To find the probability of drawing two red balls without replacement, we use the combination method:
- Total balls: 4 red + 6 blue = 10 balls.
- Calculate the number of ways to choose 2 red balls: \[ \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6. \]
- Calculate the total ways to choose 2 balls from 10: \[ \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45. \]
- Compute the probability: \[ P = \frac{\binom{4}{2}}{\binom{10}{2}} = \frac{6}{45} = \frac{2}{15}. \]
- Alternative method (sequential probability): \[ P(\text{first red}) = \frac{4}{10} = \frac{2}{5}, \] \[ P(\text{second red} \mid \text{first red}) = \frac{3}{9} = \frac{1}{3}, \] \[ P = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}. \]
- Match with options: \( \frac{2}{15} \approx 0.133 \).
Thus, the correct answer is: \[ \boxed{\frac{2}{15}} \]