$$\eqalign{ & {\text{Required Probability :}} \cr & = P\left( {5\,{\text{points}}} \right) \cr & = P\left( {{\text{two wins and one draw}}} \right) \cr & = P\left( {WWD} \right) + P\left( {WDW} \right) + P\left( {DWW} \right) \cr & = \frac{1}{3} \times \frac{1}{3} \times \frac{1}{6} + \frac{1}{3} \times \frac{1}{6} \times \frac{1}{3} + \frac{1}{6} \times \frac{1}{3} \times \frac{1}{3} \cr & = \frac{1}{{18}} \cr} $$

Let $$M$$ be the events that man will be selected and $$W$$ the events that woman will be selected. Then
$$\eqalign{ & P\left( M \right) = \frac{1}{4},\,\,{\text{so}}\,\,P\left( M \right) = 1 - \frac{1}{4} = \frac{3}{4} \cr & P\left( W \right) = \frac{1}{3},\,\,{\text{so}}\,\,P\left( W \right) = \frac{2}{3} \cr} $$
Clearly $${M_1}$$ and $$W$$ are independent events. So,
$$\eqalign{ & P\left( {M \cap W} \right) = P\left( M \right) \times P\left( W \right) \cr & = \frac{3}{4} \times \frac{2}{3} \cr & = \frac{1}{2} \cr} $$

$$\eqalign{ & {\text{Required probability :}} \cr & = \frac{{\sum\limits_{r = 0}^n {{}^n{C_r} \times {}^n{C_r}} }}{{{2^n} \times {2^n}}} \cr & = \frac{{C_0^2 + C_1^2 + C_2^2 + ..... + C_n^2}}{{{4^n}}} \cr & = \frac{{{}^{2n}{C_n}}}{{{2^{2n}}}} \cr} $$

$$\eqalign{ & {\text{Since, }}P\left( {A \cup B \cup C} \right) \cr & = P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {B \cap C} \right) - P\left( {C \cap A} \right) + P\left( {A \cap B \cap C} \right) \cr & {\text{or }}P\left( {A \cup B \cup C} \right) = 0.3 + 0.4 + 0.8 - \left( {0.08 + 0.28 + P\left( {B \cap C} \right)} \right) + 0.09 \cr & F = 1.23 - P\left( {B \cap C} \right) \cr & {\text{or }}P\left( {B \cap C} \right) = 1.23 - P\left( {A \cup B \cup C} \right) \cr & {\text{But we know that }}0 \leqslant P\left( {A \cup B \cup C} \right) \leqslant 1 \cr & {\text{Hence, }}0.23 \leqslant P\left( {B \cap C} \right) \leqslant 0.48 \cr} $$

$$P$$ (at least one hit the target)
= $$1 - P$$   (none of them hit the target)
$$\eqalign{ & = 1 - \left( {1 - \frac{1}{2}} \right)\left( {1 - \frac{1}{3}} \right)\left( {1 - \frac{1}{4}} \right)\left( {1 - \frac{1}{8}} \right) \cr & = 1 - \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{7}{8} \cr & = 1 - \frac{7}{{32}} \cr & = \frac{{25}}{{32}} \cr} $$

The two events can happen simultaneously e.g., (2, 3)
∴ not mutually exclusive.
Also are not dependent on each other.

Let $${E_1}$$ $$ \equiv $$ The Indian man is seated adjacent to his wife.
$${E_2}$$ $$ \equiv $$ Each American man is seated adjacent to his wife.
Then $$P\left( {\frac{{{E_1}}}{{{E_2}}}} \right) = \frac{{P\left( {{E_1} \cap {E_2}} \right)}}{{P\left( {{E_2}} \right)}}$$
Now $${{E_1} \cap {E_2}}$$   $$ \equiv $$ All men are seated adjacent to their wives.
∴ We can consider the 5 couples as single - single objects which can be arranged in a circle in 4! ways.
But for each couple, husband and wife can interchange their places in 2! ways.
∴ Number of ways when all men are seated adjacent to their wives $$ = 4!\, \times {\left( {2!} \right)^5}$$
Also in all 10 persons can be seated in a circle in 9! ways.
$$\therefore \,P\left( {{E_1} \cap {E_2}} \right) = \frac{{4!\, \times {{\left( {2!} \right)}^5}}}{{9!}}$$
Similarly if each American man is seated adjacent to his wife, considering each American couple as single object and Indian woman and man as seperate objects there are 6 different objects which can be arranged in a circle in 5! ways. Also for each American couple, husband and wife can interchange their places in 2! ways.
So the number of ways in which each American man is seated adjacent to his wife.
$$\eqalign{ & = 5!\,\, \times {\left( {2!} \right)^4} \cr & \therefore \,\,P\left( {{E_2}} \right) = \frac{{5!\,\, \times {{\left( {2!} \right)}^4}}}{{9!}} \cr & {\text{So, }}P\left( {\frac{{{E_1}}}{{{E_2}}}} \right) = \frac{{\frac{{\left( {4!\,\, \times {{\left( {2!} \right)}^5}} \right)}}{{9!}}}}{{\frac{{\left( {5!\,\, \times {{\left( {2!} \right)}^4}} \right)}}{{9!}}}} \cr & = \frac{2}{5} \cr} $$

$$\eqalign{ & P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right); \cr & \Rightarrow \,\,\frac{3}{4} = 1 - P\left( {\overline A } \right) + P\left( B \right) - \frac{1}{4} \cr & \Rightarrow \,\,1 = 1 - \frac{2}{3} + P\left( B \right) \cr & \Rightarrow \,\,P\left( B \right) = \frac{2}{3}; \cr & {\text{Now, }}P\left( {\overline A \cap B} \right) = P\left( B \right) - P\left( {A \cap B} \right) \cr & = \frac{2}{3} - \frac{1}{4} \cr & = \frac{5}{{12}}. \cr} $$

$$\eqalign{ & {\text{In binomial distribution}} \cr & {\text{mean}} = np = 10,\,{\text{variance}} = npq = 5 \cr & \therefore \,p = q = \frac{1}{2} \cr & {\text{Let }}x{\text{ be the mode, then}} \cr & np + p > x > np - q \cr & \therefore \,10 + \frac{1}{2} > x > 10 - \frac{1}{2} \cr & \Rightarrow \frac{{21}}{2} > x > \frac{{19}}{2} \cr & \Rightarrow 9.5 < x < 10.5 \cr & \Rightarrow x = 10 \cr} $$

$$\eqalign{ & P\left( {2{\text{ white and }}1{\text{ black}}} \right) \cr & = P\left( {{W_1}{W_2}{B_3}{\text{ or }}{W_1}{B_2}{W_3}{\text{ or }}{B_1}{W_2}{W_3}} \right) \cr & = P\left( {{W_1}{W_2}{B_3}} \right) + P\left( {{W_1}{B_2}{W_3}} \right) + P\left( {{B_1}{W_2}{W_3}} \right) \cr & = P\left( {{W_1}} \right)P\left( {{W_2}} \right)P\left( {{B_3}} \right) + P\left( {{W_1}} \right)P\left( {{B_2}} \right)P\left( {{W_3}} \right) + P\left( {{W_1}} \right)\left( {{W_2}} \right)\left( {{W_3}} \right) \cr & = \frac{3}{4}.\frac{2}{4}.\frac{3}{4} + \frac{3}{4}.\frac{2}{4}.\frac{1}{4} + \frac{1}{4}.\frac{2}{4}.\frac{1}{4} \cr & = \frac{1}{{32}}\left( {9 + 3 + 1} \right) \cr & = \frac{{13}}{{32}} \cr} $$