<p>Two red, three green and four blue balls are placed in a row at random. The probability that the two red balls are not adjacent is</p>
Step-by-Step Solution
Key Concept: P(not adjacent) = 1 - P(adjacent). Treat the two red balls as a single unit for the adjacent case.
<p>Total arrangements (balls of same colour identical): $\dfrac{9!}{2!\,3!\,4!} = 1260$.</p><p>Arrangements with both reds adjacent (treat RR as one block): $\dfrac{8!}{1!\,3!\,4!} = 280$.</p><p>$P(\text{adjacent}) = \dfrac{280}{1260} = \dfrac{2}{9}$.</p><p>$P(\text{not adjacent}) = 1 - \dfrac{2}{9} = \dfrac{7}{9}$</p>
Correct Answer: B