A box holds 4 red and 6 white balls. Two balls are drawn one after the other WITHOUT replacement. What is the probability that both are red?

Explanation

1. 01

   First ball red: $\frac{4}{10}$.

2. 02

   With one red ball gone, 3 red remain among 9: $\frac{3}{9}$.

3. 03

   P(both red) $= \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$.