We have,

$$\begin{aligned} &\text { (i) We know that, } \quad \sum_{i=1}^n P_i=1 \text {, where } P_i \geq 0\\ &\begin{array}{lr} \Rightarrow & P_1+P_2+P_3+P_4=1 \\ \Rightarrow & k+k^2+2 k^2+k=1 \\ \Rightarrow & 3 k^2+2 k-1=0 \\ \Rightarrow & 3 k^2+3 k-k-1=0 \\ \Rightarrow & 3 k(k+1)-1(k+1)=0 \\ \Rightarrow & (3 k-1)(k+1)=0 \\ \Rightarrow & k=1 / 3 \Rightarrow k=-1 \\ \text { Since, } & k \text { is } \geq 0 \Rightarrow k=1 / 3 \end{array} \end{aligned}$$

(ii) Mean of the distribution $(\mu)=E(X)=\sum_{i=1}^n x_i P_i$

$$ \begin{aligned} & =0.5(k)+1\left(k^2\right)+1.5\left(2 k^2\right)+2(k)=4 k^2+2.5 k \\ & =4 \cdot \frac{1}{9}+2.5 \cdot \frac{1}{3} \quad \left[\because k=\frac{1}{3}\right]\\ & =\frac{4+7.5}{9}=\frac{23}{18} \end{aligned}$$

(i) $\because \quad P(A)=P(A \cap B)+P(A \cap \bar{B})$

$$\begin{aligned} \therefore \quad \mathrm{RHS} & =P(A \cap B)+P(A \cap \bar{B}) \\ & =P(A) \cdot P(B)+P(A) \cdot P(\bar{B}) \\ & =P(A)[P(B)+P(\bar{B})] \\ & =P(A)[P(B)+1-P(B)] \quad [\because P(\bar{B})=1-P(B)]\\ & =P(A)=\mathrm{LHS}\quad\text{Hence proved.} \end{aligned}$$

$$\begin{aligned} &\begin{array}{lrl} \text { (ii) } \because P(A \cup B) =P(A \cap B)+P(A \cap \bar{B})+P(\bar{A} \cap B) \\ \therefore \quad R H S =P(A) \cdot P(B)+P(A) \cdot P(\bar{B})+P(\bar{A}) \cdot P(B) \end{array} \end{aligned}$$

$$ \begin{aligned} &\begin{aligned} & =P(A) \cdot P(B)+P(A) \cdot[1-P(B)]+[1-P(A)] P(B) \\ & =P(A) \cdot P(B)+P(A)-P(A) \cdot P(B)+P(B)-P(A) \cdot P(B) \\ & =P(A)+P(B)-P(A) \cdot P(B) \\ & =P(A)+P(B)-P(A \cap B) \\ & =P(A \cup B)=\mathrm{LHS} \quad\text { Hence proved. } \end{aligned}\\ \end{aligned}$$

$$\begin{aligned} &\text { Given that, random variable } X \text { is the number of tails in three tosses of a coin. }\\ &\begin{array}{l} \text { So, } & X=0,1,2,3 . \\ \Rightarrow & P(X=x)={ }^n C_x(p)^x q^{n-x}, \\ \text { where } & n=3, p=1 / 2, q=1 / 2 \text { and } x=0,1,2,3 \end{array} \end{aligned}$$

We know that, $\operatorname{Var}(X)=E\left(X^2\right)-[E(X)]^2\quad\text{.... (i)}$

where, $E\left(X^2\right)=\sum_{i=1}^n x_i^2 P\left(x_i\right)$ and $E(X)=\sum_{i=1}^n x_i P\left(x_i\right)$

$$\begin{aligned} &\begin{array}{ll} \therefore & E\left(X^2\right)=\sum_{i=1}^n x_i^2 P\left(X_i\right)=0+\frac{3}{8}+\frac{3}{2}+\frac{9}{8}=\frac{24}{8}=3 \\ \text { and } & {[E(X)]^2=\left[\sum_{i=0}^n x_i^2 P\left(x_i\right)\right]^2=\left[0+\frac{3}{8}+\frac{3}{4}+\frac{3}{8}\right]^2=\left[\frac{12}{8}\right]^2=\frac{9}{4}} \\ \therefore & \operatorname{Var}(X)=3-\frac{9}{4}=\frac{3}{4}\quad\text{[using Eq. (i)]} \end{array}\\ &\text { and standard deviation of } X=\sqrt{\operatorname{Var}(X)}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2} \end{aligned}$$

Let X is the random variable of profit per throw.

Since, she loss ₹ 1 on getting any of 2,4 or 5.

So, at $X=-1\quad,$ $$ P(X)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$$

Similarly, at $X=1\quad$, $P(X)=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}\quad$ [if die shows of either 1 or 6 ]

and at $X=4, \quad P(X)=\frac{1}{6}$ [if die shows a 3]

$\therefore$ Player's expected profit $=E(X)=\Sigma X P(X)$

$=-1 \times \frac{1}{2}+1 \times \frac{1}{3}+4 \times \frac{1}{6}$

$=\frac{-3+2+4}{6}=\frac{3}{6}=\frac{1}{2}=$ ₹ 0.50

On a throw of three dice, we have sample space $[n(S)]=6^3=216$

Let $E_1$ is the event when the sum of numbers on the dice was six and $E_2$ is the event when three two's occurs.

$$\begin{array}{r} \Rightarrow \quad E_1=\{(1,1,4),(1,2,3),(1,3,2),(1,4,1),(2,1,3),(2,2,2),(2,3,1),(3,1,2), \\ (3,2,1),(4,1,1)\} \end{array}$$

$$\begin{array}{l} \Rightarrow & n\left(E_1\right) =10 \text { and } E_2=\{2,2,2\} \\ \Rightarrow & n\left(E_2\right)=1 \\ \text { Also, } & \left(E_1 \cap E_2\right)=1 \\ \therefore & P\left(E_2 / E_1\right)=\frac{P \cdot\left(E_1 \cap E_2\right)}{P\left(E_1\right)}=\frac{1 / 216}{10 / 216}=\frac{1}{10} \end{array}$$