For a particular house being selected
Probability $$ = \frac{1}{3}$$
$$P$$ (all the persons apply for the same house)
$$\eqalign{ & = \left( {\frac{1}{3} \times \frac{1}{3} \times \frac{1}{3}} \right)3 \cr & = \frac{1}{9}. \cr} $$

The probability of $$4$$ being the minimum number $$ = \frac{{{}^6{C_2}}}{{{}^{10}{C_3}}}$$  (because, after selecting $$4$$ any two can be selected from $$5,\,6,\,7,\,8,\,9,\,10$$   ).
The probability of $$8$$ being the maximum number $$ = \frac{{{}^7{C_2}}}{{{}^{10}{C_3}}}.$$
The probability of $$4$$ being the minimum number and $$8$$ being the maximum number $$ = \frac{3}{{{}^{10}{C_3}}}$$
$$\therefore $$  the required probability
$$\eqalign{ & = P\left( {A \cup B} \right) \cr & = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) \cr & = \frac{{{}^6{C_2}}}{{{}^{10}{C_3}}} + \frac{{{}^7{C_2}}}{{{}^{10}{C_3}}} - \frac{3}{{{}^{10}{C_3}}} \cr & = \frac{{11}}{{40}}. \cr} $$

$$P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{2 \times 9 \times 9 \times 9 \times 8}}{{9 \times 10 \times 10 \times 10 \times 10}} = \frac{{9 \times 9}}{{5 \times 5 \times 5 \times 5}} = {\left( {\frac{3}{5}} \right)^4}.$$

From the given information probability that Amar will win the tournament is $$P\left( A \right) = \frac{1}{5}$$   and Bimal will win $$P\left( B \right) = \frac{1}{6}$$   and same for Chetan is $$P\left( C \right) = \frac{2}{5}.$$
Since these events are mutually exclusive.

$$n\left( S \right) = \frac{{10!}}{{\left( {7!} \right)\left( {3!} \right)}}$$
$$n\left( E \right) = {}^8{C_3} = \frac{{8!}}{{\left( {3!} \right)\left( {5!} \right)}},$$      because there are $$8$$ places for $$3$$ black balls.
$$\therefore \,P\left( E \right) = \frac{{\frac{{18!}}{{\left( {3!} \right)\left( {5!} \right)}}}}{{\frac{{10!}}{{\left( {7!} \right)\left( {3!} \right)}}}} = \frac{{\left( {8!} \right)\left( {7!} \right)}}{{\left( {10!} \right)\left( {5!} \right)}} = \frac{{7.6}}{{10.9}} = \frac{7}{{15}}.$$

$$\eqalign{ & {\text{Exhaustive number of cases}} = {12^6} \cr & {\text{Favourable cases}} = {}^{12}{C_2}\left( {{2^6} - 2} \right) \cr & \therefore \,{\text{Probabiltity}} = \frac{{{}^{12}{C_2}\left( {{2^6} - 2} \right)}}{{{{12}^6}}} = \frac{{341}}{{{{12}^5}}} \cr} $$

Total number of $$3$$-digit numbers $$ = 9 \times 8 \times 7 = 504$$
For product to be odd, we have to choose only from odd numbers.
$$\therefore $$  Total number of $$3$$-digit number whose product are odd $$ = 5 \times 4 \times 3 = 60$$
$$\therefore $$  Required probability $$ = \frac{{60}}{{504}} = \frac{5}{{42}}$$

$$p$$ (at least one failure) $$ \geqslant \frac{{31}}{{32}}$$
$$\eqalign{ & \Rightarrow \,\,1 - p\left( {{\text{no failure}}} \right) \geqslant \frac{{31}}{{32}} \cr & \Rightarrow \,\,1 - {p^5} \geqslant \frac{{31}}{{32}} \cr & \Rightarrow \,\,{p^5} \leqslant \frac{1}{{32}} \cr & \Rightarrow \,\,p \leqslant \frac{1}{2} \cr} $$
Hence $$p$$ lies in the interval $${\left[ {0,\frac{1}{2}} \right]}.$$

$$\eqalign{ & {\text{Required probability}} = P\left( {\overline A \cup \overline B } \right) \cr & = P\left( {\overline {A \cap B} } \right) \cr & = 1 - P\left( {A \cap B} \right) \cr & {\text{Again,}} \cr & = P\left( {\overline A \cup \overline B } \right) \cr & = P\left( {\overline A } \right) + P\left( {\overline B } \right) - P\left( {\overline A \cap \overline B } \right)\left[ {{\text{ By add}}{\text{. Theorem}}} \right] \cr & {\text{Again,}} \cr & = P\left( {\overline A \cup \overline B } \right) \cr & = P\left( {\overline A } \right) + P\left( {\overline B } \right) - P\left( {\overline A \cap \overline B } \right) \cr & = P\left( {\overline A } \right) + P\left( {\overline B } \right) - P\left( {\overline {A \cup B} } \right) \cr & = P\left( {\overline A } \right) + P\left( {\overline B } \right) - \left\{ {1 - P\left( {A \cup B} \right)} \right\} \cr & = P\left( {\overline A } \right) + P\left( {\overline B } \right) + P\left( {A \cup B} \right) - 1 \cr & {\text{Finally,}} \cr & P\left( {\overline A \cup \overline B } \right) = P\left( {A \cap \overline B } \right) \cup \left( {\overline A \cap B} \right) \cup \left( {\overline A \cap \overline B } \right) \cr & = P\left( {A \cap \overline B } \right) + \left( {\overline A \cap B} \right) + \left( {\overline A \cap \overline B } \right) \cr & = \left[ {\because \,A \cap \overline B ,\,\overline A \cap B{\text{ and }}\overline A \cap \overline B {\text{ are are mutually exclusive events}}} \right] \cr} $$
So, alternative (D) is the correct answer.

Let $$E$$ be the sum of the faces equals or exceeds. Then,
$$\eqalign{ & E = \left\{ {\left( {5,\,5} \right),\,\left( {4,\,6} \right),\,\left( {6,\,4} \right),\,\left( {5,\,6} \right),\,\left( {6,\,5} \right),\,\left( {6,\,6} \right)} \right\} \cr & \therefore \,n\left( E \right) = 6 \cr & {\text{Hence,}}\,\,P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{6}{{36}} = \frac{1}{6} \cr} $$