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

n(S) = number of ways of choosing 4 persons out of 9
$$ = {}^9\mathop C\nolimits_4 $$   $$ = \frac{{9 \times 8 \times 7 \times 6}}{{4 \times 3 \times 2 \times 1}}$$   = 126
n(E) = number of ways of choosing 2 children out of 4 and 2 persons out of (3 + 2) personal
n(E) $$ = \left( {{}^4\mathop C\nolimits_2 \times {}^5\mathop C\nolimits_2 } \right)$$   $$ = \left( {\frac{{4 \times 3}}{{2 \times 1}} \times \frac{{5 \times 4}}{{2 \times 1}}} \right)$$     = 60
$$\refore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{60}}{{126}} = \frac{{10}}{{21}}$$

An event is collection of one or more outcomes from an experiment.

Total number of balls = (8 + 7 + 6) = 21
Let E = event that ball drawn is neir blue nor green
= Even that ball drawn is red
∴ n(E) = 8
P(E) = $$\frac{{8}}{{21}}$$

In special rule of addition of probability events must be mutually exclusive.

In a simultaneous throw of two dice, we have n(s) = 6 × 6 = 36
Let E = event of getting two number whose sum is at least 10
Then E = {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}
refore, n(E) = 6
And p(E) = p(getting two numbers whose sum is at least 10)
$$\eqalign{ & {\text{p}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{6}{{36}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{6} \cr} $$
Hence answer is $$\frac{{1}}{{6}}$$

Let S be sample space and let E be event of selecting 2 boys and 1 girl.
Then, n(S) = number of ways of selecting 3 students out of 25
= $${}^{25}\mathop C\nolimits_3 = $$   $$\frac{{25 \times 24 \times 23}}{{3 \times 2 \times 1}}$$   = 2300
And, n(E) = $$\left( {{}^{15}\mathop C\nolimits_2 \times {}^{10}\mathop C\nolimits_1 } \right)$$   $$ = \left( {\frac{{15 \times 14}}{{2 \times 1}} \times 10} \right)$$    = 1050
$$\refore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{1050}}{{2300}} = \frac{{21}}{{46}}$$

P(X ≤ x)

p > $$\frac{{1}}{{2}}$$
If p < 0.5, n distribution is positively skewed.
If p > 0.5, distribution is negatively skewed

Total number of marbles = (4 + 5 + 3) = 12
Let E be event of drawing 3 marbles such that none is blue.
Then, n (E) = number of ways of drawing 3 marbles out of 7 = $${}^7\mathop C\nolimits_3 $$   $$ = \frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}}$$   = 35
And, $$n(S) = {}^{12}\mathop C\nolimits_3 $$   $$ = \frac{{12 \times 11 \times 10}}{{3 \times 2 \times 1}}$$   = 220
$$\refore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{35}}{{220}} = \frac{7}{{44}}$$
∴ Required probability
= 1 - P(E)
= $$\left( {1 - \frac{7}{{44}}} \right)$$
= $$ \frac{{37}}{{44}}$$