Three different numbers are selected at random from the first 15...

$$\begin{aligned} n(\mathrm{ S}) & ={ }^{15} \mathrm{C}_3 \\ 2 \times 3 & =6 \\ 2 \times 4 & =8 \\ 2 \times 5 & =10 \\ 2 \times 6 & =12 \\ 2 \times 7 & =14 \\ 3 \times 4 & =12\end{aligned}$$\therefore \quad n(\mathrm{E})=6\therefore$Required probability$=\frac{6}{{ }^{15} \mathrm{C}_3}=\frac{6}{\frac{15 \times 14 \times 13 \times 12!}{12!\times 3 \times 2}}=\frac{6}{35 \times 13}=\frac{6}{455}$