Question

A bag contains $3$ white and $3$ red balls, pairs of balls are drawn without replacement until the bag is empty. The probability that each pair consist of one white and one red ball is-

• A. $\frac{2}{5}$
• B. $\frac{4}{10}$
• C. $\frac{5}{10}$
• D. $\frac{6}{10}$

Answer 1

The correct option is A $\left(\frac{2}{5}\right)$

Given number of white balls$=3$ and number of red balls$=3$
Total number of ways to selecting $2$ balls out of total $6$ is ${}^2{C}_6$.
In first draw, probability $=\frac{{}^3{C}_1\times {}^3{C}_1}{{}^6{C}_2}=\frac{9}{15}=\frac{3}{5}$
Where, ${}^3{C}_1$ selecting $1$ ball out of three balls of each colour.
The remaining balls left in each colour would be $2$.
So, total balls$=2+2=4$.
In second draw, probability $=\frac{{}^2{C}_1\times {}^2{C}_1}{{}^4{C}_2}=\frac{4}{6}=\frac{2}{3}$

Remaining balls left in each colour would be $1$.
Total balls$=1+1=2$
In third draw, probability $=\frac{{}^1{C}_1\times {}^1{C}_1}{{}^2{C}_2}=1$
Therefore, Total probability $=\frac{3}{5}\times \frac{2}{3}\times 1=\frac{2}{5}$.