Question

A bag contains $4$ red, $3$ black, and $2$ white balls. If $2$ balls are selected at random, the probability of selecting atleast one white ball is

• A. $\frac{7}{12}$
• B. $\frac{5}{12}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

The correct option is C $\frac{1}{3}$

Bag contains $4$ red, $3$ black and, $2$ white balls

Two balls are selected at random

The total no. of ways of doing that is ${10}_{C_2}=\frac{10!}{8!\times 2!}=\frac{10\times 9\times 8!}{8!\times 2!}=\frac{10\times 9}{2}=45$

Now in the selection we need to ensure that at least on white ball is selected.

Case $1$ : $1$ white ball $+$ $1$ ball of any other color

This can be done in $2_{C_1}\times 7_{C_1}=2\times 7=14$ ways

Case $2$ : $2$ white ball

This can be done in $2_{C_2}=1$ way

$\therefore$ probability of selecting atleast one white ball is $=\frac{14+1}{45}=\frac{15}{45}=\frac{1}{3}$

Answer : Option C.