Total number of balls = (8 + 7 + 6) = 21. Let E = event that the ball drawn is neither red nor green = event that the ball drawn is blue. ∴ n(E) = 7. ∴ P(E) = n(E)/n(S) = 7/21 = 1/3. .
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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Question context
This question is part of 📏 Arithmetic Aptitude within the broader 🧠 Aptitude Mastery subject and appears in the DemoPaper-Pakistan preparation collection. Reviewing the answer together with nearby questions helps reinforce the underlying concept and identify how it may appear in competitive exams.