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)

$\frac{55}{84}$

Answer 2

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
$= \,^{3}P_1 P(E_1 )P(E_1)P(\bar{E}_1) + \,^{3}P_1P(E_2)P(E_2)P(\bar{E}_2) + \,^3P_1P(E_3)P(E_3)P(\bar{E}_3)$
$= 3\times\frac{2}{9} \times \frac{1}{8} \times \frac{7}{7}+ 3 \times \frac{3}{9}\times \frac{2}{8} \times \frac{6}{7} +3\times \frac{4}{9} \times \frac{3}{8} \times \frac{5}{7}$
$=3\left[\frac{14+36+60}{9\times 8\times 7}\right] =\frac{3\times 110}{9\times 8\times 7}=\frac{55}{84}$