Discussion
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?
Answer: A
"Neither red nor green" leaves only the blue balls.
\[ \text{total} = 8 + 7 + 6 = 21 \]
\[ P = \frac{7}{21} = \frac{1}{3} \]
Recognising the complement as a single colour turns a two-case count into one division.
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