Mean $(\overline{\mathrm{x}})=8$ (Given) $\\ \Rightarrow \frac{2+4+10+x+12+14+y}{7}=8\\$ $\Rightarrow x+y=14 \ldots(1) \ldots.\\$ Variance $\left(\sigma^{2}\right)=16$ (Given) $\\ \Rightarrow 16=\frac{2^{2}+4^{2}+10^{2}+x^{2}+12^{2}+14^{2}+y^{2}}{7}-8^{2}\\$ $\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}=100 \quad \ldots \ldots(2) \ldots.\\$ $\because(x+y)^{2}=x^{2}+y^{2}+2 x y\\$ $\Rightarrow \mathrm{xy}=48$ (sum is 14 product is 48 )$\\$ Since problem states $x>y\\$ $\therefore \mathrm{x}=8$ and $\mathrm{y}=6\\$ Now set $X=\{1,2,3,4,6,5\}\\$ Now we choose two numbers one after author without replacement total outcomes $=6 \times 5=30\\$ We want the prob. That the smaller number among the two is less than 4 $\\ \mathrm{P}($ smaller $<4)=1-\mathrm{P}($ smaller $\geq 4)\\$ $=1-\frac{6}{30}=\frac{4}{5}$