Question

A bag contains 4 red and 4 blue balls. Four balls are drawn one by one from the bag, then find the probability that the drawn balls are in alternate colour.

• A. $\frac{6}{35}$
• B. $\frac{6}{31}$
  
• C. $\frac{8}{35}$
  
• D. $\frac{8}{31}$
  

Answer 1

The correct option is A $\frac{6}{35}$

$E_1$ : Event that first drawn ball is red, second is blue and so on.
$E_2$ : Event that first drawn ball is blue, second is red and so on.
$\therefore P\left(E_1\right)=\frac{4}{8}\times \frac{4}{7}\times \frac{3}{6}\times \frac{3}{5}$ and $P\left(E_2\right)=\frac{4}{8}\times \frac{4}{7}\times \frac{3}{6}\times \frac{3}{5}$
$P\left(E\right)=P\left(E_1\right)+P\left(E_2\right)=2\times \frac{4}{8}\times \frac{4}{7}\times \frac{3}{6}\times \frac{3}{5}=\frac{6}{35}$