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The question is asking for the probability that none of the balls drawn is blue from a bag containing 4 red, 5 green, and 6 blue balls. Let's solve it step by step:

### Total number of balls:
- Red balls = 4
- Green balls = 5
- Blue balls = 6
- **Total balls = 4 + 5 + 6 = 15**

### Total number of ways to select 2 balls from 15:
The number of ways to choose 2 balls from 15 is given by the combination formula:
\[
\binom{15}{2} = \frac{15 \times 14}{2} = 105
\]

### Favorable outcomes (choosing 2 balls that are not blue):
The number of red and green balls combined is:
\[
4 + 5 = 9 \quad \text{(total non-blue balls)}
\]
The number of ways to choose 2 balls from these 9 red and green balls is:
\[
\binom{9}{2} = \frac{9 \times 8}{2} = 36
\]

### Probability that neither of the balls drawn is blue:
\[
\text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{36}{105}
\]
Simplifying the fraction:
\[
\frac{36}{105} = \frac{12}{35}
\]

### Final Answer:
The probability that none of the balls drawn is blue is **12/35**.

Let me know if you need further details or explanations.

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Here are 5 related questions to expand your understanding:
1. What is the probability that both balls drawn are blue?
2. How would the probability change if the number of blue balls was 5 instead of 6?
3. What is the probability that at least one of the balls drawn is blue?
4. How would the probability calculation change if 3 balls were drawn instead of 2?
5. What is the probability of drawing one red and one green ball?

### Tip:
Always simplify combinations and probabilities by reducing fractions to their simplest form for easier interpretation!

A bag contains 4 red, 5 green and 6 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

This problem asks for the probability of drawing two balls from a bag containing red, green, and blue balls, where neither is blue. The solution involves key probability and combination concepts, suitable for students in Grades 9-12.

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