Since, $X=$ Number of fours seen

On tossing two die, $X=0,1,2$.

Also, $$P_{(4)}=\frac{1}{10} \text { and } P_{(\text {not } 4)}=\frac{9}{10}$$

So,

$$\begin{aligned} & P(X=0)=P_{(\text {not } 4)} \cdot P_{(\text {not } 4)}=\frac{9}{10} \cdot \frac{9}{10}=\frac{81}{100} \\ & P(X=1)=P_{(\text {not } 4)} \cdot P_{(4)}+P_{(4)} \cdot P_{(\text {not } 4)}=\frac{9}{10} \cdot \frac{1}{10}+\frac{1}{10} \cdot \frac{9}{10}=\frac{18}{100} \\ & P(X=2)=P_{(4)} \cdot P_{(4)}=\frac{1}{10} \cdot \frac{1}{10}=\frac{1}{100} \end{aligned}$$

Thus, we get following table

$$\begin{aligned} \therefore \quad \operatorname{Var}(X) & =E\left(X^2\right)-[E(X)]^2=\Sigma X^2 P(X)-[\Sigma X P(X)]^2 \\ & =\left[0+\frac{18}{100}+\frac{4}{100}\right]-\left[0+\frac{18}{100}+\frac{2}{100}\right]^2 \\ & =\frac{22}{100}-\left(\frac{20}{100}\right)^2=\frac{11}{50}-\frac{1}{25} \\ & =\frac{11-2}{50}=\frac{9}{50}=\frac{18}{100}=0.18 \end{aligned}$$

We have, $X=$ number of twos seen

So, on throwing a die three times, we will have $X=0,1,2,3$.

$$\begin{aligned} & \therefore \quad P(X=0)=P_{(\text {(not 2) }} \cdot P_{(\text {(not 2) }} \cdot P_{(\text {(not 2) }}=\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6}=\frac{125}{216} \\ & P(X=1)=P_{(\text {(not 2) }} \cdot P_{(\text {not 2) }} \cdot P_{(2)}+P_{(\text {not 2) }} \cdot P_{(2)} \cdot P_{(\text {not 2) }}+P_{(2)} \cdot P_{(\text {not 2) }} \cdot P_{(\text {not 2) }} \\ & =\frac{5}{6} \cdot \frac{5}{6} \frac{1}{6}+\frac{5}{6} \cdot \frac{1}{6} \cdot \frac{5}{6}+\frac{1}{6} \cdot \frac{5}{6} \cdot \frac{5}{6}=\frac{25}{36} \cdot \frac{3}{6}=\frac{25}{72} \\ & P(X=2)=P_{(\text {not 2) }} \cdot P_{(2)} \cdot P_{(2)}+P_{(2)} \cdot P_{(2)} \cdot P_{(\text {not 2) }}+P_{(2)} \cdot P_{(\text {not 2) }}+P_{(2)} \\ & =\frac{5}{6} \cdot \frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{5}{6}+\frac{1}{6} \cdot \frac{5}{6} \cdot \frac{1}{6} \\ & =\frac{1}{36} \cdot\left[\frac{15}{6}\right]=\frac{15}{216} \\ & P(X=3)=P_{(2)} \cdot P_{(2)} \cdot P_{(2)}=\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6}=\frac{1}{216} \end{aligned}$$

We know that, $E(X)=\Sigma X P(X)=0 \cdot \frac{125}{216}+1 \cdot \frac{25}{72}+2 \cdot \frac{15}{216}+3 \cdot \frac{1}{216}$

$$=\frac{75+30+3}{216}=\frac{108}{216}=\frac{1}{2}$$

For first die, $P(6)=\frac{1}{2}$ and $P\left(6^{\prime}\right)=\frac{1}{2}$

$$\begin{array}{lr} \Rightarrow & P(1)+P(2)+P(3)+P(4)+P(5)=\frac{1}{2} \\ \Rightarrow & P(1)=\frac{1}{10} \text { and } P\left(1^1\right)=\frac{9}{10} \end{array}$$

$$[\because P(1)=P(2)=P(3)=P(4)=P(5)]$$

For second die, $\quad P(1)=\frac{2}{5}$ and $P\left(1^{\prime}\right)=1-\frac{2}{5}=\frac{3}{5}$

Let $X=$ Number of one's seen

For $X=0$,

$$\begin{aligned} & P(X=0)=P\left(1^{\prime}\right) \cdot P\left(1^{\prime}\right)=\frac{9}{10} \cdot \frac{3}{5}=\frac{27}{50}=0.54 \\ & P(X=1)=P\left(1^{\prime}\right) \cdot P\left(1^{\prime}\right)+P\left(1^{\prime}\right) \cdot P\left(1^{\prime}\right)=\frac{9}{10} \cdot \frac{2}{5}+\frac{1}{10} \cdot \frac{3}{5} \\ & \quad=\frac{18}{50}+\frac{3}{50}=\frac{21}{50}=0.42 \\ & P(X=2)=P(1) \cdot P(1)=\frac{1}{10} \cdot \frac{2}{5}=\frac{2}{50}=0.04 \end{aligned}$$

Hence, the required probability distribution is as below

$$\begin{aligned} &\text { Since, we have to prove that, } E\left(Y^2\right)=2 E(X)\\ &\begin{aligned} \therefore \quad E(X) & =\Sigma X P(X) \\ & =0 \cdot \frac{1}{5}+1 \cdot \frac{2}{5}+2 \cdot \frac{1}{5}+3 \cdot \frac{1}{5}=\frac{7}{5} \end{aligned} \end{aligned}$$

$$\begin{aligned} \Rightarrow\quad 2 E(X) & =\frac{14}{5} \quad\text{.... (i)}\\ E(Y)^2 & =\Sigma Y^2 P(Y) \\ & =0 \cdot \frac{1}{5}+1 \cdot \frac{3}{10}+4 \cdot \frac{2}{5}+9 \cdot \frac{1}{10} \\ & =\frac{3}{10}+\frac{8}{5}+\frac{9}{10}=\frac{28}{10}=\frac{14}{5} \end{aligned}$$

$$\begin{aligned} &\Rightarrow \quad E\left(Y^2\right)=\frac{14}{5}\quad\text{.... (ii)}\\ &\text { From Eqs. (i) and (ii), }\\ &E\left(Y^2\right)=2 E(X) \end{aligned}$$

Hence proved.

Let $X$ is the random variable which denotes that a bulb is defective.

Also, $n=10, p=\frac{1}{50}$ and $q=\frac{49}{50}$ and $P(X=r)={ }^n C_r p^r q^{n-r}$

(i) None of the bulbs is defective i.e., $r=0$

$$\therefore \quad p(X=r)=P_{(0)}={ }^{10} \mathrm{C}_0\left(\frac{1}{50}\right)^0\left(\frac{49}{50}\right)^{10-0}=\left(\frac{49}{50}\right)^{10}$$

(ii) Exactly two bulbs are defective i.e., $r=2$

$$\begin{aligned} \therefore \quad P(X & =r)=P_{(2)}={ }^{10} C_2\left(\frac{1}{50}\right)^2\left(\frac{49}{50}\right)^8 \\ & =\frac{10!}{8!2!}\left(\frac{1}{50}\right)^2 \cdot\left(\frac{49}{50}\right)^8=45 \times\left(\frac{1}{50}\right)^{10} \times(49)^8 \end{aligned}$$

$$\begin{aligned} &\text { (iii) More than } 8 \text { bulbs work properly i.e., there is less than } 2 \text { bulbs which are defective. }\\ &\begin{array}{lrl} \text { So, } & r<2 \Rightarrow r=0,1 \\ \therefore & P(X=r)=P(r<2)=P(0)+P(1) \end{array} \end{aligned}$$

$$\begin{aligned} & ={ }^{10} C_0\left(\frac{1}{50}\right)^0\left(\frac{49}{50}\right)^{10-0}+{ }^{10} C_1\left(\frac{1}{50}\right)^1\left(\frac{49}{50}\right)^{10-1} \\ & =\left(\frac{49}{50}\right)^{10}+\frac{10!}{1!9!} \cdot \frac{1}{50} \cdot\left(\frac{49}{50}\right)^9 \\ & =\left(\frac{49}{50}\right)^{10}+\frac{1}{5} \cdot\left(\frac{49}{50}\right)^9=\left(\frac{49}{50}\right)^9\left(\frac{49}{50}+\frac{1}{5}\right) \\ & =\left(\frac{49}{50}\right)^9\left(\frac{59}{50}\right)=\frac{59(49)^9}{(50)^{10}} \end{aligned}$$