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?
\[ \textbf{Step 1: Total balls in the box.} \] \[ \text{Total balls} = 8 + 7 + 6 = 21 \] --- \[ \textbf{Step 2: Favourable cases (neither red nor green).} \] That means the ball must be blue. \[ \text{Number of blue balls} = 7 \] --- \[ \textbf{Step 3: Probability.} \] \[ P(\text{Blue}) = \frac{\text{Favourable outcomes}}{\text{Total outcomes}} = \frac{7}{21} = \frac{1}{3} \] --- \[ \boxed{\tfrac{1}{3}} \]
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