Let's break down the question:

Problem: 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?

Step 1: Calculate the total number of balls.

• Red balls = 4
• Green balls = 5
• Blue balls = 6
• Total balls = 4 + 5 + 6 = 15 balls

Step 2: Determine the total number of possible outcomes for drawing two balls.

The total number of ways to draw 2 balls out of 15 is calculated using combinations:

$\text{Total combinations} = \binom{15}{2} = \frac{15 \times 14}{2} = 105$

Step 3: Find the number of favorable outcomes (none of the balls are blue).

To ensure no blue balls are drawn, we only consider the red and green balls. The total number of red and green balls is:

$\text{Red + Green} = 4 + 5 = 9$

The number of ways to select 2 balls from these 9 non-blue balls is:

$\text{Favorable combinations} = \binom{9}{2} = \frac{9 \times 8}{2} = 36$

Step 4: Calculate the probability.

The probability that neither ball drawn is blue is the ratio of favorable outcomes to the total outcomes:

$\text{Probability} = \frac{\text{Favorable combinations}}{\text{Total combinations}} = \frac{36}{105}$

Simplifying this:

$\frac{36}{105} = \frac{12}{35}$

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

Would you like further explanation or have more questions?

5 Related Questions:

1. What is the probability that exactly one ball drawn is blue?
2. How would the probability change if there were only 2 blue balls?
3. What if three balls were drawn? How would the probability of no blue balls change?
4. Can we compute the probability of drawing two blue balls instead?
5. How does the formula for combinations ($\binom{n}{r}$) work?

Tip:

Always ensure to count the total possibilities and favorable outcomes carefully in probability questions to avoid missing cases!