$$\begin{aligned} &\text { Referring to the above solution, we have }\\ &\begin{array}{rlrl} & A =\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\} \\ \Rightarrow & n(A) =6 \text { and } n(S)=6^2=36 & \text { [where, } S \text { is sample space] } \\ \therefore & P(A) =\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6} \end{array} \end{aligned}$$

$$\begin{array}{lrl} \text { and } & B & =\{(4,6),(6,4),(5,5),(6,5),(5,6),(6,6)\} \\ \Rightarrow & n(B) & =6 \text { and } n(S)=6^2=36 \\ \therefore & P(B) & =\frac{n(B)}{n(S)}=\frac{6}{36}=\frac{1}{6} \end{array}$$

$$\begin{array}{ll} \text { Also, } & A \cap B=\{(5,5),(6,6)\} \\ \Rightarrow & n(A \cap B)=2 \text { and } n(S)=36 \\ \therefore & P(A \cap B)=\frac{2}{36}=\frac{1}{18} \\ \text { Also, } & P(A) \cdot P(B)=\frac{1}{36} \end{array}$$

Thus, $\quad P(A \cap B) \neq P(A) \cdot P(B) \quad\left[\because \frac{1}{18} \neq \frac{1}{36}\right]$

So, we can say that both $A$ and $B$ are not independent events.

$$\begin{array}{ll} \text { Thus, } & P(A \cup B)=0.6 \text { and } P(A \cap B)=0.3 \\ \text { Also, } & P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ \Rightarrow & 0.6=P(A)+P(B)-0.3 \end{array}$$

$$\begin{array}{lrr} \Rightarrow & P(A)+P(B)=0.9 & \\ \Rightarrow & {[1-P(\bar{A})]+[1-P(\bar{B})]=0.9} & {[\because P(A)=1-P(\bar{A}) \text { and } P(B)=1-P(\bar{B})]} \\ \Rightarrow & P(\bar{A})+P(\bar{B})=2-0.9=1.1 & \end{array}$$

Let $R=\{5$ red marbles $\}$ and $B=\{3$ black marbles $\}$

For atleast one of the three marbles drawn be black, if the first marble is red, then the following three conditions will be followed

(i) Second ball is black and third is red $\left(E_1\right)$.

(ii) Second ball is black and third is also black ( $E_2$ ).

(iii) Second ball is red and third is black $\left(E_3\right)$.

$$\begin{aligned} \therefore & P\left(E_1\right)=P\left(R_1\right) \cdot P\left(B_1 / R_1\right) \cdot P\left(R_2 / R_1 B_1\right)=\frac{5}{8} \cdot \frac{3}{7} \cdot \frac{4}{6}=\frac{60}{336}=\frac{5}{28} \\ & P\left(E_2\right)=P\left(R_1\right) \cdot P\left(B_1 / R_1\right) \cdot P\left(B_2 / R_1 B_1\right)=\frac{5}{8} \cdot \frac{3}{7} \cdot \frac{2}{6}=\frac{30}{336}=\frac{5}{56} \end{aligned}$$

$$ \begin{array}{ll} \text { and } & P\left(E_3\right)=P\left(R_1\right) \cdot P\left(R_2 / R_1\right) \cdot P\left(B_1 / R_1 R_2\right)=\frac{5}{8} \cdot \frac{4}{7} \cdot \frac{3}{6}=\frac{60}{336}=\frac{5}{28} \\ \therefore & P(E)=P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)=\frac{5}{28}+\frac{5}{56}+\frac{5}{28} \end{array}$$

$=\frac{10+5+10}{56}=\frac{25}{56}$

$$\begin{aligned} &\text { Two dice are thrown together i.e., sample space }(S)=36 \Rightarrow n(S)=36\\ &E=\text { A total of } 4=\{(2,2),(3,1),(1,3)\} \end{aligned}$$

$$\begin{array}{l} \Rightarrow & n(E) =3 \\ & F =\text { A total of } 9 \text { or more } \\ & =\{(3,6),(6,3),(4,5),(4,6),(5,4),(6,4),(5,5),(5,6),(6,5),(6,6)\} \\ \Rightarrow & n(F) =10 \\ \end{array}$$

$G=$ a total divisible by $5=\{(1,4),(4,1),(2,3),(3,2),(4,6),(6,4),(5,5)\}$

$\Rightarrow \quad n(G)=7$

$$\begin{array}{lc} \text { Here, } & (E \cap F)=\phi \text { and }(E \cap G)=\phi \\ \text { Also, } & (F \cap G)=\{(4,6),(6,4),(5,5)\} \\ \Rightarrow & n(F \cap G)=3 \text { and }(E \cap F \cap G)=\phi \\ \therefore & P(E)=\frac{n(E)}{n(S)}=\frac{3}{36}=\frac{1}{12} \end{array}$$

$$\begin{aligned} P(F) & =\frac{n(F)}{n(S)}=\frac{10}{36}=\frac{5}{18} \\ P(G) & =\frac{n(G)}{n(S)}=\frac{7}{36} \\ P(F \cap G) & =\frac{3}{36}=\frac{1}{12} \\ \text{and}\quad P(F) \cdot P(G) & =\frac{5}{18} \cdot \frac{7}{36}=\frac{35}{648} \end{aligned}$$

Here, we see that $P(F \cap G) \neq P(F) \cdot P(G)$

[since, only $F$ and $G$ have common events, so only $F$ and $G$ are used here]

Hence, there is no pair which is independent.

We know that, a random variable $X$ taking values $0,1,2, \ldots, n$ is said to have a binomial distribution with parameters $n$ and $P$, if its probability distribution is given by

$$\begin{aligned} P(X & =r)={ }^n C_r p^{\prime} q^{n-r} \\ \text{where,}\quad q & =1-p \\ \text{and}\quad r & =0,1,2, \ldots, n \end{aligned}$$

Similarly, in an experiment of tossing a coin three times, we have $n=3$ and random variable $X$ can take values $r=0,1,2$ and 3 with $p=\frac{1}{2}$ and $q=\frac{1}{2}$

So, we see that in the experiment of tossing a coin three times, we have random variable $X$ which can take values $0,1,2$ and 3 with parameters $n=3$ and $P=\frac{1}{2}$. Therefore, it is said to have a Binomial distribution.