Probability of Selecting Red and Green Balls

Total balls $= 15 + 10 = 25$. Number of ways to choose 3 balls $= \binom{25}{3} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 2300$. Favorable ways to select 2 red and 1 green ball $= \binom{15}{2} \times \binom{10}{1} = 105 \times 10 = 1050$. Probability $= \frac{1050}{2300} = \frac{21}{46}$.