We can apply binomial probability distribution
We haven = 10
$$p$$ = Probability of drawing a green ball
$$ = \frac{{15}}{{25}} = \frac{3}{5}$$
Also $$q = 1 - \frac{3}{5} = \frac{2}{5}$$
Variance = $$npq$$  = $$10 \times \frac{3}{5} \times \frac{2}{5} = \frac{{12}}{5}$$

Let $${R_t}$$ be the even of drawing red ball in $${t^{th}}$$ draw and $${B_t}$$ be the event of drawing black ball in $${t^{th}}$$ draw.
Now, in the given bag there are 4 red and 6 black balls.
$$\eqalign{ & \therefore \,\,P\left( {{R_1}} \right) = \frac{4}{{10}}\,{\text{and }}P\left( {{B_1}} \right) = \frac{6}{{10}} \cr & {\text{And, }}P\left( {\frac{{{R_2}}}{{{R_1}}}} \right) = \frac{6}{{12}}\,{\text{and }}P\left( {\frac{{{R_2}}}{{{B_1}}}} \right) = \frac{4}{{12}} \cr} $$
Now, required probability
$$\eqalign{ & = P\left( {{R_1}} \right) \times P\left( {\frac{{{R_2}}}{{{R_1}}}} \right) + P\left( {{B_1}} \right) \times P\left( {\frac{{{R_2}}}{{{B_1}}}} \right) \cr & = \left( {\frac{4}{{10}} \times \frac{6}{{12}}} \right) + \left( {\frac{6}{{10}} \times \frac{4}{{12}}} \right) \cr & = \frac{2}{5} \cr} $$

There are $$6$$ ways to select $$2$$ cards of the same kind from the $$4$$ cards in the deck and there are $$13$$  different kinds of cards, so the total number of combinations possible of $$2$$ cards is $$6 \times 13 = 78.$$
There are $$4$$ ways to choose $$3$$ cards of the same kind from $$4$$ cards of the same kind, but because the $$3$$-of-a-kind suit must be different from the $$2$$-of-a-kind suit, the possible combinations of this is, $$4 \times 12 = 48.$$
So total number of ways is $$ = 48 \times 78 = 3744.$$
Sample space is $${}^{52}{C_5} = 2598560.$$
Required Probability $$ = \frac{{3744}}{{2598560}} = \frac{6}{{4165}}.$$

Let us find the probability of Heads appearing in the toss since $$P\left( H \right) = P\left( T \right) = \frac{1}{2}$$     for an unbiased coin, it will not affect or introduce any error in the result.
Thus
If a coin is tossed Once.
$$P\left( H \right) = \frac{1}{2}$$
If a coin is tossed twice.
The probability of head appearing twice is $$ = \frac{1}{2}.\frac{1}{2} = \frac{1}{4}$$
If a coin is tossed thrice.
The probability of head appearing thrice is $$ = \frac{1}{2}.\frac{1}{2}.\frac{1}{2} = \frac{1}{8}$$
If a coin is tossed 4 times.
The probability of head appearing fourth times is $$ = \frac{1}{2}.\frac{1}{2}.\frac{1}{2}.\frac{1}{2} = \frac{1}{{16}}$$
Therefore we can generalize.
If coin is tossed $$n$$ times.
The probability of head appearing $$n$$ times will $$ = \frac{1}{2}.\frac{1}{2}........ = \frac{1}{{{2^n}}}$$

$$A$$ and $$B$$ will agree in a certain statement if both speak truth or both tell a lie. We define following events
$${E_1} = A$$   and $$B$$ both speak truth
$$ \Rightarrow P\left( {{E_1}} \right) = xy$$
$${E_2} = A$$   and $$B$$ both tell a lie
$$ \Rightarrow P\left( {{E_2}} \right) = \left( {1 - x} \right)\left( {1 - y} \right)$$
$$E = A$$  and $$B$$ agree in a certain statement
Clearly, $$P\left( {\frac{E}{{{E_1}}}} \right) = 1{\text{ and }}P\left( {\frac{E}{{{E_2}}}} \right) = 1$$
The required probability is $$P\left( {\frac{{{E_1}}}{E}} \right)$$
Using Baye’s theorem
$$\eqalign{ & P\left( {\frac{{{E_1}}}{E}} \right) = \frac{{P\left( {{E_1}} \right)P\left( {\frac{E}{{{E_1}}}} \right)}}{{P\left( {{E_1}} \right)P\left( {\frac{E}{{{E_1}}}} \right) + P\left( {{E_2}} \right)P\left( {\frac{E}{{{E_2}}}} \right)}} \cr & = \frac{{xy.1}}{{xy.1 + \left( {1 - x} \right)\left( {1 - y} \right).1}} \cr & = \frac{{xy}}{{1 - x - y + 2xy}} \cr} $$

$$P\left( {X = r} \right) = \frac{{{e^{ - m}}{m^r}}}{{r!}}$$
$$P$$ (at most 1 phone call)
$$\eqalign{ & = P\left( {X \leqslant 1} \right) = P\left( {X = 0} \right) + P\left( {X = 1} \right) \cr & = {e^{ - 5}} + 5 \times {e^{ - 5}} \cr & = \frac{6}{{{e^5}}} \cr} $$

Let
$$E = $$  the event that horse $$A$$ wins
$${E_1} = $$  the event that jockey $$B$$ rides horse $$A$$
$${E_2} = $$  the event that jockey $$C$$ rides horse $$A$$
According to question odds in favour of $${E_1} = 2:1$$
$$ \therefore \,P\left( {{E_1}} \right) = \frac{2}{3}{\text{ and }}P\left( {\frac{E}{{{E_1}}}} \right) = \frac{1}{6}$$
(Since, when $$B$$ rides $$A$$, all six horses are equally likely to win)
$$\eqalign{ & P\left( {{E_2}} \right) = 1 - P\left( {{E_1}} \right) = 1 - \frac{2}{3} = \frac{1}{3} \cr & {\text{and }}P\left( {\frac{E}{{{E_2}}}} \right) = 3P\left( {\frac{E}{{{E_1}}}} \right) = \frac{1}{2} \cr & {\text{Let }}{A_1} = {E_1} \cap E{\text{ and }}{A_2} = {E_2} \cap E \cr & {\text{Now, required probability}} \cr & P\left( E \right) = P\left( {{A_1}} \right) + P\left( {{A_2}} \right) \cr & = P\left( {{E_1} \cap E} \right) + P\left( {{E_2} \cap E} \right) \cr & = P\left( {{E_1}} \right)P\left( {\frac{E}{{{E_1}}}} \right) + P\left( {{E_2}} \right)P\left( {\frac{E}{{{E_2}}}} \right) \cr & = \frac{2}{3}.\frac{1}{6} + \frac{1}{3}.\frac{1}{2} \cr & = \frac{5}{{18}} \cr & {\text{So, that odds against winning of }}A{\text{ are }}13:5 \cr} $$

Since, probabilities of failure for engines $$A,\,B$$  and $$C\,P\left( A \right),P\left( B \right)$$    and $$P\left( C \right)$$  are $$0.03,\,0.02$$   and $$0.05$$  respectively.
The aircraft will crash only when all the three engine fail.
So, probability that it crashes
$$\eqalign{ & = P\left( A \right) \times P\left( B \right) \times P\left( C \right) \cr & = 0.03 \times 0.02 \times 0.05 \cr & = 0.00003 \cr} $$
Hence, the probability that the aircraft will not crash $$ = 1 - 0.00003 = 0.99997$$

We know that $$P$$ (exactly one of $$A$$ or $$B$$ occurs)
$$ = P\left( A \right) + P\left( B \right) - 2P\left( {A \cap B} \right).$$
Therefore, $$P\left( A \right) + P\left( B \right) - 2P\left( {A \cap B} \right) = p\,\,\,.....\left( 1 \right)$$
Similarly, $$P\left( B \right) + P\left( C \right) - 2P\left( {B \cap C} \right) = P\,\,\,.....\left( 2 \right)$$
and $$P\left( C \right) + P\left( A \right) - 2P\left( {C \cap A} \right) = P\,\,\,.....\left( 3 \right)$$
Adding (1), (2) and (3) we get
$$\eqalign{ & 2\left[ {P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {B \cap C} \right) - P\left( {C \cap A} \right)} \right] = 3p \cr & \Rightarrow \,\,P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {B \cap C} \right) - P\left( {C \cap A} \right) = \frac{{3p}}{2}\,\,\,\,.....\left( 4 \right) \cr} $$
We are also given that,
$$P\left( {A \cap B \cap C} \right) = {p^2}\,\,\,\,\,.....\left( 5 \right)$$
Now, $$P$$ (at least one of $$A, B$$  and $$C$$)
$$ = P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {B \cap C} \right) - P\left( {C \cap A} \right) + P\left( {A \cap B \cap C} \right)$$
$$ = \frac{{3p}}{2} + {p^2}$$     [using (4) and (5)] $$ = \frac{{3p + 2{p^2}}}{2}$$

Prob. of one coin showing head = $$p$$
∴ Prob. of one coin showing tail = $$1 - p$$
ATQ coin is tossed 100 times and prob. of 50 coins showing head = prob. of 51 coins showing head.
Using binomial prob. distribution
$$\eqalign{ & P\left( {X = r} \right) = \,{\,^n}{C_r}{p^r}{q^{n - r}}, \cr & {\text{we get,}}{{\text{ }}^{100}}{C_{50}}{p^{50}}{\left( {1 - p} \right)^{50}} = {\,^{100}}{C_{51}}{p^{51}}{\left( {1 - p} \right)^{49}} \cr & \Rightarrow \,\,\frac{{1 - p}}{p} = \frac{{^{100}{C_{51}}}}{{^{100}{C_{50}}}} = \frac{{50!50!}}{{51!49!}} = \frac{{50}}{{51}} \cr & \Rightarrow \,\,51 - 51p = 50p \cr & \Rightarrow \,\,101p = 51 \cr & \Rightarrow \,\,p = \frac{{51}}{{101}} \cr} $$