(2)

Let us defined the events as

A : Unbiased coin is selected

B : Biased coin in selected

H : Head turns up.

$\therefore P\left(H\right)=P\left(A\right)P(H/A)+P\left(B\right)P(H/B)$

                     $=\frac{19}{20}\times \frac{1}{2}+\frac{1}{20}\times 1=\frac{21}{40}$

Now, $P(A/H)=\frac{P\left(A\right)P(H/A)}{P\left(H\right)}=\frac{{\displaystyle \frac{19}{20}}\times {\displaystyle \frac{1}{2}}}{{\displaystyle \frac{21}{40}}}=\frac{19}{21}$

$\Rightarrow \frac{m}{n}=\frac{19}{21}$

$\therefore n^2–m^2=441–361=80$.

(4)

          $P(A\cap \overline{B})=P\left(A\right)–P(A\cap B)$

$\Rightarrow P(A\cap B)=P\left(A\right)–P(A\cap \overline{B})$

                                     =0.7 – 0.5 = 0.2

Now, $P(A\cup \overline{B})=P\left(A\right)–P\left(\overline{B}\right)–P(A\cap \overline{B})$

                                     = 0.7+(1– 0.4) – 0.5 = 0.8

$P(B\cap (A\cup \overline{B}\left)\right)=P(B\cap A)\cup (B\cap \overline{B})$

                                  $=P(B\cap A) \left[P\right(B\cap \overline{B})=0]$

                                         = 0.2

Now, $P(B/(A\cup \overline{B}\left)\right)=\frac{P(B\cap (A\cup \overline{B}\left)\right)}{P(A\cup \overline{B})}=\frac{0.2}{0.8}=\frac{1}{4}$.

(4)

Required probability

          $=\frac{{\displaystyle \frac{6}{10}}\times {\displaystyle \frac{5}{9}}}{{\displaystyle \frac{4}{10}}\times {\displaystyle \frac{6}{9}}+{\displaystyle \frac{6}{10}}\times {\displaystyle \frac{5}{9}}}=\frac{30}{24+30}=\frac{30}{54}=\frac{5}{9}$

So, m = 5, n = 9          [$\because$ gcd (m, n) = 1]

$\therefore$   m + n = 14.

(4)

We have, $12x^2–7x+1=0$

$\Rightarrow x=\frac{1}{3},\frac{1}{4}$

Let $P\left(\frac{A}{B}\right)=\frac{1}{3} and P\left(\frac{B}{A}\right)=\frac{1}{4}$

$\Rightarrow \frac{P(A\cap B)}{P\left(B\right)}=\frac{1}{3} \mathrm{and} \frac{P(A\cap B)}{P\left(A\right)}=\frac{1}{4}$

$\Rightarrow P\left(B\right)=0.3 \mathrm{and} P\left(A\right)=0.4$

$\therefore P(A\cup B)=0.3+0.4–0.1=0.6$

Now, $\frac{P(\overline{A}\cup \overline{B})}{P(\overline{A}\cap \overline{B})}=\frac{P\left(\overline{A\cap B}\right)}{P\left(\overline{A\cup B}\right)}$

                                  $=\frac{1–P(A\cap B)}{1–P(A\cup B)}=\frac{1–0.1}{1–0.6}=\frac{9}{4}$.

(1)

Let x be the number shows on dice –1 and y be the number shows on dice –2.

Favorable outcomes = {(x, y) = (1, 3), (3, 1), (2, 2), (2, 3), (3, 2), (1, 4), (4, 1)}

$\therefore$   Required probability

                        $=\frac{2}{6}\times \frac{2}{6}+\frac{1}{6}\times \frac{1}{6}+\frac{2}{6}\times \frac{2}{6}+\frac{2}{6}\times \frac{2}{6}+\frac{1}{6}\times \frac{2}{6}+\frac{2}{6}\times \frac{1}{6}+\frac{1}{6}\times \frac{1}{6}$

                        $=\frac{18}{36}=\frac{1}{2}$.

(1)

Total ways for selecting any two squares $={}^{16}C_2=120$

Total ways for selecting common side squares $=\underset{\mathrm{Horizontal} \mathrm{side}}{\underbrace{3\times 4}}+\underset{\mathrm{Vertical} \mathrm{side}}{\underbrace{3\times 4}}=24$

So, required probability $=1–\frac{24}{120}=\frac{96}{120}=\frac{4}{5}$.

(2)

Let $E_1$ be the event that A get sum of 5 and $E_2$ be the event that B get sum of 8.

                            $P\left(E_1\right)=\frac{4}{36}=\frac{1}{9} \mathrm{and} P\left(E_2\right)=\frac{5}{36}$

$\therefore$   Required Probability

                               $=P\left(E_1\right)+P\left({\overline{E}}_1\right)P\left({\overline{E}}_2\right)P\left(E_1\right)+P\left({\overline{E}}_1\right)P\left({\overline{E}}_2\right)P\left({\overline{E}}_1\right)P\left({\overline{E}}_2\right)P\left(E_1\right)+...\infty$

                               $=\frac{1}{9}+\frac{8}{9}\times \frac{31}{36}\times \frac{1}{9}+\frac{8}{9}\times \frac{31}{36}\times \frac{8}{9}\times \frac{31}{36}\times \frac{1}{9}+...\infty$

                               $=\frac{{\displaystyle \frac{1}{9}}}{1–{\displaystyle \frac{62}{81}}}=\frac{9}{19}$.

(2)

Given : $A={\left[a_{ij}\right]}_{2\times 2}$ with entries 0 or 1.

Total possible ways for matrix A are $4\times 4=16$ ways.

Here E = {$\mathrm{A}|\mathrm{A}$ is invertible}

$\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right],\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right],\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right],\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right],\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \mathrm{and} \left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$ = 6 matrices

Thus, probability P(E) is given by $=\frac{6}{16}=\frac{3}{8}$.

(4)

Given, $i^{k_1}+i^{k_2}\ne 0$

Possible values of $i^{k_1}$= 4 = Possible values of $i^{k_2}$

$\therefore$  Cases are i, –1, – i, 1

$\therefore$  Total number cases $=4\times 4=16$

For $i^{k_1}+i^{k_2}=0,$

Cases are (1, –1), (–1, 1), (i, – i), (– i, i)

$\therefore$  Required probability $=1–\frac{4}{16}=\frac{12}{16}=\frac{3}{4}$.

(2)

Consider the events, $E_1$: Bag $B_1$ is selected, $E_2$: Bag $B_2$ is selected, $E_3$: Bag $B_3$ is selected

A : Ball drawn is white. Then $P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=\frac{1}{3}$

$P(A/E_1)=\frac{6}{10},P(A/E_2)=\frac{4}{10},P(A/E_3)=\frac{5}{10}$

Required probability

$P\left(\frac{E_2}{A}\right)=\frac{P\left(E_2\right)P\left({\displaystyle \frac{A}{E_2}}\right)}{P\left(E_1\right)P\left({\displaystyle \frac{A}{E_1}}\right)+P\left(E_2\right)P\left(\frac{A}{E_2}\right)+P\left(E_3\right)P\left(\frac{A}{E_3}\right)}$

                     $=\frac{{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{4}{10}}}{\left({\displaystyle \frac{1}{3}}\right)\left({\displaystyle \frac{6}{10}}\right)+\left({\displaystyle \frac{1}{3}}\right)\left({\displaystyle \frac{4}{10}}\right)+\left(\frac{1}{3}\right)\left(\frac{5}{10}\right)}=\frac{{\displaystyle \frac{4}{30}}}{{\displaystyle \frac{6}{30}}+{\displaystyle \frac{4}{30}}+{\displaystyle \frac{5}{30}}}=\frac{4}{15}$