Total number of letters in the word 'ASSASSINATION' are 13.

Out of which 3A's, 4S's, 2 I's, 2 N's, 1 T's and 10.

(i) If four S's come consecutively in the word, then we considers these 4 S's as 1 group. Now, the number of laters is 10 .

Number of words when all S's are together $=\frac{10!}{3!2!2!}$

Total number of word using letters of the word 'ASSASSINATION'

$$\begin{aligned} \quad & =\frac{13!}{3!4!2!2!} \\ \therefore \quad \text { Required probability } & =\frac{10!}{\frac{3!2!2!\times 13!}{3!4!2!2!}} \\ & =\frac{10!\times 4!}{13!}=\frac{4!}{13 \times 12 \times 11}=\frac{24}{1716}=\frac{2}{143} \end{aligned}$$

(ii) If 2 I's and 2 N 's come together, then there as 10 alphabets.

Number of word when 2 l's and 2 N's are come together

$$\begin{aligned} & =\frac{10!}{3!4!} \times \frac{4!}{2!2!} \\ \therefore \quad \text { Required probability } & =\frac{\frac{10!4!}{3!4!2!2!}}{\frac{13!}{3!4!2!2!}}=\frac{4!10!}{2!2!3!4!} \times \frac{3!4!2!2!}{13!} \\ & =\frac{4!10!}{13!}=\frac{4!}{13 \times 12 \times 11}=\frac{24}{13 \times 12 \times 11}=\frac{2}{143} \end{aligned}$$

(iii) If all A's are coming together, then there are 11 alphabets.

Number of words when all A's come together

$$=\frac{11!}{4!2!2!}$$

Probability when all A's come together

$$=\frac{\frac{11!}{4!2!2!}}{\frac{13!}{4!3!2!2!}}=\frac{11!}{4!2!2!} \times \frac{4!3!2!2!}{13!}=\frac{11!\times 3!}{13!}=\frac{6}{13 \times 12}=\frac{1}{26}$$

Required probability when all A's does not come together

$$=1-\frac{1}{26}=\frac{25}{26}$$

(iv) If no two A's are together, then first we arrange the alphabets except A's.

All the alphabets except A's are arranged in $\frac{10!}{4!2!2!}$.

There are 11 vacant places between these alphabets.

So, 3 A's can be place in 11 places in ${ }^{11} C_3$ ways $=\frac{11!}{3!8!}$

$\therefore$ Total number of words when no two A's together

$$\begin{aligned} & =\frac{11!}{3!8!} \times \frac{10!}{4!2!2!} \\ \text { Required probability } & =\frac{11!\times 10!}{3!8!4!2!2!} \times \frac{4!3!2!2!}{13!}=\frac{10!}{8!\times 13 \times 12} \\ & =\frac{10 \times 9}{13 \times 12}=\frac{90}{156}=\frac{15}{26} \end{aligned}$$

$\because$ Number of possible event $=52$

and favourable events $=4$ king +13 heart +26 red $-13-2=28$

$\therefore \quad$ Required probability $=\frac{28}{52}=\frac{7}{13}$

Given,

$$\begin{aligned} S & =\left\{E_1, E_2, E_3, E_4, E_5, E_6, E_7, E_8, E_9\right\} \\ A & =\left\{E_1, E_5, E_8\right\}, B=\left\{E_2, E_5, E_8, E_9\right\} \\ P\left(E_1\right) & =P\left(E_2\right)=0.08 \\ P\left(E_3\right) & =P\left(E_{4)}=P\left(E_5\right)=0.1\right. \\ P\left(E_6\right) & =P\left(E_7\right)=2, P\left(E_8\right)=P\left(E_9\right)=0.07 \end{aligned}$$

(i) $$\begin{aligned} P(A) & =P\left(E_1\right)+P\left(E_5\right)+P\left(E_8\right) \\ & =0.08+0.1+0.07=0.25 \end{aligned}$$

(ii)$$\begin{aligned} & P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ & \begin{aligned} \text { Now, } P(B) & =P\left(E_2\right)+P\left(E_5\right)+P\left(E_8\right)+P\left(E_9\right) \\ & =0.08+0.1+0.07+0.07=0.32 \\ A \cap B & =\left\{E_5, E_8\right\} \\ P(A \cap B) & =P\left(E_5\right)+P\left(E_8\right)=0.1+0.7=0.17 \end{aligned} \end{aligned}$$

On substituting these values in Eq.(i), we get

$$P(A \cup B)=0.25+0.32-0.17=0.40$$

(iii) $A \cup B=\left\{E_1, E_2, E_5, E_8, E_9\right\}$

$$\begin{aligned} P(A \cup B) & =P\left(E_1\right)+P\left(E_2\right)+P\left(E_5\right)+P\left(E_8\right)+P\left(E_9\right) \\ & =0.08+0.08+0.1+0.07+0.07=0.40 \end{aligned}$$

(iv) $$\begin{aligned} & \because P(\bar{B})=1-P(B)=1-0.32=0.68 \\ & \text { and } \quad \bar{B}=\left\{E_1, E_3, E_4, E_6, E_7\right\} \\ & \therefore \quad P(\bar{B})=P\left(E_1\right)+P\left(E_3\right)+P\left(E_4\right)+P\left(E_6\right)+P\left(E_7\right) \\ & =0.08+0.1+0.1+0.2+0.2=0.68 \end{aligned}$$

(i) When a die is throw the possible outcomes are

$$\begin{aligned} S & =\{1,2,3,4,5,6\} \text { out of which } 1,3,5 \text { are odd, } \\ \therefore \text { Required probability } & =\frac{3}{6}=\frac{1}{2} \end{aligned}$$

(ii) When a fair coin is tossed two times the sample space is

$$S=\{H H, H T, T H, T T\}$$

In at least one head favourable enonts are $H H, H T, T H$

$\therefore$ Required probability $=\frac{3}{4}$

(iii) Total cards $=52$

$$\begin{aligned} \text { Favourable } & =4 \text { king }+2 \text { of heart }+3 \text { of spade }=4+1+1=6 \\ \therefore \quad \text { Required probability } & =\frac{6}{52}=\frac{3}{26} \end{aligned}$$

(iv) When a pair of dice is rolled total sample parts are 36 . Out of which $(1,5),(5,1),(2,4)$, $(4,2)$ and $(3,3)$.

$\therefore$ Required probability $=\frac{5}{36}$