# A box contains $6$ red and $4$ white balls.If $3$ balls are drawn at random, the probability of getting $2$ white balls without replacement, is

$\begin{array}{1 1}(1)\frac{1}{20}&(2)\frac{18}{125}\\(3)\frac{4}{25}&(4)\frac{3}{10}\end{array}$

Total number of balls $=10$
Number of ways of choosing 3 balls from 10 balls $=10C_3$
Number of ways of choosing 2 white balls from 4 white balls $=4C_2$
Number of ways of choosing 1 red balls from 6 red balls $=6C_1$
Probability of getting 2 white balls and 1 red ball $= \large\frac{6C_1 \times 4C_2}{10C_3}$
$\qquad= \large\frac{6 \times \Large\frac{4.3}{1.2}}{\Large\frac{10.9.8}{1.2.3}}$
$\qquad= \large\frac{6 \times 6}{10 \times 12}$
$\qquad= \large\frac{3}{10}$
Hence 4 is the correct answer.