Question

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?

• A. \( \frac{3}{4} \)
• B. \( \frac{7}{19} \)
• C. \( \frac{1}{3} \)
• D. \( \frac{8}{21} \)

Hint:

To calculate probability, divide the favorable outcomes by the total outcomes.

Answer 1

The Correct Option is C

We are given that there are:
- 8 red balls,
- 7 blue balls,
- 6 green balls.
The total number of balls is:
\[ 8 + 7 + 6 = 21 \]
Now, we are asked to find the probability that the ball picked is neither red nor green. The number of balls that are neither red nor green is the number of blue balls, which is 7.
Thus, the probability of picking a ball that is neither red nor green is:
\[ \frac{\text{Number of blue balls}}{\text{Total number of balls}} = \frac{7}{21} = \frac{1}{3} \]
Thus, the correct answer is option (c).