$$\eqalign{ & {\text{Let the number selected by }}xy{\text{. Then,}} \cr & x + y = 9,\,0 \leqslant x,\,y \leqslant 9{\text{ and}} \cr & xy = 0 \Rightarrow x = 0,\,y = 9 \cr & {\text{or }}y = 0,\,x = 9 \cr & P\left( {{x_1} = \frac{9}{{{x_2}}} = 0} \right) = \frac{{P\left( {{x_1} = 9 \cap {x_2} = 0} \right)}}{{P\left( {{x_2} = 0} \right)}} \cr & {\text{Now, }}P\left( {{x_2} = 0} \right) = \frac{{19}}{{100}}{\text{ and}} \cr & P\left( {{x_1} = 9 \cap {x_2} = 0} \right) = \frac{2}{{100}} \cr & \Rightarrow P\left( {{x_1} = \frac{9}{{{x_2}}} = 0} \right) = \frac{{\frac{2}{{100}}}}{{\frac{{19}}{{100}}}} = \frac{2}{{19}} \cr} $$

Given : Probability of aeroplane $$I$$ scoring a target correctly i.e., $$P(I)$$  = 0.3 probability of scoring a target correctly by aeroplane $$II,$$ i.e. $$P(II)$$  = 0.2
$$\therefore \,\,P\left( {\overline I } \right) = 1 - 0.3 = 0.7$$
∴ The required probability
$$\eqalign{ & = P\left( {\overline I \cap II} \right) = P\left( {\overline I } \right).P\left( {II} \right) \cr & = 0.7 \times 0.2 \cr & = 0.14 \cr} $$

$$n\left( S \right) = $$  the area of the circle of radius $$r.$$
$$n\left( E \right) = $$  the area of the circle of radius $$\frac{r}{2}.$$
$$\therefore $$  the required probability $$ = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{\pi .{{\left( {\frac{r}{2}} \right)}^2}}}{{\pi {r^2}}} = \frac{1}{4}$$

$$\eqalign{ & P\left( {\overline {A \cup B} } \right) = \frac{1}{6},P\left( {A \cap B} \right) = \frac{1}{4}\,{\text{and }}P\left( {\overline A } \right) = \frac{1}{4} \cr & \Rightarrow \,\,P\left( {A \cup B} \right) = \frac{5}{6}P\left( A \right) = \frac{3}{4} \cr & {\text{Also }} \Rightarrow \,P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) \cr & \Rightarrow \,\,P\left( B \right) = \frac{5}{6} - \frac{3}{4} + \frac{1}{4} = \frac{1}{3} \cr & \Rightarrow \,\,P\left( A \right)P\left( B \right) = \frac{3}{4} - \frac{1}{3} = \frac{1}{4} \cr & = P\left( {A \cap B} \right) \cr} $$
Hence $$A$$ and $$B$$ are independent but not equally likely.

$$\eqalign{ & {\text{We have}}\,{\text{;}} \cr & np = 2 = {\text{mean }} \cr & npq = 1 = {\text{ variance}} \cr & \Rightarrow p = \frac{1}{2}\,;\,q = \frac{1}{2}\,\,\& \,\,n = 4 \cr & {\text{Required probability}} = P\left( {x > 1} \right) \cr & = 1 - P\left( {x \leqslant 1} \right) \cr & = 1 - \left[ {P\left( {x = 0} \right) + P\left( {x = 1} \right)} \right] \cr & = 1 - \left[ {{}^4{C_0}{q^4} + {}^4{C_1}{q^3}{p^1}} \right] \cr & = 1 - \frac{5}{{16}} \cr & = \frac{{11}}{{16}} \cr} $$

We know that,
$${7^1} = 7,{7^2} = 49,{7^3} = 343,{7^4} = 2401,{7^5} = 16807$$
∴ $${7^k}$$ (where $$k \in Z$$   ), results in a number whose unit’s digit 7 or 9 or 3 or 1.
Now, $${7^m} + {7^n}$$  will be divisible by 5 if unit’s place digit of resulting number is 5 or 0 clearly it can never be 5.
But it can be 0 if we consider values of $$m$$ and $$n$$ such that the sum of unit’s place digits become 0. And this can be done by choosing
\[\begin{array}{l} \left. \begin{array}{l} m = 1,5,9,.....97\\ {\rm{and \,correspondingly}}\\ n = 3,7,11,.....99 \end{array} \right\}\,\,\left( {25\,{\rm{options\, each}}} \right)\left[ {7 + 3 = 10} \right]\\ \left. \begin{array}{l} m = 2,6,10,.....98\\ {\rm{and}}\\ n = 4,8,12,.....100 \end{array} \right\}\,\,\left( {25\,{\rm{options \,each}}} \right)\left[ {9 + 1 = 10} \right] \end{array}\]
Case I : Thus $$m$$ can be chosen in 25 ways and $$n$$ can be chosen in 25 ways
Case II : $$m$$ can be chosen in 25 ways and $$n$$ can be chosen in 25 ways
∴ Total no. of selections of $$m, n$$  so that $${7^m} + {7^n}$$  is divisible by $$5 = \left( {25 \times 25 + 25 \times 25} \right) \times 2$$
Note we can interchange values of $$m$$ and $$n.$$
Also no. of total possible selections of $$m$$ and $$n$$ out of $$100 = 100 \times 100$$
∴ Req. prob. $$ = \frac{{2\left( {25 \times 25 + 25 \times 25} \right)}}{{100 \times 100}} = \frac{1}{4}.$$

$$\eqalign{ & P\,\,\left( {{\text{exactly one event out of }}A{\text{ and }}B{\text{ occurs}}} \right) \cr & = P\left[ {\left( {A \cap B'} \right) \cup \left( {A' \cap B} \right)} \right] \cr & = P\left( {A \cup B} \right) - P\left( {A \cap B} \right) \cr & = P\left( A \right) + P\left( B \right) - 2P\left( {A \cap B} \right) \cr & \therefore \,P\left( A \right) + P\left( B \right) - 2P\left( {A \cap B} \right) = 1 - a......\left( 1 \right) \cr & {\text{Similarly,}} \cr & P\left( B \right) + P\left( C \right) - 2P\left( {B \cap C} \right) = 1 - 2a......\left( 2 \right) \cr & P\left( C \right) + P\left( A \right) - 2P\left( {C \cap A} \right) = 1 - a......\left( 3 \right) \cr & P\left( {A \cap B \cap C} \right) = {a^2}......\left( 4 \right) \cr & {\text{Now,}} \cr & P\left( {A \cup B \cup C} \right) = P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {B \cap C} \right) - P\left( {C \cap A} \right) + P\left( {A \cap B \cap C} \right) \cr & = \frac{1}{2}\left[ {P\left( A \right) + P\left( B \right) - 2P\left( {B \cap C} \right) + P\left( B \right) + P\left( C \right) - 2P\left( {B \cap C} \right) + P\left( C \right) + P\left( A \right) - 2P\left( {C \cap A} \right) + P\left( {A \cap B \cap C} \right)} \right] \cr & = \frac{1}{2}\left[ {1 - a + 1 - 2a + 1 - a} \right] + {a^2}\,\,\,\,\left[ {{\text{using}}\,\left( 1 \right),\left( 2 \right),\left( 3 \right){\text{ and }}\left( 4 \right)} \right] \cr & = \frac{3}{2} - 2a + {a^2} \cr & = \frac{1}{2} + {\left( {a - 1} \right)^2} > \frac{1}{2}. \cr} $$

$$\eqalign{ & \sum {p\left( x \right)} = 1 \cr & \Rightarrow 9p + 10{p^2} = 1 \cr & \Rightarrow 10{p^2} + 9p - 1 = 0 \cr & \Rightarrow p = - 1{\text{ or }}\frac{1}{{10}} \cr & \therefore \,p = \frac{1}{{10}} \cr} $$

Let 5 horses are $${H_1},$$ $${H_2},$$ $${H_3},$$ $${H_4},$$ and $${H_5}.$$ Selected pair of horses will be one of the 10 pairs $$\left( {{\text{i}}{\text{.e}}{\text{.;}}{{\text{ }}^5}{C_2}} \right):{H_1}{H_2},$$     $${H_1}{H_3},$$   $${H_1}{H_4},$$   $${H_1}{H_5},$$   $${H_2}{H_3},$$   $${H_2}{H_4},$$   $${H_2}{H_5},$$   $${H_3}{H_4},$$   $${H_3}{H_5},$$   and $${H_4}{H_5}.$$
Any horse can win the race in 4 ways.
For example : Horses $${H_2}$$ win the race in 4 ways $${H_1}{H_2},$$   $${H_2}{H_3},$$   $${H_2}{H_4},$$   and $${H_2}{H_5}.$$
Hence required probability $$ = \frac{4}{{10}} = \frac{2}{5}$$

Let $${E_3} = $$  the event of the sum being $$3.$$ Similarly, $${E_6},\,{E_9},\,{E_{12}},\,{E_{15}},\,{E_{18}},\,{E_{21}}.n\left( S \right) = {}^{12}{C_2}$$
$$\eqalign{ & n\left( {{E_3}} \right) = 1,\,n\left( {{E_6}} \right) = 2,\,n\left( {{E_9}} \right) = 4,\,n\left( {{E_{12}}} \right) = 5, \cr & n\left( {{E_{15}}} \right) = 5,\,n\left( {{E_{18}}} \right) = 3,n\left( {{E_{21}}} \right) = 2 \cr & \therefore \,P\left( E \right) = P\left( {{E_3}} \right) + ...... + P\left( {{E_{21}}} \right) \cr & = \frac{{1 + 2 + 4 + 5 + 5 + 3 + 2}}{{{}^{12}{C_2}}} \cr & = \frac{{22 \times 2}}{{12 \times 11}} \cr & = \frac{1}{3} \cr} $$
Alternatively : The sum of the numbers for every selection is divisible by $$3$$ or leaves the remainder $$1$$ or leaves the remainder $$2.$$ These are equally probable. So, the required probability $$ = \frac{1}{3},$$  because the sum of the three probabilities is $$1.$$