1. **Problem:** A box contains 10 red, 20 blue, and 30 green marbles. 5 marbles are drawn. Find: (i) Probability all 5 are blue. (ii) Probability at least one is green. 2. **Formula and rules:** - Total marbles = $10 + 20 + 30 = 60$. - Number of ways to choose 5 marbles from 60: $\binom{60}{5}$. - Probability of an event = $\frac{\text{favorable outcomes}}{\text{total outcomes}}$. - For "at least one" problems, use complement rule: $P(\text{at least one green}) = 1 - P(\text{no green})$. 3. **Calculations:** (i) All 5 blue: - Number of blue marbles = 20. - Favorable ways = $\binom{20}{5}$. - Total ways = $\binom{60}{5}$. - Probability = $\frac{\binom{20}{5}}{\binom{60}{5}}$. Calculate values: $\binom{20}{5} = \frac{20!}{5!15!} = 15504$. $\binom{60}{5} = \frac{60!}{5!55!} = 5461512$. So, $$P(\text{all blue}) = \frac{15504}{5461512} \approx 0.00284.$$ (ii) At least one green: - Number of green marbles = 30. - No green means all 5 are chosen from red and blue only: $10 + 20 = 30$ marbles. - Favorable ways for no green = $\binom{30}{5}$. - Total ways = $\binom{60}{5}$. Calculate $\binom{30}{5} = \frac{30!}{5!25!} = 142506$. Probability no green: $$P(\text{no green}) = \frac{142506}{5461512} \approx 0.0261.$$ Therefore, $$P(\text{at least one green}) = 1 - 0.0261 = 0.9739.$$