Required probability:
$$\eqalign{ & = \frac{{^4{C_1}{ \times ^4}{C_1}{ \times ^4}{C_1}}}{{^{52}{C_3}}} \cr & = \frac{{4 \times 4 \times 4}}{{22100}} \cr & = \frac{{16}}{{5525}} \cr} $$

Total number of marbles = (6 + 4 + 2 + 3) = 15
Let E be event of drawing 1 green, 2 blue and 1 red marble.
Then,
n (E) = $${}^2\mathop C\nolimits_1 \times {}^4\mathop C\nolimits_2 \times {}^6\mathop C\nolimits_1 $$   $$ = 2 \times \frac{{4 \times 3}}{{2 \times 1}} \times 6$$     = 72
And, n (S) = $${}^{15}\mathop C\nolimits_4 = $$   $$\frac{{15 \times 14 \times 13 \times 12}}{{4 \times 3 \times 2 \times 1}}$$     = 1365
$$\refore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{72}}{{1365}}$$      $$ = \frac{{24}}{{455}}$$

Number greater than 5 is 6, so only 1 number
Total cases of dice = [1, 2, 3, 4, 5, 6]
So probability = $$\frac{{1}}{{6}}$$

Arithmetic Mean (AM)

We know that in a simultaneous throw of two dice,
n(S) = 6 × 6 = 36
Let E = event of getting a total of 7
= {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
$$\refore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{6}{{36}} = \frac{1}{6}$$

Total number of erasers in box = 3 + 4 + 7 = 14
Let S be sample space
Then, n(S) = number of ways of taking 5 out of 14
Therefore,
$$\eqalign{ & {\text{n}}\left( {\text{S}} \right) = {}^{14}{C_5} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{14 \times 13 \times 12 \times 11 \times 10}}{{2 \times 3 \times 4 \times 5}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 14 \times 13 \times 11 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2002 \cr} $$
Let E be event of getting all 5 blue erasers
Therefore,
$$\eqalign{ & {\text{n}}\left( {\text{E}} \right) = {}^7{C_5} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{7 \times 6 \times 5 \times 4 \times 3}}{{2 \times 3 \times 4 \times 5}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 21 \cr} $$
Now, required probability
$$\eqalign{ & \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} = \frac{{21}}{{2002}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{{286}} \cr} $$

Here, S = {1, 2, 3, 4, ....., 19, 20}
Let E = event of getting a multiple of 3 or 5 = {3, 6, 9, 12, 15, 18, 5, 10, 20}
$$\refore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{9}{{20}}$$

$$\eqalign{ & P\left( A \right) = \frac{2}{5} \cr & P\left( B \right) = \frac{4}{7} \cr} $$
E = {A and B both get selected}
$$\eqalign{ & P\left( E \right) = {\text{ }}P\left( A \right) \times P\left( B \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{5} \times \frac{4}{7} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{8}{{35}} \cr} $$

Total number of balls = (10 + 10) = 20
Let E be event of drawing 2 balls of same colour.
n (E) = number of ways of drawing 2 black balls or 2 white balls.
n (E) =   $$\left( {{}^{10}\mathop C\nolimits_2 \times {}^{10}\mathop C\nolimits_2 } \right)$$   $$ = 2 \times {}^{10}\mathop C\nolimits_2 $$   $$ = 2 \times \frac{{10 \times 9}}{{2 \times 1}}$$   = 90
n (S) = number of ways of drawing 2 balls out of 20 balls
$$ = {}^{20}\mathop C\nolimits_2 = \frac{{20 \times 19}}{{2 \times 1}}$$   = 190
$$\refore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{90}}{{190}} = \frac{9}{{19}}$$

Probability of occurrence of an event lies between 0 and 1