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

Then \(\mathrm{p}(x \geq 2)\) is equal to

\(\begin{aligned} & \text { Since } \sum_{x=1}^5 \mathrm{P}(\mathrm{X}=x)=1 \\ & \mathrm{k}^2+2 \mathrm{k}+\mathrm{k}+2 \mathrm{k}+5 \mathrm{k}^2=1 \\ & \Rightarrow 6 \mathrm{k}^2+5 \mathrm{k}-1=0 \\ & \Rightarrow(\mathrm{k}+1)(6 \mathrm{k}-1)=0 \\ & \Rightarrow \mathrm{k}=\frac{1}{6} \quad \ldots[\because \mathrm{k} \geq 0]\end{aligned}\)
\(\begin{aligned} & \mathrm{P}(\mathrm{X} \geq 2)=\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4) \\ &+\mathrm{P}(\mathrm{X}=5)\end{aligned}\)
\(\begin{aligned} & =\frac{2}{6}+\frac{1}{6}+\frac{2}{6}+\frac{5}{36} \\ & =\frac{12+6+12+5}{36} \\ & =\frac{35}{36}\end{aligned}\)