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

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

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

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., $r=2$

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

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

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

Let $E_1=$ Event that fair coin is drawn

$E_2=$ Event that 2 headed coin is drawn

$E=$ Event that tossed coin get a head

$\therefore \quad P\left(E_1\right)=1 / 2, P\left(E_2\right)=1 / 2, P\left(E / E_1\right)=1 / 2$ and $P\left(E / E_2\right)=1$

$$\begin{aligned} &\text { Now, using Baye's theorem } P\left(E_1 / E\right)=\frac{P\left(E_1\right) \cdot P\left(E / E_1\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)}\\ &=\frac{\frac{1}{2} \cdot \frac{1}{2}}{\frac{1}{2} \cdot \frac{1}{2}+\frac{1}{2} \cdot 1}=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{2}}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3} \end{aligned}$$

$$\begin{aligned} & E_1=\text { Event that the person selected is of blood group O } \\ & E_2=\text { Event that the person selected is of other than blood group O } \\ & \left(E_3\right)=\text { Event that selected person is left handed } \\ & \therefore \quad \begin{aligned} P\left(E_1\right)=0.30, P\left(E_2\right)=0.70 \\ P\left(E_3 / E_1\right)=0.06 \text { and } P\left(E_3 / E_2\right)=0.10 \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { By using Baye's theorem, }\\ &\begin{aligned} P\left(E_1 / E_3\right) & =\frac{P\left(E_1\right) \cdot P\left(E_3 / E_1\right)}{P\left(E_1\right) \cdot P\left(E_3 / E_1\right)+P\left(E_2\right) \cdot P\left(E_3 / E_2\right)} \\ & =\frac{0.30 \times 0.06}{0.30 \cdot 0.06+0.70 \cdot 0.10} \\ & =\frac{0.0180}{0.0180+0.0700} \\ & =\frac{0.0180}{0.0880}=\frac{180}{880}=\frac{9}{44} \end{aligned} \end{aligned}$$

$$\begin{aligned} & \because \text { Set } S=\{1,2,3, \ldots, n\} \\ & \therefore \quad P(r \leq p / S \leq p)=\frac{P(p \cap S)}{P(S)} \\ & \\ \end{aligned}$$

$=\frac{p-1}{n} \times \frac{n}{n-1}=\frac{p-1}{n-1}$