Question

A bag contains $6$ red, $4$ white, and $8$ blue balls. If three bals are drawn at random, then the probability that $2$ are white and $1$is red is

• A. $\frac{5}{204}$
• B. $\frac{7}{102}$
• C. $\frac{3}{68}$
• D. $\frac{1}{13}$

Answer 1

The correct option is C

$\frac{3}{68}$

Explanation for correct option:

Calculate the Probability :

Total number of balls $=6+4+8=18$

Out of $18$ balls, three balls are drawn at random.

Thus, the total cases $={}^{18}C_3=816$

Let $A$ be the event that the drawn balls contain two white and one red balls. then,

$n\left(A\right)={}^{4}C_2\times {}^{6}C_1$

$=6\times 6=36$

Therefore, the required probability $=\frac{36}{816}=\frac{3}{68}$ $\left[\because \mathrm{Probability}=\frac{\mathrm{Favourable}\mathrm{cases}}{\mathrm{Total}\mathrm{cases}}\right]$

Hence the correct option is (C).