p = 0.5

Here, n(S) = 52
There are 26 red cards (including 2 kings) and re are 2 more kings.
Let E = event of getting a red card or a king.
Then, n(E) = 28
$$\refore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{28}}{{52}} = \frac{7}{{13}}$$

A dice has 6 faces
So re are 6 possible outcomes
Dice is rolled once and n again
So total possibilities = 6 × 6 = 36
The sum should be 8 of 2 throws
So which combination of numbers from 1 to 6 will yield us a sum of 8
They are - (2, 6); (6, 2); (3, 5); (5, 3); (4, 4);
So re are total 5 possibilities where addition is 8.
But only 1 possibility where first throw of dice is 4.
So, Probability for first throw to be 4 and sum to be 8 = $$\frac{{1}}{{36}}$$

Total number of students = 20
Let S be sample space
Then, n(S) = number of ways of three scored first mark
$$\eqalign{ & {\text{n}}\left( {\text{S}} \right) = {}^{20}{C_3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{20 \times 19 \times 18}}{{2 \times 3}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 20 \times 19 \times 3 \cr} $$
Let, E be event of 1 girl and 2 boys
Therefore, n(E) = number of possible of 1 girl out of 8 and 2 boys out of 12
$$\eqalign{ & {\text{n}}\left( {\text{E}} \right) = {}^8{C_1} \times {}^{12}{C_2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{8 \times 12 \times 11}}{{1 \times 2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8 \times 6 \times 11 \cr} $$
Now, required probability
$$\eqalign{ & = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} \cr & = \frac{{8 \times 6 \times 11}}{{20 \times 19 \times 3}} \cr & = \frac{{44}}{{95}} \cr} $$

The variance of binomial distribution is always Less than mean

In two throws of dice, n(S) = (6 × 6) = 36
Let E = event of getting a sum 9 = [(3, 6), (4, 5), (5, 4), (6, 3)]
$$\refore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{4}{{36}} = \frac{1}{9}$$

n(S) = 6 × 6 = 36
E = {(1, 6), (6, 1), (5, 2), (2, 5), (4, 3), (3, 4)}
n(E) = 6
P(E) = $$\frac{{6}}{{36}}$$ = $$\frac{{1}}{{6}}$$

Dependent

Out of mangoes, 4 mangoes are rotten
∴ Required probability
$$\eqalign{ & = \frac{{{}^6{C_2}}}{{{}^{10}{C_2}}} \cr & = \frac{{\frac{{6!}}{{2!\left( {6 - 2} \right)!}}}}{{\frac{{10!}}{{2!\left( {10 - 2} \right)!}}}} \cr & = \frac{{\frac{{6!}}{{2!4!}}}}{{\frac{{10!}}{{2! \times 8!}}}} \cr & = \frac{{\frac{{6 \times 5}}{{1 \times 2}}}}{{\frac{{10 \times 9}}{{1 \times 2}}}} \cr & = \frac{{6 \times 5}}{{10 \times 9}} \cr & = \frac{1}{3} \cr} $$

Let X be event that cards are in a club which is not king and or is king of club.
Let Y be event that one is any club card and or is a non-club king.
Hence, required probability:
$$\eqalign{ & = P(A) + P(B) \cr & = \frac{{^{12}{C_1}{ \times ^1}{C_1}}}{{^{52}{C_2}}} + \frac{{^{13}{C_1}{ \times ^3}{C_1}}}{{^{52}{C_2}}} \cr & = \left( {\frac{{2 \times \left( {12 \times 1} \right)}}{{52 \times 51}}} \right) + \left( {\frac{{2\left( {13 \times 3} \right)}}{{52 \times 51}}} \right) \cr & = \left( {\frac{{24 + 78}}{{52 \times 51}}} \right) \cr & = \frac{1}{{26}} \cr} $$