A random varibale x has its range {0,1,2} and the probabilities are given by P(x=0) = 3k³; P(x=1) = 4k–10k²; P(x=2)= 5k–1 where k is constant then k = 
 

Given 

Since $\sum _{\mathrm{i}=0}^2\mathrm{P}\left({\mathrm{x}}_{\mathrm{i}}\right)=1$ then

 $$\begin{gathered} \Rightarrow 3{\mathrm{k}}^3+4\mathrm{k}-10{\mathrm{k}}^2+5\mathrm{k}-1=1 \\ \Rightarrow 3{\mathrm{k}}^3-10{\mathrm{k}}^2+9\mathrm{k}-2=0 \end{gathered}$$

Here k=1 is a root $\Rightarrow \mathrm{k}-1$ is a factor

$$\begin{gathered} \Rightarrow (\mathrm{k}-1)(3{\mathrm{k}}^2-7\mathrm{k}+2)=0 \\ \Rightarrow (\mathrm{k}-1)(3{\mathrm{k}}^2-6\mathrm{k}-\mathrm{k}+2)=0 \\ \Rightarrow (\mathrm{k}-1)(\mathrm{k}-2)(3\mathrm{k}-1)=0 \\ \Rightarrow \mathrm{k}=1 \mathrm{or} \mathrm{k}=2 \mathrm{or} \mathrm{k}=\frac{1}{3} \\ \therefore \mathrm{k}=\frac{1}{3} (\therefore 0\le \mathrm{P}(x)\le 1) \end{gathered}$$