Answer 1

Total number of balls = 6 + 4 + 8 = 18
Total number of elementary events, n(S) = ¹⁸C₃
Let E be the event of favourable outcomes.
Here, E = getting one red, one white and one blue ball
So, favourable number of elementary events, n(E) = ⁶C₁ ×⁴C₁ × ⁸C₁
Hence, required probability = \[\frac{n\left( E \right)}{n\left( S \right)} = \frac{^{6}{}{C}_1 \times ^{4}{}{C}_1 \times^{8}{}{C}_1}{^{18}{}{C}_3}\]

\[= \frac{6 \times 4 \times 8}{\frac{18 \times 17 \times 16}{3 \times 2}}\]
\[ = \frac{6 \times 4 \times 8}{6 \times 17 \times 8}\]
\[ = \frac{4}{17}\]