Binomial distribution Questions in English

(A) The repeated tossing of a die are Bernoulli trials. Let $X$ represent the number of times of getting $5$ in $7$ throws of the die.
Probability of getting $5$ in a single throw of the die,$p = \frac{1}{6}$.
Therefore,$q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.
Clearly,$X$ follows a binomial distribution with $n = 7$ and $p = \frac{1}{6}$.
The probability mass function is given by $P(X = x) = ^{n}C_{x} \cdot q^{n-x} \cdot p^{x} = ^{7}C_{x} \cdot (\frac{5}{6})^{7-x} \cdot (\frac{1}{6})^{x}$.
We need to find the probability of getting $5$ exactly twice,i.e.,$P(X = 2)$.
$P(X = 2) = ^{7}C_{2} \cdot (\frac{5}{6})^{7-2} \cdot (\frac{1}{6})^{2}$.
$= \frac{7 \times 6}{2 \times 1} \cdot (\frac{5}{6})^{5} \cdot (\frac{1}{6})^{2}$.
$= 21 \cdot \frac{5^5}{6^5} \cdot \frac{1}{6^2} = 21 \cdot \frac{5^5}{6^7}$.

(A) The repeated tossing of the die are Bernoulli trials. Let $X$ represent the number of times of getting sixes in $6$ throws of the die.
Probability of getting a six in a single throw of a die,$p = \frac{1}{6}$.
$\therefore q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.
Clearly,$X$ follows a binomial distribution with $n = 6$.
$\therefore P(X = x) = ^{n}C_{x} q^{n-x} p^{x} = ^{6}C_{x} \left(\frac{5}{6}\right)^{6-x} \left(\frac{1}{6}\right)^{x}$.
$P(\text{at most } 2 \text{ sixes}) = P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$.
$P(X = 0) = ^{6}C_{0} \left(\frac{5}{6}\right)^{6} = \left(\frac{5}{6}\right)^{6}$.
$P(X = 1) = ^{6}C_{1} \left(\frac{5}{6}\right)^{5} \left(\frac{1}{6}\right) = 6 \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^{5} = \left(\frac{5}{6}\right)^{5}$.
$P(X = 2) = ^{6}C_{2} \left(\frac{5}{6}\right)^{4} \left(\frac{1}{6}\right)^{2} = 15 \cdot \frac{1}{36} \cdot \left(\frac{5}{6}\right)^{4} = \frac{15}{36} \left(\frac{5}{6}\right)^{4} = \frac{5}{12} \left(\frac{5}{6}\right)^{4}$.
Summing these probabilities:
$P(X \leq 2) = \left(\frac{5}{6}\right)^{4} \left[ \left(\frac{5}{6}\right)^{2} + \left(\frac{5}{6}\right) + \frac{5}{12} \right]$.
$= \left(\frac{5}{6}\right)^{4} \left[ \frac{25}{36} + \frac{30}{36} + \frac{15}{36} \right] = \left(\frac{5}{6}\right)^{4} \left[ \frac{70}{36} \right]$.
$= \frac{35}{18} \left(\frac{5}{6}\right)^{4}$.

(B) The repeated selections of articles in a random sample are Bernoulli trials. Let $X$ denote the number of defective articles in a random sample of $12$ articles.
Clearly,$X$ follows a binomial distribution with $n=12$ and $p=10 \% = \frac{10}{100} = \frac{1}{10}$.
Therefore,$q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10}$.
The probability mass function is given by $P(X=x) = ^{n}C_{x} p^{x} q^{n-x}$.
Substituting the values,we get $P(X=9) = ^{12}C_{9} \left(\frac{1}{10}\right)^{9} \left(\frac{9}{10}\right)^{12-9}$.
$P(X=9) = ^{12}C_{3} \left(\frac{1}{10}\right)^{9} \left(\frac{9}{10}\right)^{3}$.
Since $^{12}C_{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$,we have:
$P(X=9) = 220 \times \frac{1}{10^{9}} \times \frac{9^{3}}{10^{3}} = 220 \times \frac{9^{3}}{10^{12}} = \frac{22 \times 9^{3}}{10^{11}}$.

(D) The selection of bulbs from a box can be modeled as Bernoulli trials. Let $X$ denote the number of defective bulbs in a sample of $5$ bulbs.
The probability of selecting a defective bulb is $p = \frac{10}{100} = \frac{1}{10}$.
Therefore,the probability of selecting a non-defective bulb is $q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10}$.
Since the sample size $n = 5$ is small relative to the total population,we can treat this as a binomial distribution $X \sim B(n, p)$ with $n = 5$ and $p = \frac{1}{10}$.
The probability mass function is given by $P(X = x) = ^{n}C_{x} \cdot p^{x} \cdot q^{n-x} = ^{5}C_{x} \cdot \left(\frac{1}{10}\right)^{x} \cdot \left(\frac{9}{10}\right)^{5-x}$.
We need to find the probability that none of the bulbs are defective,which is $P(X = 0)$.
$P(X = 0) = ^{5}C_{0} \cdot \left(\frac{1}{10}\right)^{0} \cdot \left(\frac{9}{10}\right)^{5-0}$.
$P(X = 0) = 1 \cdot 1 \cdot \left(\frac{9}{10}\right)^{5} = \left(\frac{9}{10}\right)^{5}$.
The correct answer is $D$.

(A) The selection of students who are swimmers follows a binomial distribution. Let $X$ be the number of students who are swimmers out of $n=5$ students.
The probability that a student is not a swimmer is $q = \frac{1}{5}$.
The probability that a student is a swimmer is $p = 1 - q = 1 - \frac{1}{5} = \frac{4}{5}$.
The probability distribution is given by $P(X=x) = ^{n}C_{x} p^{x} q^{n-x}$.
For $n=5$ and $x=4$,we have:
$P(X=4) = ^{5}C_{4} \left(\frac{4}{5}\right)^{4} \left(\frac{1}{5}\right)^{5-4} = ^{5}C_{4} \left(\frac{4}{5}\right)^{4} \left(\frac{1}{5}\right)$.
Thus,the correct option is $A$.

(B) For a Binomial distribution $B(n, p)$,the mean is given by the formula $\mu = np$.
Given the distribution $B\left(4, \frac{1}{3}\right)$,we have:
$n = 4$
$p = \frac{1}{3}$
Therefore,the mean $\mu$ is:
$\mu = n \times p = 4 \times \frac{1}{3} = \frac{4}{3}$.

(C) Let the shooter fire $n$ times. These $n$ fires are $n$ Bernoulli trials.
In each trial,the probability of hitting the target is $p = \frac{3}{4}$ and the probability of not hitting the target is $q = 1 - p = \frac{1}{4}$.
The probability of hitting the target at least once is given by $P(X \geq 1) = 1 - P(X = 0)$.
Here,$P(X = 0)$ is the probability of not hitting the target in any of the $n$ trials,which is $q^n = (\frac{1}{4})^n$.
We are given that $P(X \geq 1) > 0.99$.
Therefore,$1 - (\frac{1}{4})^n > 0.99$.
This simplifies to $1 - 0.99 > (\frac{1}{4})^n$,which means $0.01 > (\frac{1}{4})^n$.
This is equivalent to $\frac{1}{100} > \frac{1}{4^n}$,or $4^n > 100$.
Testing values for $n$:
For $n = 3$,$4^3 = 64 < 100$.
For $n = 4$,$4^4 = 256 > 100$.
Thus,the minimum number of times the shooter must fire is $4$.

(B) Let $X$ be the number of right-handed people in a sample of $n = 10$.
The probability of a person being right-handed is $p = 0.9$,and the probability of being left-handed is $q = 1 - 0.9 = 0.1$.
$X$ follows a binomial distribution $B(n, p) = B(10, 0.9)$.
The probability that at most $6$ people are right-handed is $P(X \le 6)$.
Using the probability mass function of the binomial distribution,$P(X = r) = {^{n}C_r} p^r q^{n-r}$.
Thus,$P(X \le 6) = \sum_{r=0}^{6} {^{10}C_r} (0.9)^r (0.1)^{10-r}$.

(A) Total number of balls in the urn $= 25$.
Balls bearing mark $'X'$ $= 10$.
Balls bearing mark $'Y'$ $= 15$.
Probability of drawing a ball with mark $'X'$ is $p = \frac{10}{25} = \frac{2}{5}$.
Since the ball is replaced after each draw,the trials are independent Bernoulli trials.
We draw $n = 6$ balls.
Let $X$ be the random variable representing the number of balls with mark $'X'$.
This follows a binomial distribution $B(n, p)$ where $n = 6$ and $p = \frac{2}{5}$.
The probability that all $6$ balls bear mark $'X'$ is $P(X = 6)$.
Using the binomial probability formula $P(X = k) = ^{n}C_{k} p^{k} q^{n-k}$,where $q = 1 - p = \frac{3}{5}$:
$P(X = 6) = ^{6}C_{6} \left(\frac{2}{5}\right)^{6} \left(\frac{3}{5}\right)^{0} = 1 \times \left(\frac{2}{5}\right)^{6} \times 1 = \left(\frac{2}{5}\right)^{6}$.

(NONE) Total number of balls in the urn $= 25$.
Balls bearing mark $'X'$ $= 10$.
Balls bearing mark $'Y'$ $= 15$.
Let $p$ be the probability of drawing a ball with mark $'Y'$.
$p = P(\text{ball bearing mark } 'Y') = \frac{15}{25} = \frac{3}{5}$.
Let $q$ be the probability of drawing a ball with mark $'X'$.
$q = P(\text{ball bearing mark } 'X') = \frac{10}{25} = \frac{2}{5}$.
Since $6$ balls are drawn with replacement,the number of trials $n = 6$.
Let $Z$ be the random variable representing the number of balls with $'Y'$ mark. $Z$ follows a binomial distribution $B(n, p)$ where $n = 6$ and $p = \frac{3}{5}$.
The probability is given by $P(Z = z) = ^nC_z p^z q^{n-z}$.
We need to find $P(Z \leq 2) = P(Z = 0) + P(Z = 1) + P(Z = 2)$.
$P(Z = 0) = ^6C_0 (\frac{3}{5})^0 (\frac{2}{5})^6 = 1 \times 1 \times (\frac{2}{5})^6 = \frac{64}{15625}$.
$P(Z = 1) = ^6C_1 (\frac{3}{5})^1 (\frac{2}{5})^5 = 6 \times \frac{3}{5} \times \frac{32}{3125} = \frac{576}{15625}$.
$P(Z = 2) = ^6C_2 (\frac{3}{5})^2 (\frac{2}{5})^4 = 15 \times \frac{9}{25} \times \frac{16}{625} = \frac{2160}{15625}$.
$P(Z \leq 2) = \frac{64 + 576 + 2160}{15625} = \frac{2800}{15625} = \frac{112}{625}$.
Note: The original options provided were incorrect. The calculated result is $\frac{112}{625}$.

(A) Total number of balls in the urn $= 25$.
Number of balls bearing mark $'X'$ $= 10$.
Number of balls bearing mark $'Y'$ $= 15$.
Let $p$ be the probability of drawing a ball with mark $'X'$ and $q$ be the probability of drawing a ball with mark $'Y'$.
$p = P(\text{mark } 'X') = \frac{10}{25} = \frac{2}{5}$.
$q = P(\text{mark } 'Y') = \frac{15}{25} = \frac{3}{5}$.
Since $6$ balls are drawn with replacement,the trials are Bernoulli trials. Let $Z$ be the random variable representing the number of balls with mark $'Y'$.
$Z$ follows a binomial distribution with $n = 6$ and probability of success $q = \frac{3}{5}$ (since we are looking for mark $'Y'$).
The probability of getting at least one $'Y'$ mark is $P(Z \geq 1) = 1 - P(Z = 0)$.
$P(Z = 0) = ^{6}C_{0} \cdot p^{6} \cdot q^{0} = 1 \cdot (\frac{2}{5})^6 \cdot 1 = (\frac{2}{5})^6$.
Therefore,$P(Z \geq 1) = 1 - (\frac{2}{5})^6$.

(A) Total number of balls in the urn $= 25$.
Balls bearing mark $'X'$ $= 10$.
Balls bearing mark $'Y'$ $= 15$.
Let $p$ be the probability of drawing a ball with mark $'X'$: $p = \frac{10}{25} = \frac{2}{5}$.
Let $q$ be the probability of drawing a ball with mark $'Y'$: $q = \frac{15}{25} = \frac{3}{5}$.
Since $6$ balls are drawn with replacement,this follows a binomial distribution with $n = 6$.
We want the number of balls with $'X'$ mark and $'Y'$ mark to be equal,which means we need $3$ balls of mark $'X'$ and $3$ balls of mark $'Y'$.
Using the binomial probability formula $P(Z = k) = ^{n}C_{k} p^{k} q^{n-k}$:
$P(Z = 3) = ^{6}C_{3} \times (\frac{2}{5})^{3} \times (\frac{3}{5})^{3}$.
$P(Z = 3) = 20 \times \frac{8}{125} \times \frac{27}{125}$.
$P(Z = 3) = \frac{20 \times 216}{15625} = \frac{4320}{15625} = \frac{864}{3125}$.

(B) Let $p$ be the probability of clearing a hurdle and $q$ be the probability of knocking down a hurdle.
Given $p = \frac{5}{6}$,so $q = 1 - p = 1 - \frac{5}{6} = \frac{1}{6}$.
Let $X$ be the number of hurdles knocked down. Since there are $n = 10$ hurdles,$X$ follows a binomial distribution $B(n, q)$ where $n = 10$ and $q = \frac{1}{6}$ is the probability of success (knocking down a hurdle).
The probability of knocking down fewer than $2$ hurdles is $P(X < 2) = P(X = 0) + P(X = 1)$.
$P(X = x) = ^{10}C_x q^x p^{10-x}$.
$P(X = 0) = ^{10}C_0 (\frac{1}{6})^0 (\frac{5}{6})^{10} = (\frac{5}{6})^{10}$.
$P(X = 1) = ^{10}C_1 (\frac{1}{6})^1 (\frac{5}{6})^9 = 10 \times \frac{1}{6} \times (\frac{5}{6})^9 = \frac{10}{6} \times (\frac{5}{6})^9 = \frac{5}{3} \times (\frac{5}{6})^9$.
$P(X < 2) = (\frac{5}{6})^{10} + \frac{5}{3} \times (\frac{5}{6})^9 = (\frac{5}{6})^9 [\frac{5}{6} + \frac{5}{3}] = (\frac{5}{6})^9 [\frac{5+10}{6}] = (\frac{5}{6})^9 [\frac{15}{6}] = (\frac{5}{6})^9 [\frac{5}{2}] = \frac{5^{10}}{2 \times 6^9}$.

(A) The probability of getting a six in a single throw of a die is $p = \frac{1}{6}$,and the probability of not getting a six is $q = 1 - p = \frac{5}{6}$.
For the third six to occur exactly on the $6^{th}$ throw,there must be exactly $2$ sixes in the first $5$ throws,and the $6^{th}$ throw must be a six.
The probability of getting exactly $2$ sixes in the first $5$ throws follows the binomial distribution formula $P(X=k) = ^{n}C_{k} p^{k} q^{n-k}$,where $n=5$ and $k=2$.
$P(\text{2 sixes in 5 throws}) = ^{5}C_{2} \left(\frac{1}{6}\right)^{2} \left(\frac{5}{6}\right)^{3} = 10 \times \frac{1}{36} \times \frac{125}{216} = \frac{1250}{7776}$.
The probability that the $6^{th}$ throw results in a six is $\frac{1}{6}$.
Thus,the required probability is $P = \left(\frac{1250}{7776}\right) \times \frac{1}{6} = \frac{1250}{46656} = \frac{625}{23328}$.

(A) Let $p$ be the probability of success and $q$ be the probability of failure. Given that the experiment succeeds twice as often as it fails,we have $p = 2q$. Since $p + q = 1$,we get $2q + q = 1$,which implies $3q = 1$,so $q = \frac{1}{3}$ and $p = \frac{2}{3}$.
For $n = 6$ trials,the number of successes $X$ follows a binomial distribution $B(n, p) = B(6, \frac{2}{3})$.
The probability of at least $4$ successes is $P(X \geq 4) = P(X=4) + P(X=5) + P(X=6)$.
Using the formula $P(X=k) = \binom{n}{k} p^k q^{n-k}$:
$P(X=4) = \binom{6}{4} (\frac{2}{3})^4 (\frac{1}{3})^2 = 15 \times \frac{16}{81} \times \frac{1}{9} = \frac{240}{729}$.
$P(X=5) = \binom{6}{5} (\frac{2}{3})^5 (\frac{1}{3})^1 = 6 \times \frac{32}{243} \times \frac{1}{3} = \frac{192}{729}$.
$P(X=6) = \binom{6}{6} (\frac{2}{3})^6 (\frac{1}{3})^0 = 1 \times \frac{64}{729} \times 1 = \frac{64}{729}$.
Summing these probabilities: $P(X \geq 4) = \frac{240 + 192 + 64}{729} = \frac{496}{729}$.

(C) Let the man toss the coin $n$ times. The $n$ tosses are $n$ Bernoulli trials.
Probability $(p)$ of getting a head at the toss of a coin is $\frac{1}{2}$.
$p = \frac{1}{2}, q = \frac{1}{2}$.
Therefore,$P(X = x) = ^{n}C_{x} p^{x} q^{n-x} = ^{n}C_{x} (\frac{1}{2})^{x} (\frac{1}{2})^{n-x} = ^{n}C_{x} (\frac{1}{2})^{n}$.
It is given that $P(\text{getting at least one head}) > \frac{90}{100}$.
$P(X \geq 1) > 0.9$.
$1 - P(X = 0) > 0.9$.
$1 - ^{n}C_{0} \cdot (\frac{1}{2})^{n} > 0.9$.
$^{n}C_{0} \cdot (\frac{1}{2})^{n} < 0.1$.
$\frac{1}{2^{n}} < 0.1$.
$2^{n} > \frac{1}{0.1} = 10$.
Since $2^{3} = 8$ and $2^{4} = 16$,the minimum value of $n$ that satisfies $2^{n} > 10$ is $4$.
Thus,the man should toss the coin $4$ or more times.

(A) The total number of questions is $n = 6$.
We need to choose $4$ questions to be answered correctly out of $6$,which can be done in ${}^{6}C_{4}$ ways.
For each of the $4$ correct questions,there is only $1$ way to choose the correct answer.
For each of the remaining $2$ questions $(6 - 4 = 2)$,the candidate must choose an incorrect answer. Since there are $4$ alternatives and only $1$ is correct,there are $3$ incorrect options for each question.
Thus,the number of ways is ${}^{6}C_{4} \times (1)^4 \times (3)^2$.
${}^{6}C_{4} = \frac{6 \times 5}{2 \times 1} = 15$.
Number of ways = $15 \times 1 \times 9 = 135$.

(B) Let $n$ be the number of shots fired.
The probability of hitting the target in a single shot is $p = \frac{1}{10}$.
The probability of missing the target in a single shot is $q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10}$.
The probability of missing the target in all $n$ shots is $q^n = \left(\frac{9}{10}\right)^n$.
The probability of hitting the target at least once is $1 - P(\text{missing all}) = 1 - \left(\frac{9}{10}\right)^n$.
We are given that this probability is greater than $\frac{1}{4}$:
$1 - \left(\frac{9}{10}\right)^n > \frac{1}{4}$
$\Rightarrow \frac{3}{4} > \left(\frac{9}{10}\right)^n$.
For $n = 1$: $\left(\frac{9}{10}\right)^1 = 0.9 > 0.75$ (False).
For $n = 2$: $\left(\frac{9}{10}\right)^2 = 0.81 > 0.75$ (False).
For $n = 3$: $\left(\frac{9}{10}\right)^3 = 0.729 < 0.75$ (True).
Thus,the least number of shots required is $3$.

(A) Let $n$ be the number of bombs dropped. The probability of a bomb hitting the target is $p = \frac{1}{2}$,and the probability of missing is $q = 1 - p = \frac{1}{2}$.
The target is destroyed if there are at least $2$ hits. Let $X$ be the number of hits. We want $P(X \geq 2) \geq 0.99$.
This is equivalent to $1 - P(X < 2) \geq 0.99$,which means $1 - [P(X=0) + P(X=1)] \geq 0.99$.
Using the binomial distribution,$P(X=k) = {}^{n}C_{k} p^k q^{n-k}$.
$1 - [{}^{n}C_{0} (\frac{1}{2})^n + {}^{n}C_{1} (\frac{1}{2})^n] \geq 0.99$
$1 - \frac{1 + n}{2^n} \geq 0.99$
$\frac{1 + n}{2^n} \leq 0.01 = \frac{1}{100}$
$2^n \geq 100(n + 1)$.
Testing values for $n$:
For $n=10$: $2^{10} = 1024$,$100(11) = 1100$. $1024 \geq 1100$ is false.
For $n=11$: $2^{11} = 2048$,$100(12) = 1200$. $2048 \geq 1200$ is true.
Thus,the minimum number of bombs required is $11$.

(A) Let $X$ be the number of dice showing a $3$ or a $5$ in a single throw of $4$ dice. The probability of success (getting $3$ or $5$) for one die is $p = \frac{2}{6} = \frac{1}{3}$.
The probability of failure is $q = 1 - p = \frac{2}{3}$.
Since there are $n = 4$ dice,the number of successes $X$ follows a binomial distribution $B(4, \frac{1}{3})$.
We want the probability that at least two dice show a $3$ or a $5$,which is $P(X \ge 2) = 1 - P(X = 0) - P(X = 1)$.
$P(X = 0) = \binom{4}{0} (\frac{1}{3})^0 (\frac{2}{3})^4 = 1 \times 1 \times \frac{16}{81} = \frac{16}{81}$.
$P(X = 1) = \binom{4}{1} (\frac{1}{3})^1 (\frac{2}{3})^3 = 4 \times \frac{1}{3} \times \frac{8}{27} = \frac{32}{81}$.
$P(X \ge 2) = 1 - (\frac{16}{81} + \frac{32}{81}) = 1 - \frac{48}{81} = \frac{81 - 48}{81} = \frac{33}{81}$.
The experiment is performed $N = 27$ times. The expected number of times is $E = N \times P(X \ge 2) = 27 \times \frac{33}{81} = \frac{33}{3} = 11$.

(A) In a Binomial distribution with $n=5$ trials,the probability of $k$ successes is given by $P(X=k) = {}^{5}C_{k} \cdot p^{k} \cdot q^{n-k}$,where $p+q=1$.
Given $P(X=1) = {}^{5}C_{1} \cdot p \cdot q^{4} = 5pq^{4} = 0.4096$.
Given $P(X=2) = {}^{5}C_{2} \cdot p^{2} \cdot q^{3} = 10p^{2}q^{3} = 0.2048$.
Dividing the two equations: $\frac{10p^{2}q^{3}}{5pq^{4}} = \frac{0.2048}{0.4096} \Rightarrow \frac{2p}{q} = 0.5 \Rightarrow q = 4p$.
Since $p+q=1$,we have $p+4p=1 \Rightarrow 5p=1 \Rightarrow p = \frac{1}{5} = 0.2$ and $q = \frac{4}{5} = 0.8$.
Now,the probability of exactly $3$ successes is $P(X=3) = {}^{5}C_{3} \cdot p^{3} \cdot q^{2}$.
$P(X=3) = 10 \cdot (\frac{1}{5})^{3} \cdot (\frac{4}{5})^{2} = 10 \cdot \frac{1}{125} \cdot \frac{16}{25} = \frac{160}{3125} = \frac{32}{625}$.

(D) Let $n$ be the number of times the dice is rolled. The probability of getting an odd number in a single throw is $p = \frac{1}{2}$,and the probability of getting an even number is $q = 1 - p = \frac{1}{2}$.
The probability of getting an odd number $2$ times is given by the binomial distribution: $P(X=2) = {}^{n}C_{2} (\frac{1}{2})^{2} (\frac{1}{2})^{n-2} = {}^{n}C_{2} (\frac{1}{2})^{n}$.
The probability of getting an even number $3$ times is equivalent to getting an odd number $(n-3)$ times: $P(Y=3) = {}^{n}C_{3} (\frac{1}{2})^{3} (\frac{1}{2})^{n-3} = {}^{n}C_{3} (\frac{1}{2})^{n}$.
Given that $P(X=2) = P(Y=3)$,we have ${}^{n}C_{2} = {}^{n}C_{3}$.
Using the property ${}^{n}C_{r} = {}^{n}C_{n-r}$,we get $2 + 3 = n$,so $n = 5$.
We need to find the probability of getting an odd number an odd number of times,which is $P(X=1) + P(X=3) + P(X=5)$.
$P(X=1) + P(X=3) + P(X=5) = {}^{5}C_{1} (\frac{1}{2})^{5} + {}^{5}C_{3} (\frac{1}{2})^{5} + {}^{5}C_{5} (\frac{1}{2})^{5} = \frac{1}{2^{5}} (5 + 10 + 1) = \frac{16}{32} = \frac{1}{2}$.

(A) Let the coin be tossed $n$ times.
Since the coin is fair,the probability of getting a head is $p = \frac{1}{2}$ and the probability of getting a tail is $q = \frac{1}{2}$.
The probability of getting $r$ heads in $n$ tosses is given by the binomial distribution formula: $P(X=r) = {}^{n}C_{r} p^{r} q^{n-r} = {}^{n}C_{r} (\frac{1}{2})^{n}$.
Given that $P(X=7) = P(X=9)$,we have:
${}^{n}C_{7} (\frac{1}{2})^{n} = {}^{n}C_{9} (\frac{1}{2})^{n}$
${}^{n}C_{7} = {}^{n}C_{9}$
Using the property ${}^{n}C_{x} = {}^{n}C_{y} \implies x+y=n$ (if $x \neq y$),we get $n = 7 + 9 = 16$.
Now,we need to find the probability of getting $2$ heads,which is $P(X=2)$:
$P(X=2) = {}^{16}C_{2} (\frac{1}{2})^{16}$
$P(X=2) = \frac{16 \times 15}{2 \times 1} \times \frac{1}{2^{16}}$
$P(X=2) = 8 \times 15 \times \frac{1}{2^{16}} = \frac{15}{2^{3} \times 2^{16-3}} = \frac{15}{2^{13}}$.

(C) Let $X$ be the number of heads obtained by person $A$ and $Y$ be the number of heads obtained by person $B$. Both $X$ and $Y$ follow a binomial distribution $B(n=3, p=1/2)$.
The probability of getting $k$ heads in $3$ tosses is given by $P(X=k) = \binom{3}{k} (1/2)^3 = \binom{3}{k} / 8$.
We want to find $P(X=Y) = P(X=0, Y=0) + P(X=1, Y=1) + P(X=2, Y=2) + P(X=3, Y=3)$.
Since $A$ and $B$ are independent,$P(X=k, Y=k) = P(X=k) \times P(Y=k) = [P(X=k)]^2$.
$P(X=0) = \binom{3}{0}/8 = 1/8 \implies P(X=0, Y=0) = (1/8)^2 = 1/64$.
$P(X=1) = \binom{3}{1}/8 = 3/8 \implies P(X=1, Y=1) = (3/8)^2 = 9/64$.
$P(X=2) = \binom{3}{2}/8 = 3/8 \implies P(X=2, Y=2) = (3/8)^2 = 9/64$.
$P(X=3) = \binom{3}{3}/8 = 1/8 \implies P(X=3, Y=3) = (1/8)^2 = 1/64$.
Total probability $= 1/64 + 9/64 + 9/64 + 1/64 = 20/64 = 5/16$.

(C) The total number of ways to distribute $9$ distinct balls into $4$ boxes is $4^{9}$.
The number of ways to choose $3$ balls for box $B_{3}$ is ${}^{9}C_{3}$.
The remaining $6$ balls can be distributed among the other $3$ boxes $(B_{1}, B_{2}, B_{4})$ in $3^{6}$ ways.
Thus,the probability that $B_{3}$ contains exactly $3$ balls is $P = \frac{{}^{9}C_{3} \cdot 3^{6}}{4^{9}}$.
We can rewrite this as $P = \frac{{}^{9}C_{3} \cdot 3^{6}}{4^{9}} = \frac{84 \cdot 3^{6}}{4^{9}}$.
Since we want the form $k \left(\frac{3}{4}\right)^{9}$,we write $P = k \cdot \frac{3^{9}}{4^{9}}$.
Equating the two expressions: $k \cdot \frac{3^{9}}{4^{9}} = \frac{84 \cdot 3^{6}}{4^{9}}$.
$k = \frac{84 \cdot 3^{6}}{3^{9}} = \frac{84}{3^{3}} = \frac{84}{27} = \frac{28}{9} \approx 3.11$.
Checking the options for $k = \frac{28}{9} \approx 3.11$:
$|x-3| < 1 \Rightarrow 2 < x < 4$. Since $3.11$ lies in this interval,option $C$ is correct.

(B) The probability of getting a head in a single toss is $P(H) = \frac{1}{2}$.
The probability of getting no heads in $n$ tosses is $P(X=0) = \left(\frac{1}{2}\right)^n$.
The probability of getting at least one head is $P(X \geq 1) = 1 - P(X=0) = 1 - \left(\frac{1}{2}\right)^n$.
Given that $P(X \geq 1) \geq 0.9$,we have:
$1 - \left(\frac{1}{2}\right)^n \geq 0.9$
Rearranging the inequality:
$1 - 0.9 \geq \left(\frac{1}{2}\right)^n$
$0.1 \geq \left(\frac{1}{2}\right)^n$
$\frac{1}{10} \geq \frac{1}{2^n}$
$2^n \geq 10$
Testing values for $n$:
For $n=3$,$2^3 = 8 < 10$.
For $n=4$,$2^4 = 16 \geq 10$.
Thus,the minimum value of $n$ is $4$.

(A) Let $X$ be the number of correct answers. Since the student guesses,$X$ follows a binomial distribution $B(n=8, p=1/2)$.
We want to find the smallest $n$ such that $P(X \geq n) < \frac{1}{2}$.
$P(X \geq n) = \sum_{r=n}^{8} {}^{8}C_{r} (\frac{1}{2})^{r} (\frac{1}{2})^{8-r} = \frac{1}{2^{8}} \sum_{r=n}^{8} {}^{8}C_{r} < \frac{1}{2}$.
$\sum_{r=n}^{8} {}^{8}C_{r} < 2^{7} = 128$.
We know that $\sum_{r=0}^{8} {}^{8}C_{r} = 2^{8} = 256$.
So,$\sum_{r=n}^{8} {}^{8}C_{r} = 256 - \sum_{r=0}^{n-1} {}^{8}C_{r} < 128$.
$\sum_{r=0}^{n-1} {}^{8}C_{r} > 128$.
For $n=5$,$\sum_{r=0}^{4} {}^{8}C_{r} = {}^{8}C_{0} + {}^{8}C_{1} + {}^{8}C_{2} + {}^{8}C_{3} + {}^{8}C_{4} = 1 + 8 + 28 + 56 + 70 = 163$.
Since $163 > 128$,the condition holds for $n=5$.
For $n=4$,$\sum_{r=0}^{3} {}^{8}C_{r} = 1 + 8 + 28 + 56 = 93$,which is not greater than $128$.
Thus,the smallest value of $n$ is $5$.

(D) Let $S_1$ be the set of $4$ questions with probability of success $p_1 = \frac{3}{4}$ and $S_2$ be the set of $6$ questions with probability of success $p_2 = \frac{1}{4}$.
To get exactly $8$ questions correct,we consider the following cases $(x, y)$ where $x$ is the number of correct answers from $S_1$ and $y$ is the number of correct answers from $S_2$,such that $x+y=8$:
Case $1$: $x=4, y=4$. Probability $= \binom{4}{4} (\frac{3}{4})^4 (\frac{1}{4})^0 \times \binom{6}{4} (\frac{1}{4})^4 (\frac{3}{4})^2 = 1 \times \frac{81}{256} \times 15 \times \frac{9}{4096} = \frac{10935}{4^{10}}$.
Case $2$: $x=3, y=5$. Probability $= \binom{4}{3} (\frac{3}{4})^3 (\frac{1}{4})^1 \times \binom{6}{5} (\frac{1}{4})^5 (\frac{3}{4})^1 = 4 \times \frac{27}{64} \times \frac{1}{4} \times 6 \times \frac{3}{4096} = \frac{1944}{4^{10}}$.
Case $3$: $x=2, y=6$. Probability $= \binom{4}{2} (\frac{3}{4})^2 (\frac{1}{4})^2 \times \binom{6}{6} (\frac{1}{4})^6 (\frac{3}{4})^0 = 6 \times \frac{9}{16} \times \frac{1}{16} \times 1 \times \frac{1}{4096} = \frac{54}{4^{10}}$.
Total probability $= \frac{10935 + 1944 + 54}{4^{10}} = \frac{12933}{4^{10}}$.
Given $\frac{27k}{4^{10}} = \frac{12933}{4^{10}}$,we have $27k = 12933 \Rightarrow k = \frac{12933}{27} = 479$.

(A) Given $n = 33$,let the probability of success be $p$ and $q = 1 - p$.
Given $3P(X=0) = P(X=1)$.
Using the Binomial distribution formula $P(X=k) = {}^{n}C_{k} p^{k} q^{n-k}$,we have:
$3 \cdot {}^{33}C_{0} p^{0} q^{33} = {}^{33}C_{1} p^{1} q^{32}$
$3q = 33p \implies q = 11p$.
Since $p + q = 1$,we have $p + 11p = 1 \implies 12p = 1 \implies p = \frac{1}{12}$ and $q = \frac{11}{12}$.
Thus,$\frac{q}{p} = 11$.
Now,consider the expression $\frac{P(X=15)}{P(X=18)} - \frac{P(X=16)}{P(X=17)}$.
Using the formula $\frac{P(X=k)}{P(X=k+1)} = \frac{{}^{n}C_{k} p^{k} q^{n-k}}{{}^{n}C_{k+1} p^{k+1} q^{n-k-1}} = \frac{k+1}{n-k} \cdot \frac{q}{p}$,we evaluate the terms:
$\frac{P(X=15)}{P(X=18)} = \frac{P(X=15)}{P(X=16)} \cdot \frac{P(X=16)}{P(X=17)} \cdot \frac{P(X=17)}{P(X=18)} = \left(\frac{16}{18} \cdot 11\right) \cdot \left(\frac{17}{17} \cdot 11\right) \cdot \left(\frac{18}{16} \cdot 11\right) = 11^3 = 1331$.
Alternatively,$\frac{P(X=k)}{P(X=n-k)} = \frac{{}^{n}C_{k} p^{k} q^{n-k}}{{}^{n}C_{n-k} p^{n-k} q^{k}} = \left(\frac{q}{p}\right)^{n-2k}$.
For the first term: $\frac{P(X=15)}{P(X=18)} = \frac{{}^{33}C_{15} p^{15} q^{18}}{{}^{33}C_{18} p^{18} q^{15}} = \left(\frac{q}{p}\right)^{18-15} = \left(\frac{q}{p}\right)^3 = 11^3 = 1331$.
For the second term: $\frac{P(X=16)}{P(X=17)} = \frac{{}^{33}C_{16} p^{16} q^{17}}{{}^{33}C_{17} p^{17} q^{16}} = \left(\frac{q}{p}\right)^{17-16} = \left(\frac{q}{p}\right)^1 = 11$.
Therefore,the value is $1331 - 11 = 1320$.

(D) Let $P(H) = x$ and $P(T) = 1 - x$.
Given $P(4H, 1T) = P(5H)$.
Using the binomial distribution formula,${}^{5}C_{4} x^{4}(1-x)^{1} = {}^{5}C_{5} x^{5}$.
$5(1-x) = x$.
$5 - 5x = x \implies 6x = 5 \implies x = \frac{5}{6}$.
Thus,$P(H) = \frac{5}{6}$ and $P(T) = \frac{1}{6}$.
We need to find $P(\text{at most } 2H) = P(0H) + P(1H) + P(2H)$.
$P(0H) = {}^{5}C_{0} (\frac{5}{6})^{0} (\frac{1}{6})^{5} = \frac{1}{6^{5}}$.
$P(1H) = {}^{5}C_{1} (\frac{5}{6})^{1} (\frac{1}{6})^{4} = 5 \times \frac{5}{6^{5}} = \frac{25}{6^{5}}$.
$P(2H) = {}^{5}C_{2} (\frac{5}{6})^{2} (\frac{1}{6})^{3} = 10 \times \frac{25}{6^{5}} = \frac{250}{6^{5}}$.
Summing these,$P(\text{at most } 2H) = \frac{1 + 25 + 250}{6^{5}} = \frac{276}{6^{5}}$.
Simplifying,$\frac{276}{6^{5}} = \frac{46 \times 6}{6^{5}} = \frac{46}{6^{4}}$.

(B) Given that $X$ follows a binomial distribution $B(n, p)$ with $n = 7$.
The probability mass function is given by $P(X=k) = {}^{n}C_{k} p^{k} (1-p)^{n-k}$.
Given the condition $P(X=3) = 5P(X=4)$:
${}^{7}C_{3} p^{3} (1-p)^{4} = 5 \times {}^{7}C_{4} p^{4} (1-p)^{3}$.
Since ${}^{7}C_{3} = {}^{7}C_{4} = 35$,we can simplify the equation:
$35 p^{3} (1-p)^{4} = 5 \times 35 p^{4} (1-p)^{3}$.
Dividing both sides by $35 p^{3} (1-p)^{3}$ (assuming $p \neq 0, 1$):
$(1-p) = 5p$.
$1 = 6p \Rightarrow p = \frac{1}{6}$.
Thus,$q = 1 - p = \frac{5}{6}$.
Mean $= np = 7 \times \frac{1}{6} = \frac{7}{6}$.
Variance $= npq = 7 \times \frac{1}{6} \times \frac{5}{6} = \frac{35}{36}$.
Sum of mean and variance $= \frac{7}{6} + \frac{35}{36} = \frac{42 + 35}{36} = \frac{77}{36}$.

(C) For a binomial distribution,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $p+q=1$.
Given $np + npq = 24$ and $(np)(npq) = 128$.
Let $A = np$ and $B = npq$. Then $A+B=24$ and $AB=128$.
The quadratic equation $t^2 - 24t + 128 = 0$ has roots $t = 8$ and $t = 16$.
Case $1$: $np = 16$ and $npq = 8$. Then $q = \frac{8}{16} = \frac{1}{2}$,so $p = \frac{1}{2}$. Thus $n(\frac{1}{2}) = 16 \implies n = 32$.
Case $2$: $np = 8$ and $npq = 16$. Then $q = \frac{16}{8} = 2$,which is impossible since $q \leq 1$.
So,$n=32, p=\frac{1}{2}, q=\frac{1}{2}$.
The probability of one or two successes is $P(X=1) + P(X=2) = {}^{32}C_1 (\frac{1}{2})^1 (\frac{1}{2})^{31} + {}^{32}C_2 (\frac{1}{2})^2 (\frac{1}{2})^{30}$.
$= 32 \cdot (\frac{1}{2})^{32} + \frac{32 \cdot 31}{2} \cdot (\frac{1}{2})^{32} = (32 + 496) \cdot \frac{1}{2^{32}} = 528 \cdot \frac{1}{2^{32}} = \frac{33 \cdot 16}{2^{32}} = \frac{33}{2^{28}}$.

(B) For a binomial distribution,the mean is $np = 4$ and the variance is $npq = \frac{4}{3}$.
Dividing the variance by the mean,we get $q = \frac{4/3}{4} = \frac{1}{3}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{3} = \frac{2}{3}$.
Substituting $p$ into $np = 4$,we get $n \times \frac{2}{3} = 4$,which gives $n = 6$.
The probability mass function is $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k} = {}^{6}C_{k} (\frac{2}{3})^{k} (\frac{1}{3})^{6-k}$.
We need to calculate $54 P(X \leq 2) = 54 [P(X=0) + P(X=1) + P(X=2)]$.
$P(X=0) = {}^{6}C_{0} (\frac{2}{3})^{0} (\frac{1}{3})^{6} = 1 \times 1 \times \frac{1}{729} = \frac{1}{729}$.
$P(X=1) = {}^{6}C_{1} (\frac{2}{3})^{1} (\frac{1}{3})^{5} = 6 \times \frac{2}{3} \times \frac{1}{243} = \frac{12}{729}$.
$P(X=2) = {}^{6}C_{2} (\frac{2}{3})^{2} (\frac{1}{3})^{4} = 15 \times \frac{4}{9} \times \frac{1}{81} = \frac{60}{729}$.
Summing these,$P(X \leq 2) = \frac{1 + 12 + 60}{729} = \frac{73}{729}$.
Finally,$54 P(X \leq 2) = 54 \times \frac{73}{729} = \frac{2 \times 73}{27} = \frac{146}{27}$.

(C) For a binomial distribution,the mean is $np = \alpha$ and the variance is $npq = \frac{\alpha}{3}$.
Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{\alpha/3}{\alpha} = \frac{1}{3}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{3} = \frac{2}{3}$.
Given $P(X=1) = \frac{4}{243}$,we use the formula $P(X=k) = {}^{n}C_{k} p^{k} q^{n-k}$.
${}^{n}C_{1} (\frac{2}{3})^{1} (\frac{1}{3})^{n-1} = \frac{4}{243} \implies n \cdot \frac{2}{3} \cdot \frac{1}{3^{n-1}} = \frac{4}{243} \implies \frac{2n}{3^{n}} = \frac{4}{243} \implies \frac{n}{3^{n}} = \frac{2}{243}$.
Since $3^{5} = 243$,we have $n=6$.
Now,we calculate $P(X=4 \text{ or } 5) = P(X=4) + P(X=5)$.
$P(X=4) = {}^{6}C_{4} (\frac{2}{3})^{4} (\frac{1}{3})^{2} = 15 \cdot \frac{16}{81} \cdot \frac{1}{9} = \frac{240}{729} = \frac{80}{243}$.
$P(X=5) = {}^{6}C_{5} (\frac{2}{3})^{5} (\frac{1}{3})^{1} = 6 \cdot \frac{32}{243} \cdot \frac{1}{3} = \frac{192}{729} = \frac{64}{243}$.
$P(X=4 \text{ or } 5) = \frac{80}{243} + \frac{64}{243} = \frac{144}{243} = \frac{16}{27}$.

(B) Let $\mu = np$ be the mean and $\sigma^2 = npq$ be the variance of the binomial distribution $X \sim B(n, p)$.
Given $\mu + \sigma^2 = 24$ and $\mu \sigma^2 = 128$.
Let $x = \mu$ and $y = \sigma^2$. Then $x + y = 24$ and $xy = 128$.
The quadratic equation $t^2 - 24t + 128 = 0$ has roots $t = 16$ and $t = 8$.
Since $\mu > \sigma^2$ (as $q < 1$),we have $\mu = 16$ and $\sigma^2 = 8$.
Thus,$np = 16$ and $npq = 8$. Dividing these gives $q = \frac{8}{16} = \frac{1}{2}$.
Since $p = 1 - q$,we have $p = \frac{1}{2}$.
Then $n \times \frac{1}{2} = 16$,so $n = 32$.
We need to find $P(X > n - 3) = P(X > 29) = P(X = 30) + P(X = 31) + P(X = 32)$.
$P(X = r) = {}^{n}C_r p^r q^{n-r} = {}^{32}C_r (\frac{1}{2})^r (\frac{1}{2})^{32-r} = \frac{{}^{32}C_r}{2^{32}}$.
So,$P(X > 29) = \frac{{}^{32}C_{30} + {}^{32}C_{31} + {}^{32}C_{32}}{2^{32}} = \frac{k}{2^{32}}$.
Thus,$k = {}^{32}C_{30} + {}^{32}C_{31} + {}^{32}C_{32} = {}^{32}C_2 + {}^{32}C_1 + {}^{32}C_0$.
$k = \frac{32 \times 31}{2} + 32 + 1 = 496 + 32 + 1 = 529$.

(D) Let the mean be $m = np$ and the variance be $v = npq$,where $p + q = 1$.
Given,the sum $m + v = 82.5 = \frac{165}{2}$ and the product $mv = 1350$.
Since $m$ and $v$ are roots of the quadratic equation $x^2 - (m+v)x + mv = 0$,we have:
$x^2 - \frac{165}{2}x + 1350 = 0$
$2x^2 - 165x + 2700 = 0$
Solving this quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{165 \pm \sqrt{165^2 - 4(2)(2700)}}{2(2)} = \frac{165 \pm \sqrt{27225 - 21600}}{4} = \frac{165 \pm \sqrt{5625}}{4} = \frac{165 \pm 75}{4}$
Thus,$x_1 = \frac{240}{4} = 60$ and $x_2 = \frac{90}{4} = 22.5$.
Since $m > v$ for a binomial distribution (as $q < 1$),we have $m = 60$ and $v = 22.5$.
Now,$v = mq \implies 22.5 = 60q \implies q = \frac{22.5}{60} = \frac{225}{600} = \frac{3}{8}$.
Since $p = 1 - q$,we have $p = 1 - \frac{3}{8} = \frac{5}{8}$.
Finally,$m = np \implies 60 = n \times \frac{5}{8} \implies n = 60 \times \frac{8}{5} = 12 \times 8 = 96$.

(D) Let $p = \frac{1}{4}$ be the probability of hitting the target and $q = 1 - p = \frac{3}{4}$ be the probability of missing the target.
For the shooter to fire exactly $6$ bullets and hit the target exactly $3$ times,the $6^{th}$ bullet must be the $3^{rd}$ successful hit.
This means in the first $5$ shots,the shooter must have hit the target exactly $2$ times and missed $3$ times.
The probability of this event is given by the negative binomial distribution logic:
$P = \binom{5}{2} p^2 q^3 \times p = \binom{5}{2} p^3 q^3$.
Substituting the values:
$P = 10 \times \left(\frac{1}{4}\right)^3 \times \left(\frac{3}{4}\right)^3 = 10 \times \frac{1}{64} \times \frac{27}{64} = \frac{270}{4096}$.
Calculating the decimal value:
$P = \frac{270}{4096} \approx 0.0659179$.
This value lies in the interval $(0.0658, 0.0660)$.

(A) For event $E_1$,six fair dice are rolled. The probability that no die shows a six is $(\frac{5}{6})^6$. Thus,$p_1 = 1 - (\frac{5}{6})^6 = 1 - 0.3349 = 0.6651$.
For event $E_2$,twelve fair dice are rolled. Let $X$ be the number of dice showing a six. $X$ follows a binomial distribution $B(n=12, p=\frac{1}{6})$.
$p_2 = P(X \geq 2) = 1 - [P(X=0) + P(X=1)]$.
$P(X=0) = (\frac{5}{6})^{12} \approx 0.1122$.
$P(X=1) = \binom{12}{1} (\frac{1}{6})^1 (\frac{5}{6})^{11} = 12 \times \frac{1}{6} \times 0.1346 \approx 0.2692$.
$p_2 = 1 - (0.1122 + 0.2692) = 1 - 0.3814 = 0.6186$.
Since $0.6651 > 0.6186$,we have $p_1 > p_2$.

(D) Let $X_i$ be the number of balls added to box $B_1$ in the $i$-th trial.
In each trial,$D_1$ shows $j \in \{1, 2, 3, 4, 5, 6\}$ and $D_2$ shows $k \in \{1, 2, 3, 4, 5, 6\}$.
Box $B_1$ receives $j$ balls if $k=1$,and $0$ balls if $k \neq 1$.
For $B_1$ to contain at most one ball after $n$ trials,either zero balls are added in all $n$ trials,or exactly one ball is added in one trial and zero in the remaining $n-1$ trials.
Case $1$: Zero balls are added in all $n$ trials. This happens if $k \neq 1$ in every trial. The probability is $(\frac{5}{6})^n$.
Case $2$: Exactly one ball is added in one trial and zero in the others. This happens if in one trial $j=1$ and $k=1$ (probability $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$),and in the other $n-1$ trials $k \neq 1$ (probability $(\frac{5}{6})^{n-1}$).
Since there are $n$ choices for the trial in which the ball is added,the probability is $n \times \frac{1}{36} \times (\frac{5}{6})^{n-1} = n \times \frac{5^{n-1}}{6^{n-1}} \times \frac{1}{6^2}$.
Total probability = $(\frac{5}{6})^n + n \times \frac{5^{n-1}}{6^{n-1}} \times \frac{1}{6^2}$.

(B) Let $H$ be the number of heads and $T$ be the number of tails. Since the coin is tossed $10$ times,$H + T = 10$.
The total score $S$ is given by $S = 1 \times H + 2 \times T = H + 2(10 - H) = 20 - H$.
We want to find the largest $K$ such that $P(S \geq K) > \frac{1}{2}$.
Substituting $S = 20 - H$,we get $P(20 - H \geq K) = P(H \leq 20 - K) > \frac{1}{2}$.
Let $m = 20 - K$. We need $P(H \leq m) > \frac{1}{2}$.
The probability distribution of $H$ is binomial with $n = 10$ and $p = \frac{1}{2}$.
$P(H \leq m) = \frac{1}{2^{10}} \sum_{i=0}^{m} \binom{10}{i}$.
We know that $\sum_{i=0}^{10} \binom{10}{i} = 2^{10} = 1024$. Since the distribution is symmetric,$P(H \leq 4) = \sum_{i=0}^{4} \binom{10}{i} / 1024 = (1 + 10 + 45 + 120 + 210) / 1024 = 386 / 1024 < \frac{1}{2}$.
$P(H \leq 5) = (386 + \binom{10}{5}) / 1024 = (386 + 252) / 1024 = 638 / 1024 > \frac{1}{2}$.
Thus,the largest $m$ is $5$.
Since $m = 20 - K$,we have $5 = 20 - K$,which gives $K = 15$.

(D) The total number of questions is $n = 8$.
Each question has $4$ options,meaning there is $1$ correct option and $3$ incorrect options.
The student needs to choose exactly $5$ correct answers out of $8$.
The number of ways to select $5$ questions to be correct is given by the combination formula $\binom{8}{5}$.
For the $5$ correct questions,there is only $1$ way to choose the correct option.
For the remaining $8 - 5 = 3$ incorrect questions,each can be answered in $3$ different ways (since there are $4$ options and $1$ is correct,$4 - 1 = 3$ are incorrect).
Thus,the total number of ways is $\binom{8}{5} \times (1)^5 \times (3)^3$.
Calculating this: $\binom{8}{5} = \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
Total ways $= 56 \times 1 \times 27 = 1512$.

(B) The faces are $\{-2, -1, 0, 1, 2, 3\}$. The product is positive if no outcome is $0$ and the number of negative outcomes is even.
Let $P$ be the event of getting a positive number $\{1, 2, 3\}$,$N$ be the event of getting a negative number $\{-2, -1\}$,and $Z$ be the event of getting $0$.
$P(P) = \frac{3}{6} = \frac{1}{2}$,$P(N) = \frac{2}{6} = \frac{1}{3}$,$P(Z) = \frac{1}{6}$.
For the product to be positive,we must have no $0$s and an even number of negative outcomes ($0, 2, 4$ negatives).
Case $1$: $0$ negatives,$5$ positives: $\binom{5}{0} (\frac{1}{2})^5 = \frac{1}{32} = \frac{81}{2592}$.
Case $2$: $2$ negatives,$3$ positives: $\binom{5}{2} (\frac{1}{3})^2 (\frac{1}{2})^3 = 10 \times \frac{1}{9} \times \frac{1}{8} = \frac{10}{72} = \frac{360}{2592}$.
Case $3$: $4$ negatives,$1$ positive: $\binom{5}{4} (\frac{1}{3})^4 (\frac{1}{2})^1 = 5 \times \frac{1}{81} \times \frac{1}{2} = \frac{5}{162} = \frac{80}{2592}$.
Total probability $= \frac{81 + 360 + 80}{2592} = \frac{521}{2592}$.

(B) Total balls = $6$. Since balls are drawn with replacement,the total number of outcomes for drawing $k$ balls is $6^k$.
For $p$: Two balls are drawn. Both are the same colour. There are $6$ choices for the colour,so the number of favorable outcomes is $6$. Thus,$p = \frac{6}{6^2} = \frac{6}{36} = \frac{1}{6}$.
For $q$: Four balls are drawn. Exactly three are of the same colour.
Step $1$: Choose the colour that appears $3$ times ($6$ ways).
Step $2$: Choose the position of these $3$ balls in $4$ draws ($^4C_3 = 4$ ways).
Step $3$: Choose the colour of the remaining $1$ ball ($5$ ways).
Number of favorable outcomes = $6 \times 4 \times 5 = 120$.
Thus,$q = \frac{120}{6^4} = \frac{120}{1296} = \frac{5}{54}$.
Ratio $p : q = \frac{1}{6} : \frac{5}{54} = \frac{9}{54} : \frac{5}{54} = 9 : 5$.
Here $m = 9$ and $n = 5$,which are coprime.
Therefore,$m + n = 9 + 5 = 14$.

(B) For a binomial distribution $B(n, p)$,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $q = 1-p$.
Given that the sum of mean and variance is $5$: $np + npq = 5 \Rightarrow np(1+q) = 5$.
Given that the product of mean and variance is $6$: $np \cdot npq = 6 \Rightarrow n^2p^2q = 6$.
From the first equation,$np = \frac{5}{1+q}$. Substituting this into the second equation:
$(\frac{5}{1+q})^2 \cdot q = 6 \Rightarrow 25q = 6(1+q)^2$.
$25q = 6(1 + 2q + q^2) \Rightarrow 6q^2 + 12q + 6 = 25q$.
$6q^2 - 13q + 6 = 0$.
Solving the quadratic equation: $6q^2 - 9q - 4q + 6 = 0 \Rightarrow 3q(2q-3) - 2(2q-3) = 0$.
$(3q-2)(2q-3) = 0$. Since $q < 1$,we have $q = \frac{2}{3}$.
Then $p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$.
Using $np(1+q) = 5$: $n(\frac{1}{3})(1 + \frac{2}{3}) = 5 \Rightarrow n(\frac{1}{3})(\frac{5}{3}) = 5 \Rightarrow n(\frac{5}{9}) = 5 \Rightarrow n = 9$.
Finally,$6(n+p-q) = 6(9 + \frac{1}{3} - \frac{2}{3}) = 6(9 - \frac{1}{3}) = 54 - 2 = 52$.

(B) The total number of outcomes when throwing two dice is $6 \times 6 = 36$.
The outcomes resulting in a sum of $5$ are $(1, 4), (2, 3), (3, 2), (4, 1)$,which are $4$ outcomes.
The probability of success $p = \frac{4}{36} = \frac{1}{9}$.
The probability of failure $q = 1 - p = 1 - \frac{1}{9} = \frac{8}{9}$.
We use the binomial distribution formula $P(X = r) = {}^nC_r p^r q^{n-r}$ with $n = 5$.
$P(\text{at least } 4 \text{ successes}) = P(X = 4) + P(X = 5)$.
$P(X = 4) = {}^5C_4 \times (\frac{1}{9})^4 \times (\frac{8}{9})^1 = 5 \times \frac{1}{9^4} \times \frac{8}{9} = \frac{40}{9^5} = \frac{40}{3^{10}}$.
$P(X = 5) = {}^5C_5 \times (\frac{1}{9})^5 \times (\frac{8}{9})^0 = 1 \times \frac{1}{9^5} = \frac{1}{3^{10}}$.
Total probability $= \frac{40}{3^{10}} + \frac{1}{3^{10}} = \frac{41}{3^{10}} = \frac{41 \times 3}{3^{11}} = \frac{123}{3^{11}}$.
Comparing this with $\frac{k}{3^{11}}$,we get $k = 123$.

(D) Let $p$ be the probability of getting an odd number,$p = \frac{1}{2}$.
Given $P(\text{odd } 7 \text{ times}) = P(\text{odd } 9 \text{ times})$.
Using the binomial distribution formula $P(X=r) = {}^{n}C_{r} p^r q^{n-r}$:
${}^{n}C_{7} (\frac{1}{2})^7 (\frac{1}{2})^{n-7} = {}^{n}C_{9} (\frac{1}{2})^9 (\frac{1}{2})^{n-9}$
${}^{n}C_{7} = {}^{n}C_{9}$
Since ${}^{n}C_{a} = {}^{n}C_{b}$ implies $a+b=n$ or $a=b$,we have $n = 7+9 = 16$.
Now,we need the probability of getting even numbers twice in $16$ rolls. The probability of an even number is $q = \frac{1}{2}$.
$P(\text{even } 2 \text{ times}) = {}^{16}C_{2} (\frac{1}{2})^2 (\frac{1}{2})^{16-2} = {}^{16}C_{2} (\frac{1}{2})^{16}$
$= \frac{16 \times 15}{2} \times \frac{1}{2^{16}} = 120 \times \frac{1}{2^{16}} = \frac{120}{2 \times 2^{15}} = \frac{60}{2^{15}}$.
Comparing with $\frac{k}{2^{15}}$,we get $k = 60$.

(C) For a binomial distribution $B(n, p)$,the mean is $np$ and the variance is $npq$,where $q = 1-p$.
Given $np - npq = 1$,we have $np(1-q) = 1$,which implies $np^2 = 1$.
Given $2 P(X=2) = 3 P(X=1)$,we substitute the binomial probability formula $P(X=k) = \binom{n}{k} p^k q^{n-k}$:
$2 \binom{n}{2} p^2 q^{n-2} = 3 \binom{n}{1} p^1 q^{n-1}$
$2 \cdot \frac{n(n-1)}{2} p^2 q^{n-2} = 3n p q^{n-1}$
$(n-1) p = 3q$
Since $q = 1-p$,we have $(n-1)p = 3(1-p) \Rightarrow np - p = 3 - 3p \Rightarrow np + 2p = 3$.
From $np^2 = 1$,we have $n = \frac{1}{p^2}$. Substituting this into $np + 2p = 3$:
$\frac{1}{p^2} \cdot p + 2p = 3 \Rightarrow \frac{1}{p} + 2p = 3 \Rightarrow 2p^2 - 3p + 1 = 0$.
Solving the quadratic equation: $(2p-1)(p-1) = 0$. Since $p < 1$,we have $p = \frac{1}{2}$.
Then $n = \frac{1}{(1/2)^2} = 4$.
We need to find $n^2 P(X > 1) = 16(1 - (P(X=0) + P(X=1)))$.
$P(X=0) = \binom{4}{0} (1/2)^4 = 1/16$.
$P(X=1) = \binom{4}{1} (1/2)^1 (1/2)^3 = 4/16 = 1/4$.
$P(X > 1) = 1 - (1/16 + 4/16) = 1 - 5/16 = 11/16$.
Thus,$n^2 P(X > 1) = 16 \times \frac{11}{16} = 11$.

(A) Let the probability of getting a tail be $P(T) = p$. Then the probability of getting a head is $P(H) = 2p$.
Since $P(H) + P(T) = 1$,we have $2p + p = 1$,which gives $3p = 1$,so $p = \frac{1}{3}$.
Thus,$P(T) = \frac{1}{3}$ and $P(H) = \frac{2}{3}$.
We need the probability of getting $2$ tails and $1$ head in $3$ tosses.
The number of ways to arrange $2$ tails and $1$ head is given by the binomial coefficient $\binom{3}{1} = 3$ (the arrangements are $TTH, THT, HTT$).
The probability for each arrangement is $P(T) \times P(T) \times P(H) = \left(\frac{1}{3}\right)^2 \times \frac{2}{3} = \frac{2}{27}$.
Therefore,the total probability is $3 \times \frac{2}{27} = \frac{6}{27} = \frac{2}{9}$.

(C) $P(W) = \frac{1}{3}, P(L) = \frac{2}{3}$. Let $x$ be the number of wins and $y$ be the number of losses. Given $x+y=10$ and $|x-y| \leq 2$.
Case $I$: $|x-y|=0 \Rightarrow x=y$. Since $x+y=10$,we have $x=5, y=5$. The probability is $P(x=5) = {}^{10}C_5 (\frac{1}{3})^5 (\frac{2}{3})^5 = {}^{10}C_5 \frac{2^5}{3^{10}}$.
Case $II$: $|x-y|=1$. Since $x+y=10$,$x-y = \pm 1$ implies $2x = 11$ or $2x = 9$,which has no integer solutions. Thus,$P(|x-y|=1) = 0$.
Case $III$: $|x-y|=2$. This implies $x-y=2$ or $x-y=-2$.
If $x-y=2$ and $x+y=10$,then $x=6, y=4$.
If $x-y=-2$ and $x+y=10$,then $x=4, y=6$.
$P(|x-y|=2) = P(x=6) + P(x=4) = {}^{10}C_6 (\frac{1}{3})^6 (\frac{2}{3})^4 + {}^{10}C_4 (\frac{1}{3})^4 (\frac{2}{3})^6 = {}^{10}C_6 \frac{2^4}{3^{10}} + {}^{10}C_4 \frac{2^6}{3^{10}}$.
Total probability $p = P(|x-y|=0) + P(|x-y|=2) = \frac{{}^{10}C_5 2^5 + {}^{10}C_6 2^4 + {}^{10}C_4 2^6}{3^{10}}$.
$3^9 p = \frac{{}^{10}C_5 2^5 + {}^{10}C_6 2^4 + {}^{10}C_4 2^6}{3} = \frac{252 \times 32 + 210 \times 16 + 210 \times 64}{3} = \frac{8064 + 3360 + 13440}{3} = \frac{24864}{3} = 8288$.

(B) For $C_1$,$P(H) = \frac{2}{3}$. The number of heads $\alpha$ follows a binomial distribution $B(2, \frac{2}{3})$.
$P(\alpha = 0) = (\frac{1}{3})^2 = \frac{1}{9}$
$P(\alpha = 1) = 2 \times \frac{2}{3} \times \frac{1}{3} = \frac{4}{9}$
$P(\alpha = 2) = (\frac{2}{3})^2 = \frac{4}{9}$
For $C_2$,$P(H) = \frac{1}{3}$. The number of heads $\beta$ follows a binomial distribution $B(2, \frac{1}{3})$.
$P(\beta = 0) = (\frac{2}{3})^2 = \frac{4}{9}$
$P(\beta = 1) = 2 \times \frac{1}{3} \times \frac{2}{3} = \frac{4}{9}$
$P(\beta = 2) = (\frac{1}{3})^2 = \frac{1}{9}$
The roots of $x^2 - \alpha x + \beta = 0$ are real and equal if the discriminant $D = \alpha^2 - 4\beta = 0$,i.e.,$\alpha^2 = 4\beta$.
Possible pairs $(\alpha, \beta)$ satisfying this are $(0, 0)$ and $(2, 1)$.
Probability $= P(\alpha=0)P(\beta=0) + P(\alpha=2)P(\beta=1)$
$= (\frac{1}{9} \times \frac{4}{9}) + (\frac{4}{9} \times \frac{4}{9}) = \frac{4}{81} + \frac{16}{81} = \frac{20}{81}$.