A box contains 10 red marbles, 20 blue marbles nad 30 green marbles. 5 marbles are drawn from the box, what is the probability that: (i) all will be blue? (ii) at least one will be green?

Explanation

To solve the problem, we need to determine the probability of two scenarios when drawing 5 marbles from a box containing 10 red marbles, 20 blue marbles, and 30 green marbles.

### Total Marbles Calculation
First, we calculate the total number of marbles in the box:
- Red marbles = 10
- Blue marbles = 20
- Green marbles = 30

Total marbles = 10 + 20 + 30 = 60 marbles.

### (i) Probability that all 5 marbles drawn are blue

To find the probability that all 5 marbles drawn are blue, we can use the combination formula to determine the number of ways to choose the marbles.

1. **Calculate the total ways to choose 5 marbles from 60:**
\[ \text{Total ways} = \binom{60}{5} \]

2. **Calculate the ways to choose 5 blue marbles from 20:**
\[ \text{Ways to choose 5 blue} = \binom{20}{5} \]

3. **Calculate the probability:**
\[ P(\text{All blue}) = \frac{\text{Ways to choose 5 blue}}{\text{Total ways}} = \frac{\binom{20}{5}}{\binom{60}{5}} \]

### (ii) Probability that at least one marble drawn is green

To find the probability that at least one marble is green, we can use the complementary approach:

1. **Calculate the probability that none of the marbles drawn are green (i.e., all are either red or blue):**
- Total red and blue marbles = 10 + 20 = 30 marbles.
- Calculate the total ways to choose 5 marbles from these 30:
\[ \text{Total ways (no green)} = \binom{30}{5} \]

2. **Calculate the probability that none of the marbles drawn are green:**
\[ P(\text{No green}) = \frac{\text{Total ways (no green)}}{\text{Total ways}} = \frac{\binom{30}{5}}{\binom{60}{5}} \]

3. **Calculate the probability that at least one marble is green:**
\[ P(\text{At least one green}) = 1 - P(\text{No green}) = 1 - \frac{\binom{30}{5}}{\binom{60}{5}} \]

### Final Answers
- (i) The probability that all 5 marbles drawn are blue is:
\[ P(\text{All blue}) = \frac{\binom{20}{5}}{\binom{60}{5}} \]

- (ii) The probability that at least one marble drawn is green is:
\[ P(\text{At least one green}) = 1 - \frac{\binom{30}{5}}{\binom{60}{5}} \]