Given equation
$$\eqalign{ & x + \frac{{100}}{x} > 50 \cr & \Rightarrow {x^2} - 50x + 100 > 0\,\, \Rightarrow {\left( {x - 25} \right)^2} > 525 \cr & \Rightarrow x - 25 < - \sqrt {\left( {525} \right)} {\text{ or }}x - 25 > \sqrt {\left( {525} \right)} \cr & \Rightarrow x < 25 - \sqrt {\left( {525} \right)} {\text{ or }}x > 25 + \sqrt {\left( {525} \right)} \cr} $$
As $$x$$ is positive integer and $$\sqrt {\left( {525} \right)} = 22.91,$$    we must have
$$x \leqslant 2{\text{ or }}x \geqslant 48$$
Let $$E$$ be the event for favourable cases and $$S$$ be the sample space.
$$\eqalign{ & \therefore \,E = \left\{ {1,\,2,\,48,\,49,\,......,\,100} \right\} \cr & \therefore \,n\left( E \right) = 55{\text{ and }}n\left( S \right) = 100 \cr} $$
Hence the required probability
$$P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{55}}{{100}} = \frac{{11}}{{20}}$$

The probability of getting an odd number in a throw $$ = \frac{3}{6} = \frac{1}{2}$$
The probability of getting an odd number $$2$$ times in $$n$$ trials $$ = {}^n{C_2}.{\left( {\frac{1}{2}} \right)^2}.{\left( {\frac{1}{2}} \right)^{n - 2}}$$
Similarly, the probability of getting an even number $$3$$ times in $$n$$ trials $$ = {}^n{C_3}.{\left( {\frac{1}{2}} \right)^3}.{\left( {\frac{1}{2}} \right)^{n - 3}}$$
From the question, $$ {}^n{C_2}.{\left( {\frac{1}{2}} \right)^n} = {}^n{C_3}.{\left( {\frac{1}{2}} \right)^n}\,\,\,\,\, \Rightarrow n = 5$$
$$\therefore $$  the required probability
$$\eqalign{ & = {}^5{C_1}.\left( {\frac{1}{2}} \right).{\left( {\frac{1}{2}} \right)^4} + {}^5{C_3}.{\left( {\frac{1}{2}} \right)^3}.{\left( {\frac{1}{2}} \right)^2} + {}^5{C_5}.{\left( {\frac{1}{2}} \right)^5} \cr & = \left( {5 + 10 + 1} \right).\frac{1}{{{2^5}}} \cr & = \frac{{16}}{{32}} \cr & = \frac{1}{2} \cr} $$

Required number of elementary events $$ = {}^8{C_2} \times 2! = 56$$

Possible outcomes are (Head, 1), (Head, 2), (Head, 3), (Head, 4), (Head, 5), (Head, 6), Tail.

$$\eqalign{ & n\left( S \right) = \frac{{9!}}{{4!3!2!}}{\text{ and }}n\left( E \right) = 3! \cr & \therefore \,\,P\left( E \right) = \frac{{\left( {3!} \right)\left( {4!} \right)\left( {3!} \right)\left( {2!} \right)}}{{9!}} = \frac{1}{{210}}. \cr} $$

Probability of hitting the target $$ = 0.3$$
If $$'n'$$ is the number of times that the Missile is fired.
$$\therefore $$  Probability of hitting at least once
$$\eqalign{ & \Rightarrow 1 - {\left[ {1 - 0.3} \right]^n} = 0.8 \cr & \Rightarrow {0.7^n} = 0.2 \cr & \Rightarrow n\,\log \,0.7 = \log \,0.2 \cr & \Rightarrow n = 4.512 \cr & {\text{for }}n = 4\,;\,p < 0.8 \cr & {\text{taken }}n = 5 \cr & \boxed{n = 5} \cr} $$
Hence $$5$$ missiles should be fired so that there is at least $$80\% $$  probability that the target is hit.

Let $$A,\,B$$  and $$C$$ be the events that the student is successful in tests I, II and III respectively.
Then $$P$$ (The student is successful)
$$\eqalign{ & = P\left( A \right)P\left( B \right)\left\{ {1 - P\left( C \right)} \right\} + P\left( A \right)\left\{ {1 - P\left( B \right)} \right\}P\left( C \right) + P\left( A \right)P\left( B \right)P\left( C \right) \cr & = p.q\left( {1 - \frac{1}{2}} \right) + p\left( {1 - q} \right)\frac{1}{2} + p.q\frac{1}{2} \cr & = \frac{1}{2}pq + \frac{1}{2}p\left( {1 - q} \right) + \frac{1}{2}pq \cr & = \frac{1}{2}\left( {pq + p - pq + pq} \right) \cr & = \frac{1}{2}\left( {pq + p} \right) \cr & \therefore \,\frac{1}{2} = \frac{1}{2}\left( {pq + p} \right) \Rightarrow 1 = pq + p \cr} $$

Let $${E_1}$$ denote the event "a coin with head on both sides is selected" and $${E_2}$$ denotes the event "a fair coin is selected".
Let $$A$$ be the event "he toss, results in heads".
$$\eqalign{ & \therefore \,P\left( {{E_1}} \right) = \frac{1}{{n + 1}},\,P\left( {{E_2}} \right) = \frac{n}{{n + 1}}{\text{ and}} \cr & P\left( {\frac{A}{{{E_1}}}} \right) = 1,\,P\left( {\frac{A}{{{E_2}}}} \right) = \frac{1}{2} \cr & \therefore \,P\left( A \right) = P\left( {{E_1}} \right)P\left( {\frac{A}{{{E_1}}}} \right) + P\left( {{E_2}} \right)P\left( {\frac{A}{{{E_2}}}} \right) \cr & \Rightarrow \frac{7}{{12}} = \frac{1}{{n + 1}} \times 1 + \frac{n}{{n + 1}} \times \frac{1}{2} \cr & \Rightarrow n = 5 \cr} $$

There are two cases when Abhay will say ‘Yes’ :
$${\bf{Case}}\,\left( {\bf{i}} \right)\,{\bf{:}}$$   The number that came out is a prime and Abhay is speaking truth, probability for this case is $$P\left( P \right) \times P\left( T \right)$$
Here $$P\left( P \right) = $$  probability of getting a prime $$ = \frac{3}{6} = \frac{1}{2} = 0.5$$
$$P\left( T \right)$$  is probability that Abhay is speaking truth and $$P\left( T \right) = 0.6$$
So probability for this case is $$ = 0.5 \times 0.6 = 0.3$$
$${\bf{Case}}\,\left( {{\bf{ii}}} \right)\,{\bf{:}}$$   The number that came out is not a prime and Abhay is not speaking truth, probability for this case is
$$P\left( {P'} \right) \times P\left( {T'} \right) = 0.5 \times 0.4 = 0.2$$
So total probability for the given case is $$ = 0.3 + 0.2 = 0.5$$
New sample space is $$0.5$$  and we have to find the probability of case $$\left( {\text{i}} \right)$$ which is $$ = \frac{{0.3}}{{0.5}} = 0.6$$

Favourable cases $$ = 6;\left\{ {\left( {1,1,1} \right),\left( {2,2,2} \right),.....\left( {6,6,6} \right)} \right\}$$
Total cases $$ = 6 \times 6 \times 6 = 216$$
$$\therefore \,\,{\text{Req}}{\text{. prob}}{\text{. = }}\frac{6}{{216}} = \frac{1}{{36}}$$