An urn has 9 red, 7 white, and 4 black balls (total 20). If one ball is drawn at random, what is the probability that it is not red?

Introduction / Context:This is a direct single-draw probability problem with a finite equally likely sample space. We want the chance that the colour of the drawn ball is not red.

Given Data / Assumptions:

• Red: 9, White: 7, Black: 4.
• Total balls = 9 + 7 + 4 = 20.
• Single draw, each ball equally likely.

Concept / Approach:P(not red) = 1 − P(red) = (non-red count) / (total count). Non-red balls are white + black.

Step-by-Step Solution:Non-red balls = 7 + 4 = 11.Total = 20.Probability = 11/20.

Verification / Alternative check:Compute P(red) = 9/20, then 1 − 9/20 = 11/20, consistent.

Why Other Options Are Wrong:1/11 and 2/11 invert counts; 9/20 is P(red); 9/20 ≠ P(not red).

Common Pitfalls:Forgetting to add both white and black when counting non-red outcomes.

Final Answer:11/20