A number ‘k’ selected from the set $\left\{1,2,3,4,....,10\right\}$ . Then the probability If  ${\mathrm{k}}^2-4\mathrm{k}+3<0$  is

A number ‘k’ selected from the set $\left\{1,2,3,4,....,10\right\}$

n(S)=10

$E=\left\{k/{\mathrm{k}}^2-4\mathrm{k}+3<0 and1\le k\le 10\right\}$

$\Rightarrow \begin{array}{l}{\mathrm{k}}^2-4\mathrm{k}+3<0\\ \Rightarrow {\mathrm{k}}^2-3\mathrm{k}-\mathrm{k}+3<0\\ \Rightarrow \mathrm{k}(\mathrm{k}-3)-1(\mathrm{k}-3)<0\\ (\mathrm{k}-3)(\mathrm{k}-1)<0\\ \Rightarrow \mathrm{k}\in (1,3)\\ \therefore \mathrm{k}=2\\ \Rightarrow n\left(E\right)=1,n\left(S\right)=10\end{array}$
Probability $=\frac{1}{10}$