If a random variable \(X\) has the following probability distribution values

Then \(P(X \geq 6)\) has the value

\(\begin{aligned} & \text { Since } \sum_{x=0}^7 \mathrm{P}(\mathrm{X}=x)=1 \\ & \begin{array}{l}0+\mathrm{k}+2 \mathrm{k}+2 \mathrm{k}+3 \mathrm{k}+\mathrm{k}^2+2 \mathrm{k}^2+7 \mathrm{k}^2+\mathrm{k}=1 \\ \begin{aligned} & \Rightarrow 10 \mathrm{k}^2+9 \mathrm{k}-1=0 \\ & \Rightarrow(\mathrm{k}+1)(10 \mathrm{k}-1)=0 \\ & \Rightarrow \mathrm{k}=\frac{1}{10}\end{aligned} \\ \begin{aligned} \mathrm{P}(\mathrm{X} \geq 6) & =\mathrm{P}(\mathrm{X}=6)+\mathrm{P}(\mathrm{X}=7) \\ & =2\left(\frac{1}{10}\right)^2+7\left(\frac{1}{10}\right)^2+\frac{1}{10} \\ & =\frac{2}{100}+\frac{7}{100}+\frac{10}{100} \\ & =\frac{19}{100}\end{aligned}\end{array} \begin{array}{l}\text {. }\end{array}\end{aligned}\)
\(\ldots[\because \mathrm{k} \geq 0]\)