In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?
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3
3
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7
19
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21
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21

To solve the problem, let's first identify the total number of balls and then the number of favorable outcomes.

1. **Total Number of Balls:**
   - Red balls = 8
   - Blue balls = 7
   - Green balls = 6
   - Total = 8 + 7 + 6 = 21 balls

2. **Favorable Outcomes:**
   - We want the probability of picking a ball that is **neither red nor green**, which means the ball must be blue.
   - Number of blue balls = 7

3. **Probability Calculation:**
   - Probability of picking a blue ball (neither red nor green) = Number of blue balls / Total number of balls
   \[
   \text{Probability} = \frac{7}{21} = \frac{1}{3}
   \]

So, the probability that the ball is neither red nor green is **\(\frac{1}{3}\)**.

Would you like more details or have any other questions?

Here are 5 related questions you might consider:
1. What is the probability of picking a red ball?
2. What is the probability of picking a green ball?
3. What is the probability of picking either a red or green ball?
4. What is the probability of not picking a blue ball?
5. If two balls are picked randomly, what is the probability that both are blue?

**Tip:** Remember that probability is always a fraction of the favorable outcomes over the total outcomes.

Understand probability with a problem involving picking balls from a box containing 8 red, 7 blue, and 6 green balls. Learn how to calculate the probability of picking a ball that is neither red nor green, with a detailed step-by-step solution. Suitable for students in Grades 6-8.

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