Solution:
Let S be the sample space
Then, n(S) = number of ways of drawing 3 balls out of 15
$$\eqalign{ = {}^{15}{C_3} \cr = \frac{{ {15 \times 14 \times 13} }}{{ {3 \times 2 \times 1} }} \cr = 455 \cr} $$
Let E = event of getting all the 3 red balls
$$\eqalign{ \therefore n\left( E \right) = {}^5{C_3} = {}^5{C_2} \cr = \frac{{ {5 \times 4} }}{{ {2 \times 1} }} = 10 \cr \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr = \frac{{10}}{{455}} \cr = \frac{2}{{91}} \cr} $$

Solution:
Let S be the sample space
Total number of students in the class = 12 boys + 16 girls = 28
Then, n(S) = 28
Let E be the event of calling one of them by enroll number
Given that, number of girls = 16
Then, n(E) = 16
The probability that the one called is a girl
$$\eqalign{ = \frac{{{\text{n}}\left( {\text{S}} \right)}}{{{\text{n}}\left( {\text{E}} \right)}} \cr = \frac{{16}}{{28}} \cr = \frac{4}{7} \cr} $$

Solution:
In a simultaneous throw of two dice, we have n (S) = (6 × 6) = 36
Let E = event of getting two numbers whose product is even.
Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n (E) = 27
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{27}}{{36}} = \frac{3}{4}$$

Solution:
As probability of a both the teams (Arrogant and Overconfident) winning simultaneously is zero.
$$\eqalign{ P\left( {A \cap O} \right) = 0 \cr P\left( {A \cap B} \right) = P\left( A \right) + P\left( B \right) \cr = \frac{5}{8} + \frac{1}{5} \cr = \frac{{33}}{{40}} \cr} $$
So required odds will be 33 : 7

Solution:
Drawing two balls of same color from seven green balls can be done in $${}^7{C_2}$$ ways.
Similarly from eight white balls two can be drawn in $${}^8{C_2}$$ ways.
$$\eqalign{ P = \frac{{{}^7{C_2}}}{{{}^{15}{C_2}}} + \frac{{{}^8{C_2}}}{{{}^{15}{C_2}}} \cr \,\,\,\,\,\,\, = \frac{7}{{15}} \cr} $$

Solution:
Here, n(S) = 52
Let E = event of getting a queen of club or a king of heart
Then, n(E) = 2
$$\eqalign{ \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr = \frac{2}{{52}} \cr = \frac{1}{{26}} \cr} $$

Solution:
Hypergeometric distribution has 3 parameters