9-4)!}=\dfrac{9(8)(7)(6)}{1(2)(3)(4)}=126$

$\dbinom{6}{0}=1$

$$P(0R, 4W,0B)=\dfrac{126}{7315}=\dfrac{18}{1045}$$

$P(0R, 0W,4B)=\dfrac{\dbinom{7}{0}\dbinom{9}{0}\dbinom{6}{4}}{\dbinom{22}{4}}$

$\dbinom{22}{4}=7315$

$\dbinom{7}{0}=1$

$\dbinom{9}{0}=1$

$\dbinom{6}{4}=\dfrac{6!}{4!(6-4)!}=\dfrac{6(5)}{1(2)}=15$

$$P(0R, 0W,4B)=\dfrac{15}{7315}=\dfrac{3}{1463}$$

$$P(all\ the \ same\ colour)=\dfrac{1}{209}+\dfrac{18}{1045}+\dfrac{3}{1463}=$$

$$=\dfrac{16}{665}$$

(d) $P(at\ least \ one\ ball\ at\ each\ colour)=$

$=P(2R, 1W, 1B)+P(1R, 2W, 1B)+P(1R, 1W, 2B)=$

$\dbinom{7}{2}=21, \dbinom{7}{1}=7$

$\dbinom{9}{2}=36, \dbinom{9}{1}=9$

$\dbinom{6}{2}=15, \dbinom{6}{1}=6$

$$P(at\ least \ one\ ball\ at\ each\ colour)=$$

$$=\dfrac{21(9)(6)}{7315}+\dfrac{7(36)(6)}{7315}+\dfrac{7(9)(15)}{7315}=$$

$$=\dfrac{27}{55}$$