Question

A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that:
(i) All the three balls are white.
(ii) All the three balls are red.
(iii) One ball is red and two balls are white.

Answer 1

Total number of balls = 8 + 5 = 13
Total number of events for drawing 3 balls = ${}^{13}{C}_3$
(i) Total number of events for getting white balls = ${}^5{C}_3$
P(all 3 balls white) = $\frac{\mathrm{favourable}\mathrm{outcomes}}{\mathrm{total}\mathrm{outcomes}}={\frac{{}^5{\displaystyle {\mathrm{C}}_3}}{{}^{13}{\displaystyle \mathrm{C}}}}_3$
$=\frac{5!}{3!\times 2!}\times \frac{3!\times 10!}{13!}$
$\Rightarrow$P(all 3 balls white)$=\frac{5}{143}$


(ii) Favourable number of events for getting red balls = ${}^8{C}_3$
P(all 3 balls red) = $\frac{\mathrm{favourable}\mathrm{outcomes}}{\mathrm{total}\mathrm{outcomes}}={\frac{{}^8{\displaystyle {\mathrm{C}}_3}}{{}^{13}{\displaystyle \mathrm{C}}}}_3$
$=\frac{8!}{3!\times 5!}\times \frac{3!\times 10!}{13!}$
$\Rightarrow$P(all 3 balls red)$=\frac{28}{143}$

(iii) Favourable number of events for getting 1 red ball = ${}^8{C}_1$
Favourable number of events for getting 2 white balls = ${}^5{C}_2$
P(1 red and 2 white balls) = $\frac{\mathrm{favourable}\mathrm{outcomes}}{\mathrm{total}\mathrm{outcomes}}=\frac{{}^8{\displaystyle {\mathrm{C}}_1}{\times }^5{C}_2}{{}^{13}{\displaystyle {\mathrm{C}}_3}}$
$=\frac{8!}{1!\times 7!}\times \frac{5!}{2!\times 3!}\times \frac{3!\times 10!}{13!}$
$\Rightarrow$P(all 3 balls red)$=\frac{40}{143}$