Question

An urn contains $9$ balls, $2$ of which are white, $3$ blue and $4$ black. $3$ balls are drawn at random from the urn. The chance that $2$ balls will be of the same colour and the third of a different colour is

• A. $\frac{45}{84}$
• B. $\frac{55}{84}$
• C. $\frac{35}{84}$
• D. $\frac{25}{84}$

Answer 1

Answer: (B)

Let $E_1,E_2$ and $E_3$ be the events such that drawn balI is white, blue and black respectively.
$\therefore$ The required probability
$={}^3{P}_{1}P(E_1)P(E_1)P({\bar{E}}_1)+{}^3{P}_{1}P(E_2)P(E_2)P({\bar{E}}_2)+{}^3{P}_{1}P(E_3)P(E_3)P({\bar{E}}_3)$
$=3\times \frac{2}{9}\times \frac{1}{8}\times \frac{7}{7}+3\times \frac{3}{9}\times \frac{2}{8}\times \frac{6}{7}+3\times \frac{4}{9}\times \frac{3}{8}\times \frac{5}{7}$
$=3\left[\frac{14+36+60}{9\times 8\times 7}\right]=\frac{3\times 110}{9\times 8\times 7}=\frac{55}{84}$