We can solve each part of the problem using the concepts of probability and combinatorics. The total number of balls in the bag is:

$3 \text{ (blue)} + 4 \text{ (green)} + 5 \text{ (red)} = 12 \text{ balls}.$

The total number of ways to select 3 balls from 12 is given by the combination formula $C(n, k) = \frac{n!}{k!(n-k)!}$, which represents the number of ways to choose $k$ objects from $n$ objects without replacement.

Thus, the total number of ways to choose 3 balls from 12 is:

$C(12, 3) = \frac{12!}{3!(12-3)!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220.$

Now, let's solve each part:

(i) Probability that the balls are of different colors:

We need to select one ball of each color: one blue, one green, and one red. The number of ways to do this is:

$C(3, 1) \times C(4, 1) \times C(5, 1) = 3 \times 4 \times 5 = 60.$

So, the probability that the three balls are of different colors is:

$P(\text{different colors}) = \frac{60}{220} = \frac{3}{11}.$

(ii) Probability that two are blue and one is green:

The number of ways to select 2 blue balls and 1 green ball is:

$C(3, 2) \times C(4, 1) = \frac{3 \times 2}{2 \times 1} \times 4 = 3 \times 4 = 12.$

Thus, the probability is:

$P(2 \text{ blue, } 1 \text{ green}) = \frac{12}{220} = \frac{6}{110} = \frac{3}{55}.$

(iii) Probability that the balls are of the same color:

For the balls to be of the same color, they must all be blue, or all green, or all red. The number of ways to select 3 blue balls, 3 green balls, or 3 red balls are:

• Blue: $C(3, 3) = 1$
• Green: $C(4, 3) = \frac{4 \times 3 \times 2}{3 \times 2 \times 1} = 4$
• Red: $C(5, 3) = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$

Thus, the total number of ways to select 3 balls of the same color is:

$1 + 4 + 10 = 15.$

Therefore, the probability is:

$P(\text{same color}) = \frac{15}{220} = \frac{3}{44}.$

Final Answers:

1. The probability that the balls are of different colors is $\frac{3}{11}$.
2. The probability that two are blue and one is green is $\frac{3}{55}$.
3. The probability that the balls are of the same color is $\frac{3}{44}$.

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Would you like further clarification on any part of the solution?

Here are some related questions:

1. What is the probability of drawing exactly one red ball and two other balls?
2. How would the probabilities change if there were 6 red balls instead of 5?
3. What is the probability of drawing at least one green ball?
4. How does the combination formula help in solving probability problems?
5. What is the probability of drawing no blue balls at all?

Tip: Always use the combination formula for problems where order doesn't matter, especially in probability questions related to choosing groups.