Let $${E_A} = $$  the event of $$A$$ becoming secretary. Similarly, $${E_B}$$ and $${E_C}.$$
$$\,\,\,\,\,\,\,\,\,\,\,{E_L} = $$   the event of admitting lady members.
Here, $$P\left( {{E_A}} \right) = \frac{1}{3},\,\,P\left( {{E_B}} \right) = \frac{2}{9},\,\,P\left( {{E_C}} \right) = \frac{4}{9}$$
Clearly, $${E_A},\,{E_B},\,{E_C}$$    are mutually exclusive and exhaustive.
Also, $$P\left( {\frac{{{E_L}}}{{{E_A}}}} \right) = 0.6,\,\,\,P\left( {\frac{{{E_L}}}{{{E_B}}}} \right) = 0.7,\,\,\,\,P\left( {\frac{{{E_L}}}{{{E_C}}}} \right) = 0.5$$
$$\therefore $$  the required probability
$$\eqalign{ & = P\left( {{E_A}} \right).P\left( {\frac{{{E_L}}}{{{E_A}}}} \right) + P\left( {{E_B}} \right).P\left( {\frac{{{E_L}}}{{{E_B}}}} \right) + P\left( {{E_C}} \right).P\left( {\frac{{{E_L}}}{{{E_C}}}} \right) \cr & = \frac{1}{3} \times \frac{3}{5} + \frac{2}{9} \times \frac{7}{{10}} + \frac{4}{9} \times \frac{5}{{10}} \cr & = \frac{{26}}{{45}} \cr} $$

The faulty machines can be identified in two tests only if both the tested machines are either all defective or all non-defective. See the following tree diagram.

(Here $$D$$ is for Defective & $$ND$$  is for Non-Defective)
Required Probability $$ = \frac{2}{4} \times \frac{1}{3} + \frac{2}{4} \times \frac{1}{3} = \frac{1}{3}$$
$$\because $$  The probability that first machine is defective (or non-defective) is $$\frac{2}{4}$$ and the probability that second machine is also defective (or non-defective) is $$\frac{1}{3}$$ as $$1$$ defective machine remains in total three machines.

$$\eqalign{ & P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) \cr & \Rightarrow \,\frac{5}{6} = \left( {1 - \frac{1}{2}} \right) + P\left( B \right) - \frac{1}{3} \cr & \Rightarrow \,P\left( B \right) = \frac{2}{3} \cr & P\left( A \right).P\left( B \right) = \frac{1}{2}.\frac{2}{3} = \frac{1}{3} = P\left( {A \cap B} \right) \cr} $$
So, $$A$$ and $$B$$ are independent and therefore, not mutually exclusive

$$n\left( S \right) = \frac{{7!}}{{3!}},\,n\left( E \right) = 5!.{\text{ So, }}P\left( E \right) = \frac{{5!}}{{\frac{{7!}}{{3!}}}}.$$

$$P\left( A \right) = \frac{1}{4},P\left( {\frac{A}{B}} \right) = \frac{1}{2},P\left( {\frac{B}{A}} \right) = \frac{2}{3}$$
By conditional probability,
$$\eqalign{ & P\left( {A \cap B} \right) = P\left( A \right)P\left( {\frac{B}{A}} \right) = P\left( B \right)P\left( {\frac{A}{B}} \right) \cr & \Rightarrow \,\,\frac{1}{4} \times \frac{2}{3} = P\left( B \right) \times \frac{1}{2} \cr & \Rightarrow \,\,P\left( B \right) = \frac{1}{3} \cr} $$

$$\eqalign{ & P\left( {{E^c} \cap \frac{{{F^c}}}{G}} \right) = \frac{{P{{\left( {E \cup F} \right)}^c}}}{G} \cr & 1 - P\left( {\frac{{E \cup F}}{G}} \right) \cr & = 1 - P\left( {\frac{E}{G}} \right) - P\left( {\frac{F}{G}} \right) + P\left( {E \cap \frac{F}{G}} \right) \cr & = 1 - P\left( E \right) - P\left( F \right) + O \cr} $$
(∵ $$E, F, G$$   are pairwise independent and
$$P\left( {E \cap F \cap G} \right) = 0$$
$$ \Rightarrow \,\,P\left( E \right).P\left( F \right) = 0\,\,{\text{as }}P\left( G \right) > 0)$$
$$ = P\left( {{E^C}} \right) - P\left( F \right)$$

$$q$$ is an integer, then number of possible outcomes in $$\left[ { - 10,\,10} \right] = 21$$
Now, for real roots, discriminant $$ \geqslant 0$$
$$\eqalign{ & \Rightarrow \left( {q - 4} \right)\left( {q + 1} \right) \geqslant 0 \cr & \Rightarrow q \geqslant 4,\,\,q \leqslant - 1 \cr} $$
Then, number of favourable outcomes $$= 7 + 10 = 17$$
Hence required probability $$ = \frac{{17}}{{21}}$$

$$\eqalign{ & {\text{Here, }}p = \frac{1}{2},\,\,q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2} \cr & n = 6,\,N = 64 \cr & {\text{Then,}} \cr & p\left( r \right) = {}^n{C_r}\,{p^r}{q^{n - r}} \cr & = {}^6{C_r}{\left( {\frac{1}{2}} \right)^r}.{\left( {\frac{1}{2}} \right)^{6 - r}} \cr & = {}^6{C_r}{\left( {\frac{1}{2}} \right)^6} \cr & \therefore \,f\left( r \right) = Np\left( r \right) \cr & = 64.{}^6{C_r}.\frac{1}{{64}} \cr & = {}^6{C_r} \cr & {\text{Now, }}\sum\limits_3^6 {p\left( r \right)} = p\left( 3 \right) + p\left( 4 \right) + p\left( 5 \right) + p\left( 6 \right) \cr & = \left( {{}^6{C_3} + {}^6{C_4} + {}^6{C_5} + {}^6{C_6}} \right)\frac{1}{{{2^6}}} \cr & = \left( {{2^6} - {}^6{C_0} - {}^6{C_1} - {}^6{C_2}} \right)\frac{1}{{{2^6}}} \cr & = \left( {64 - 1 - 6 - 15} \right)\frac{1}{{{2^6}}} \cr & = \frac{{42}}{{64}} \cr & = \frac{{21}}{{32}} \cr & \therefore \,f{\left( r \right)_{r \geqslant 3}} = N\sum\limits_3^6 {p\left( r \right)} = 64.\frac{{21}}{{32}} = 42 \cr} $$

$$n\left( S \right) = {}^{10}{C_3}.$$   Clearly, $$E = \left\{ {\left( {10,\,5,\,2} \right),\,\left( {8,\,2,\,4} \right),\,\left( {6,\,2,\,3} \right)} \right\}.\,{\text{So, }}n\left( E \right) = 3$$
$$\therefore \,P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{3}{{{}^{10}{C_3}}} = \frac{1}{{40}}.$$

$$n\left( S \right) = {\,^9}{C_3},n\left( E \right) = {\,^3}{C_1} \times {\,^4}{C_1} \times {\,^2}{C_1}$$
Probability $$ = \frac{{3 \times 4 \times 2}}{{^9{C_3}}}$$
$$\eqalign{ & = \frac{{24 \times 3!}}{{9!}} \times 6! \cr & = \frac{{24 \times 6}}{{9 \times 8 \times 7}} \cr & = \frac{2}{7} \cr} $$