A coin is tossed and a die is thrown. The probability that the outcome will be head or a number greater than 4 or both, is

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

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

The p.m.f. of a random variable X is $$\mathrm{P(X = x) = {1 \over {{2^5}}}\left( {_x^5} \right),x = 0,1,2,3,4,5}=0$$ then

The variance of the following probability distribution is,