If \(\mathrm{x}\) is a random variable with p.m.f. as follows.
\(P(X=x)=\frac{5}{16}, x=0,1 =\frac{k x}{48}, x=2,\) \(\text { then } E(x)=\frac{1}{4}, x=3\)

From given data, we write
When \(\mathrm{x}=0, \mathrm{P}=\frac{5}{16}=\frac{15}{48}\)
\(
\begin{aligned}
& x=1, P=\frac{5}{16}=\frac{15}{48} \\
& x=2, P=\frac{2 k}{48} \\
& x=3, P=\frac{1}{4}=\frac{12}{48}
\end{aligned}
\)
Here \(\sum \mathrm{Pi}=1\)
\(
\begin{aligned}
& \therefore \frac{15}{48}+\frac{15}{48}+\frac{2 \mathrm{k}}{48}+\frac{12}{48}=1 \Rightarrow \mathrm{k}=3 \\
& \therefore \text { When } \mathrm{x}=2, \mathrm{P}=\frac{6}{48}=\frac{1}{8}
\end{aligned}
\)
\(
\begin{aligned}
& \text { Now } \mathrm{E}=\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} \end{aligned}\)
\(=\left[\left(\frac{5}{16}\right)(0)\right]+\left[\left(\frac{5}{16}\right)(1)\right]+\left[\left(\frac{1}{8}\right)(2)\right]+\left[\left(\frac{1}{4}\right)(3)\right]=0\) \(+~\frac{5}{16}+\frac{1}{4}+\frac{3}{4}=\frac{21}{16}=1.3125\)