Question

A bag contains 5 white and 3 black balls, and 4 balls are successively drawn out and not replaced. What's the chance of getting different colours alternatively?

• A. $\frac{1}{6}$
• B. $\frac{1}{5}$
• C. $\frac{1}{4}$
• D. $\frac{1}{7}$

Answer 1

The correct option is D $\frac{1}{7}$

Total number of balls = 8. Let the first drawn ball be white, so required probability $=\frac{5}{8}\times \frac{3}{7}\times \frac{4}{6}\times \frac{2}{5}=\frac{1}{14}$.
But here we had started with a white ball. When we start with a black ball, the required probability
$=\frac{3}{8}\times \frac{5}{7}\times \frac{2}{6}\times \frac{4}{5}=\frac{1}{14}$.
Since these two cases are mutually exclusive.
Total probability = $\frac{1}{14}+\frac{1}{14}=\frac{2}{14}=\frac{1}{7}$.