Question

Three balls are drawn from a bag containing $4$ red, $3$ black and $5$ white balls. The probability of drawing $2$ balls of same colour and one of different colour is

• A. $\frac{29}{44}$
• B. $\frac{119}{144}$
• C. $\frac{19}{24}$
• D. $\frac{25}{49}$

Answer 1

The correct option is A $\frac{29}{44}$

Total number of ways of selecting $3$ balls out of $12$ balls is ${}^{12}{C}_3=220$
Favourable cases of drawing two balls of same colour and one of different colour is :
$\left(i\right)$ $2$ balls of red colour and remaining $1$ of different colour $={}^4{C}_2\times {}^8{C}_1$
$\left(ii\right)$ $2$ balls of black colour and remaining $1$ of different colour $={}^3{C}_2\times {}^9{C}_1$
$\left(iii\right)$ $2$ balls of white colour and remaining $1$ of different colour $={}^5{C}_2\times {}^7{C}_1$
$\therefore$ Total number of favourable cases $={}^4{C}_2\times {}^8{C}_1+{}^3{C}_2\times {}^9{C}_1+{}^5{C}_2\times {}^7{C}_1=145$
Required probability $=\frac{145}{220}=\frac{29}{44}$