Probability

If \(\mathrm{X}\) is a random variable with p.m.f. as follows.

$$\begin{aligned} \mathrm{P}(\mathrm{X}=\mathrm{x}) & =\frac{5}{16}, \mathrm{x}=0,1 \\ & =\frac{\mathrm{kx}}{48}, \mathrm{x}=2, \quad \text { then } \mathrm{E}(\mathrm{x})= \\ & =\frac{1}{4}, \mathrm{x}=3 \end{aligned}$$