In a box there are 8 red balls, 7 blue balls and 6 green balls, for a total of 21 balls. One ball is picked at random. What is the probability that the chosen ball is neither red nor green, that is, that it is blue?

Introduction / Context:
This question deals with a simple color based probability scenario. The box contains balls of three different colors, and we are interested in the probability that a randomly chosen ball is neither red nor green. Since the only remaining color is blue, the problem effectively asks for the probability of drawing a blue ball from the mixture.

Given Data / Assumptions:

• Number of red balls = 8.
• Number of blue balls = 7.
• Number of green balls = 6.
• Total number of balls = 8 + 7 + 6 = 21.
• One ball is drawn at random and all balls are equally likely to be selected.

Concept / Approach:
The event of interest is that the ball is neither red nor green. Since the only other color present is blue, this is simply the event that the ball is blue. The probability is found by dividing the number of blue balls by the total number of balls in the box. There is no conditioning or replacement involved because the selection consists of a single draw.

Step-by-Step Solution:
Step 1: Compute the total number of balls: 8 red + 7 blue + 6 green = 21 balls.Step 2: Identify the number of favorable outcomes, which is the number of blue balls. There are 7 blue balls.Step 3: The event that a ball is neither red nor green is exactly the same as the event that it is blue.Step 4: Use the formula probability = favorable outcomes / total outcomes.Step 5: Substitute values to get probability = 7 / 21.Step 6: Simplify 7 / 21 by dividing both numerator and denominator by 7 to obtain 1 / 3.
Verification / Alternative check:
We can verify this answer by also computing the probabilities of drawing red or green balls and ensuring that all probabilities sum to 1. The probability of red is 8 / 21, of blue is 7 / 21 and of green is 6 / 21. Adding them gives (8 + 7 + 6) / 21 = 21 / 21 = 1. This confirms that our probability distribution is consistent and that 1 / 3 for blue is correct.

Why Other Options Are Wrong:
3/5 would require 12 or 13 favorable outcomes out of 21, which is inconsistent with the count of 7 blue balls. 8/21 corresponds to the probability of drawing a red ball, not a blue one. 7/21 is the unsimplified form of the correct answer, but 1/3 is the reduced and standard form. 2/7 does not correspond to any natural color count in this scenario, and therefore cannot be right.

Common Pitfalls:
Students sometimes misread neither red nor green and incorrectly attempt to exclude blue instead. Another mistake is to forget to simplify the fraction, although leaving it unsimplified still represents the same probability. It is helpful to systematically list the counts of each color and then directly relate the desired event to one of those counts.

Final Answer:
The probability that the chosen ball is neither red nor green, that is, that it is blue, is 1/3.