The probability of getting at least one head
$$ = 1 - $$  probability of getting no heads
$$\eqalign{ & = 1 - {}^n{C_0}{\left( {\frac{1}{2}} \right)^0}.{\left( {\frac{1}{2}} \right)^n} \cr & = 1 - \frac{1}{{{2^n}}} \cr} $$
The probability of getting at least two tails
$$ = 1 - $$  probability of getting no tails $$-$$ probability of getting $$1$$ tail
$$\eqalign{ & = 1 - {}^n{C_n}{\left( {\frac{1}{2}} \right)^n}.{\left( {\frac{1}{2}} \right)^0} - {}^n{C_{n - 1}}{\left( {\frac{1}{2}} \right)^{n - 1}}.\frac{1}{2} \cr & = 1 - \frac{1}{{{2^n}}} - n\frac{1}{{{2^n}}} \cr} $$
From the question,
$$\eqalign{ & \left( {1 - \frac{1}{{{2^n}}}} \right) - \left( {1 - \frac{{1 + n}}{{{2^n}}}} \right) = \frac{5}{{12}} \cr & {\text{or }}\frac{{n + 1}}{{{2^n}}} - \frac{1}{{{2^n}}} = \frac{5}{{32}} \cr & {\text{or }}\frac{n}{{{2^n}}} = \frac{5}{{32}}\,\,\,\,\,\,\, \Rightarrow n = 5 \cr} $$

Total number of students in four schools $$ = 12 + 20 + 13 + 17 = 62$$
Now, one student is selected at random.
$$\therefore $$  Total outcomes $$ = {}^{62}{C_1}$$
Now, number of students in school $${B_2} = 20$$
Number of ways to select a student from $${B_2} = {}^{20}{C_1}$$
$$\therefore $$  Required probability $$ = \frac{{{}^{20}{C_1}}}{{{}^{62}{C_1}}} = \frac{{20}}{{62}} = \frac{{10}}{{31}}$$

$$\eqalign{ & n\left( S \right) = {}^{10}{C_4} = 210 \cr & n\left( E \right) = {}^5{C_2} \times {}^3{C_1} \times {}^2{C_1} + {}^5{C_1} \times {}^3{C_2} \times {}^2{C_1} + {}^5{C_1} \times {}^3{C_1} \times {}^2{C_2} = 105 \cr & \therefore \,P\left( E \right) = \frac{{105}}{{210}} = \frac{1}{2}. \cr} $$

$$\eqalign{ & P\left( {{T_1}} \right) = \frac{{20}}{{100}},P\left( {{T_2}} \right) = \frac{{80}}{{100}},P\left( D \right) = \frac{7}{{100}} \cr & {\text{Let }}P\left( {\frac{D}{{{T_2}}}} \right) = x,\,{\text{then }}P\left( {\frac{D}{{{T_1}}}} \right) = 10x \cr & {\text{Also }}P\left( D \right) = P\left( {{T_1}} \right)P\left( {\frac{D}{{{T_1}}}} \right) + P\left( {{T_2}} \right)P\left( {\frac{D}{{{T_2}}}} \right) \cr & \Rightarrow \,\,\frac{7}{{100}} = \frac{{20}}{{100}} \times 10x + \frac{{80}}{{100}} \times x \cr & \Rightarrow \,\,\frac{7}{{280}} = x\,\,{\text{or }}x = \frac{1}{{40}} \cr & P\left( {\frac{D}{{{T_1}}}} \right) = \frac{{10}}{{40}}\,\,{\text{and }}P\left( {\frac{D}{{{T_2}}}} \right) = \frac{1}{{40}} \cr & \Rightarrow \,\,P\left( {\frac{{\overline D }}{{{T_1}}}} \right) = \frac{{30}}{{40}}\,\,{\text{and}}{\text{ }}P\left( {\frac{{\overline D }}{{{T_2}}}} \right) = \frac{{39}}{{40}} \cr & P\left( {\frac{{{T_2}}}{{\overline D }}} \right) = \frac{{P\left( {\frac{{\overline D }}{{{T_2}}}} \right)P\left( {{T_2}} \right)}}{{P\left( {\frac{{\overline D }}{{{T_1}}}} \right)P\left( {{T_1}} \right) + P\left( {\frac{{\overline D }}{{{T_2}}}} \right)P\left( {{T_2}} \right)}} \cr & = \frac{{\frac{{80}}{{100}} \times \frac{{39}}{{40}}}}{{\frac{{20}}{{100}} \times \frac{{30}}{{40}} + \frac{{80}}{{100}} \times \frac{{39}}{{40}}}} \cr & = \frac{{156}}{{186}} \cr & = \frac{{26}}{{31}} \cr & {\text{Also }}\frac{{78}}{{93}} = \frac{{26}}{{31}} \cr} $$

Note :- The question should state ‘3 different’ boxes instead of ‘3 identical boxes’ and one particular box has 3 balls. Then the solution would be:
Required probability $$ = \frac{{^{12}{C_3} \times {2^9}}}{{{3^{12}}}} = \frac{{55}}{3}{\left( {\frac{2}{3}} \right)^{11}}$$

$$n\left( S \right) = {}^{52}{C_2}{\text{ and}}$$
$$n\left( E \right) = $$   the number of selections of $$1$$ spade, $$1$$ ace from $$3$$ aces or selections of the ace of spade and $$1$$ other spade
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {}^{13}{C_1} \times {}^3{C_1} + {}^{12}{C_1} \times {}^1{C_1} = 51 \cr & \therefore \,\,\,\,P\left( E \right) = \frac{{51}}{{{}^{52}{C_2}}} = \frac{1}{{26}}. \cr} $$

$$\eqalign{ & P\left( A \right) = \frac{6}{{12}} = \frac{1}{2},\,P\left( B \right) = \frac{4}{{12}} = \frac{1}{3} \cr & {\text{Required probability}} \cr & = \frac{1}{2}.\frac{1}{3}.\frac{1}{2} + \frac{1}{3}.\frac{1}{2}.\frac{1}{3} \cr & = \frac{1}{{12}} + \frac{1}{{18}} \cr & = \frac{5}{{36}} \cr} $$

We know $${7^k},\,k\, \in \,N,$$   has $$1,\,3,\,9,\,7$$   at the units place for $$k = 4p,\,4p - 1,\,4p - 2,\,4p - 3$$       respectively, where $$p = 1,\,2,\,3,.....$$
Clearly, $${7^m} + {7^n}$$   will be divisible by $$5$$ if $${7^m}$$ has $$3$$ or $$7$$ in the units place and $${7^n}$$ has $$7$$ or $$3$$ in the units place or $${7^m}$$ has $$1$$ or $$9$$ in the units place and $${7^n}$$ has $$9$$ or $$1$$ in the units place.
$$\therefore $$  for any choice of $$m,\,n$$  the digit in the units place of $${7^m} + {7^n}$$   is $$2,\,4,\,6,\,0$$   or $$8.$$ It is divisible by $$5$$ only when this digit is $$0.$$
$$\therefore $$  the required probability $$ = \frac{1}{5}.$$

Five tickets out of $$50$$  can drawn in $${{}^{50}{C_5}}$$  ways.
Since $${{x_1} < {x_2} < {x_3} < {x_4} < {x_5}}$$     and $${x_3} = 30,\,{x_1},\,{x_2} < 30,$$     i.e., $${x_1}$$ and $${x_2}$$ should come from tickets numbered $$1$$ and $$29$$  and this may happen in $${{}^{29}{C_2}}$$  ways.
Remaining ways, i.e., $${x_4},\,{x_5} > 30,$$   should come from $$20$$ tickets numbered $$31$$  to $$50$$  in $${{}^{20}{C_2}}$$  ways.
So, favourable number of cases $$ = {}^{29}{C_2}{}^{29}{C_2}$$
Hence, required probability $$ = \frac{{{}^{20}{C_2} \times {}^{29}{C_2}}}{{{}^{50}{C_5}}}$$

Let the number of red and blue balls be $$r$$ and $$b$$, respectively. Then, the probability of drawing two red balls
$$ = {p_1} = \frac{{{}^r{C_2}}}{{{}^{r + b}{C_2}}} = \frac{{r\left( {r - 1} \right)}}{{\left( {r + b} \right)\left( {r + b - 1} \right)}}$$
The probability of drawing two blue balls is
$$ = {p_2} = \frac{{{}^b{C_2}}}{{{}^{r + b}{C_2}}} = \frac{{b\left( {b - 1} \right)}}{{\left( {r + b} \right)\left( {r + b - 1} \right)}}$$
The probability of drawing one red and one blue ball
$$ = {p_3} = \frac{{{}^r{C_1}{}^b{C_1}}}{{{}^{r + b}{C_2}}} = \frac{{2br}}{{\left( {r + b} \right)\left( {r + b - 1} \right)}}$$
By hypothesis, $${p_1} = 5{p_2}$$   and $${p_3} = 6{p_2}.$$
Therefore, $$r\left( {r - 1} \right) = 5b\left( {b - 1} \right)$$     and $$2br = 6b\left( {b - 1} \right) \Rightarrow r = 6,\,b = 3.$$