Question

A bag contains $5$ white and $3$ black balls, and $4$ are successively drawn out and not replaced; what is the chance that they are alternately of different colours?

Answer 1

Probability of getting alternating colours when the firat ball is white $=\frac{5}{8}\times \frac{3}{7}\times \frac{4}{6}\times \frac{2}{5}=\frac{1}{14}$

Probability of getting alternating colours when the first ball is black$=\frac{3}{8}\times \frac{5}{7}\times \frac{2}{6}\times \frac{4}{5}=\frac{1}{14}$

$\Rightarrow$Probability of getting balls of alternating colour $=\frac{1}{14}+\frac{1}{14}=\frac{1}{7}$