Question

A bag contains $3$ red, $4$ white and $5$ blue balls. If two balls are drawn at random, then the probability that they are of different colours, is

• A. $\frac{47}{66}$
• B. $\frac{23}{33}$
• C. $\frac{47}{132}$
• D. $\frac{47}{33}$
• E. $\frac{70}{33}$

Answer 1

Answer: (A)

$\frac{47}{66}$

Total number of bassl $=3+4+5=12$

Favourable number of cases are total minus the balls drawn from same colour

${{}^{12}C}_2-({{}^{3}C}_2+{{}^{4}C}_2+{{}^{5}C}_2)$

$=66-(3+6+10)=66-19=47$

Total number of cases $={{}^{12}C}_2=66$

Hence the required probability

$=\frac{Favourablecases}{Totalnumberofcases}=\frac{47}{66}$