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 =

$$\begin{gathered} \mathrm{A} \mathrm{random} \mathrm{varibale} \mathrm{X} \mathrm{has} \mathrm{its} \mathrm{range} \left\{0,1,2\right\}. \\ P(x=0) +P(x=1)+P(x=2)==1 \\ \mathrm{where} P(x=0) = 3k^3; P(x=1) = 4k–10k^2; P(x=2)= 5k-1 \\ \therefore 3k^3+ 4k–10k^2+5k-1=1 \\ \Rightarrow 3k^3–10k^2+9K-2=0 \\ \mathrm{Since} \mathrm{sum} \mathrm{of} \mathrm{coefficents} \mathrm{is} \mathrm{zero} \mathrm{k}=1 \mathrm{is} \mathrm{a} \mathrm{root} \\ \Rightarrow (\mathrm{k}-1)(3{\mathrm{k}}^2-7\mathrm{k}+2)=0 \\ \Rightarrow (\mathrm{k}-1)(k-2)(3k-1)=0 \\ \Rightarrow k=1,2,\frac{1}{3} \\ \mathrm{If} k=1\mathrm{then} \mathrm{P}(\mathrm{x}=0) = 3>1 \mathrm{not} \mathrm{possible} \\ \mathrm{If} k=2 \mathrm{then} \mathrm{P}(\mathrm{x}=0) = 12>1 \mathrm{not} \mathrm{possible} \end{gathered}$$

 K=1,2 not possible .$\Rightarrow \mathrm{k}=\frac{1}{3}$