Three fair coins numbered 1,0 are tossed.̌/p>

$$\begin{array}{ll} \therefore \quad & \text { Sample space }=\{111,110,101,011,100,010, \\ & 001,000\} \\ \therefore \quad & n(S)=8 \end{array}$$

$$\mathrm{X}$$ represents the sum of numbers on upper most face

$$\begin{array}{r} \mathrm{P}(\mathrm{X}=0)=\frac{1}{8} \\ \mathrm{P}(\mathrm{X}=1)=\frac{3}{8} \\ \mathrm{P}(\mathrm{X}=2)=\frac{3}{8}, \\ \mathrm{P}(\mathrm{X}=3)=\frac{1}{8} \end{array}$$

$$\therefore \quad$$ Probability distribution of $$\mathrm{X}$$ is

$$\mathrm{X}$$	0	1	2	3
$$\mathrm{P(X)}$$	$$\frac{1}{8}$$	$$\frac{3}{8}$$	$$\frac{3}{8}$$	$$\frac{1}{8}$$

$$\begin{aligned} & \mathrm{E}(\mathrm{X})=\sum_{x=0}^3 x_{\mathrm{i}} \mathrm{P}\left(x_{\mathrm{i}}\right) \\ & \therefore \quad \mathrm{E}(\mathrm{X})= 0 \times \frac{1}{8}+1 \times \frac{3}{8}+2 \times \frac{3}{8}+3 \times \frac{1}{8} \\ &= \frac{12}{8}=\frac{3}{2} \\ & \mathrm{E}\left(\mathrm{X}^2\right)=\sum_{x=0}^3 x_{\mathrm{i}}^2 \mathrm{P}\left(x_{\mathrm{i}}\right)=3 \\ & \text { Variance of } \mathrm{X}=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\ &=3-\left(\frac{3}{2}\right)^2 \\ &=3-\frac{9}{4} \\ &=\frac{3}{4}=0.75 \end{aligned}$$

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