If the p. m. f. of a random variable $\mathrm{X}$ is $\begin{array}{|c|c|c|c|c|c|} \hline \mathrm{X} & 1 & 2…
If the p. m. f. of a random variable $\mathrm{X}$ is
$\begin{array}{|c|c|c|c|c|c|}
\hline \mathrm{X} & 1 & 2 & 3 & 4 & 5 \\
\hline \mathrm{P}(\mathrm{X}=x) & k & \frac{k}{3} & \frac{k}{4} & \frac{k}{2} & \frac{k}{2} \\
\hline
\end{array}$
then $k=$
$\frac{15}{31}$
$\frac{1}{12}$
$\frac{11}{12}$
$\frac{12}{31}$
Solution
Here $\mathrm{k}+\frac{\mathrm{k}}{3}+\frac{\mathrm{k}}{4}+\frac{\mathrm{k}}{2}+\frac{\mathrm{k}}{2}=1$
$\therefore \mathrm{k}\left(\frac{12+4+3+6+6}{12}\right)=1 \Rightarrow \mathrm{k}\left(\frac{31}{12}\right)=1 \Rightarrow \mathrm{k}=\frac{12}{31}$