A random variable \(\mathrm{X}\) has the following probability distribution

then \(\mathrm{P}(\mathrm{X} < 2)\) is

\(\begin{aligned} & \text { Since } \sum_{x=0}^2 \mathrm{P}(\mathrm{X}=x)=1 \\ & 4 \mathrm{k}-10 \mathrm{k}^2+5 \mathrm{k}-1+3 \mathrm{k}^3=1 \\ & \Rightarrow 3 \mathrm{k}^3-10 \mathrm{k}^2+9 \mathrm{k}-2=0 \\ & \Rightarrow(\mathrm{k}-1)(\mathrm{k}-2)(3 \mathrm{k}-1)=0 \\ & \Rightarrow \mathrm{k}=1 \text { or } \mathrm{k}=2 \text { or } \mathrm{k}=\frac{1}{3} \\ & \begin{aligned} & \text { For } \mathrm{k}=1 \text { or } \mathrm{k}=2 \\ & \mathrm{P}(\mathrm{X}=0) < 0, \text { which is not possible } \\ & \mathrm{k}=\frac{1}{3} \quad\end{aligned} \\ & \begin{aligned} \text { Now, } \mathrm{P}(\mathrm{X} < 2) & =\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1) \\ & =4 \mathrm{k}-10 \mathrm{k}^2+5 \mathrm{k}-1 \\ & =9 \mathrm{k}-10 \mathrm{k}^2-1 \\ & =9\left(\frac{1}{3}\right)-10\left(\frac{1}{9}\right)-1 \\ & =\frac{8}{9}\end{aligned}\end{aligned}\)