A random variable X takes values 0, 1, 2,….with probability proportional to $(x+1){\left(\frac{1}{5}\right)}^x$ then $5\left[P{(x<1)}^{1/2}\right]$ is

$P\left(x\right)=K(x+1){\left(\frac{1}{5}\right)}^x$

$\therefore P\left(0\right)=K,P\left(1\right)=\frac{2K}{5},P\left(2\right)=\frac{3K}{5^2}$

$\therefore Xhas\text{\hspace{0.17em}}aP.d.f$

$\begin{array}{l}\therefore Sumofallprobabilites=1\\ \therefore k[1+2\left(\frac{1}{5}\right)+3{\left(\frac{1}{5}\right)}^2+\mathrm{........}]=1\end{array}$

$\begin{array}{l}\Rightarrow K[\frac{a}{1-r}+\frac{dr}{{(1-r)}^2}]=1\\ \Rightarrow k[\frac{1}{1-\frac{1}{5}}+\frac{1\left(\frac{1}{5}\right)}{{(1-\frac{1}{5})}^2}]=1\\ \Rightarrow k\left[\frac{25}{16}\right]=1\Rightarrow k=\frac{16}{25}\end{array}$

$\begin{array}{l}\therefore 5\left[P{(X<1)}^{1/2}\right]=5{\left[P(x=0)\right]}^{1/2}=5{\left(k\right)}^{1/2}\\ =5{\left[\frac{15}{25}\right]}^{1/2}=5\left(\frac{4}{5}\right)=4\end{array}$