Binomial distribution Questions in English

(B) Given that $\lim _{t \rightarrow 0}(1+5t)^{\frac{1}{t}} = K$.
Using the standard limit $\lim _{t \rightarrow 0}(1+at)^{\frac{1}{t}} = e^a$,we get $K = e^5$.
For the random variable $X$ representing the number of successes in $n = 100$ independent trials with probability of success $p = 0.05$,the distribution follows a Binomial distribution $B(n, p)$.
Since $n$ is large and $p$ is small,we can approximate this using a Poisson distribution with parameter $\lambda = np = 100 \times 0.05 = 5$.
The probability of getting at least one success is $P(X \geq 1) = 1 - P(X = 0)$.
Using the Poisson formula $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$,we have $P(X = 0) = e^{-5}$.
Therefore,$P(X \geq 1) = 1 - e^{-5} = 1 - \frac{1}{e^5} = 1 - \frac{1}{K} = \frac{K-1}{K}$.

(B) Let $p$ be the probability of success and $q$ be the probability of failure. Given $p = 3q$. Since $p + q = 1$,we have $3q + q = 1$,which implies $4q = 1$,so $q = \frac{1}{4}$ and $p = \frac{3}{4}$.
For a binomial distribution with $n = 5$ trials,the probability of $x$ successes is given by $P(X = x) = {}^nC_x p^x q^{n-x}$.
We need to find the probability of at least $4$ successes,which is $P(X \ge 4) = P(X = 4) + P(X = 5)$.
$P(X = 4) = {}^5C_4 (\frac{3}{4})^4 (\frac{1}{4})^1 = 5 \cdot \frac{81}{256} \cdot \frac{1}{4} = \frac{405}{1024}$.
$P(X = 5) = {}^5C_5 (\frac{3}{4})^5 (\frac{1}{4})^0 = 1 \cdot \frac{243}{1024} \cdot 1 = \frac{243}{1024}$.
Thus,$P(X \ge 4) = \frac{405}{1024} + \frac{243}{1024} = \frac{648}{1024} = \frac{81}{128}$.

(C) Given that the mean $\mu = np = \frac{5}{2}$ and the variance $\sigma^2 = npq = \frac{5}{4}$.
Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{5/4}{5/2} = \frac{1}{2}$.
Since $p = 1 - q$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $np = \frac{5}{2}$,we get $n \times \frac{1}{2} = \frac{5}{2}$,which implies $n = 5$.
We need to find $P(X > 1) = 1 - \{P(X = 0) + P(X = 1)\}$.
Using the Binomial probability formula $P(X = k) = {}^nC_k p^k q^{n-k}$:
$P(X = 0) = {}^5C_0 (\frac{1}{2})^0 (\frac{1}{2})^5 = 1 \times 1 \times \frac{1}{32} = \frac{1}{32}$.
$P(X = 1) = {}^5C_1 (\frac{1}{2})^1 (\frac{1}{2})^4 = 5 \times \frac{1}{2} \times \frac{1}{16} = \frac{5}{32}$.
Therefore,$P(X > 1) = 1 - (\frac{1}{32} + \frac{5}{32}) = 1 - \frac{6}{32} = 1 - \frac{3}{16} = \frac{13}{16}$.

(C) Given that the person fails $4$ times in $9$ games,the number of successes is $9 - 4 = 5$.
Thus,the probability of success in a single game is $p = \frac{5}{9}$.
Consequently,the probability of failure is $q = 1 - p = 1 - \frac{5}{9} = \frac{4}{9}$.
For $n = 15$ trials,we want the probability of at most one success,i.e.,$P(X \leq 1) = P(X = 0) + P(X = 1)$.
Using the binomial distribution formula $P(X = r) = {}^{n}C_{r} p^r q^{n-r}$:
$P(X = 0) = {}^{15}C_{0} \left(\frac{5}{9}\right)^0 \left(\frac{4}{9}\right)^{15} = \left(\frac{4}{9}\right)^{15}$.
$P(X = 1) = {}^{15}C_{1} \left(\frac{5}{9}\right)^1 \left(\frac{4}{9}\right)^{14} = 15 \times \frac{5}{9} \times \left(\frac{4}{9}\right)^{14} = \frac{75}{9} \times \left(\frac{4}{9}\right)^{14}$.
Adding these probabilities:
$P(X \leq 1) = \left(\frac{4}{9}\right)^{15} + \frac{75}{9} \times \left(\frac{4}{9}\right)^{14} = \left(\frac{4}{9}\right)^{14} \left[ \frac{4}{9} + \frac{75}{9} \right] = \frac{79}{9} \left(\frac{4}{9}\right)^{14}$.

(D) Let $p$ denote the probability that a ship departed at $A$ reaches $B$ safely.
Then,$p = \frac{9}{10} = 0.9$.
Therefore,$q = 1 - p = 1 - 0.9 = 0.1$.
Let $X$ denote the number of ships that reach $B$ safely out of $n = 5$ ships.
This follows a binomial distribution $B(n, p) = B(5, 0.9)$.
The required probability is $P(X \geq 4) = P(X = 4) + P(X = 5)$.
Using the formula $P(X = k) = {}^nC_k p^k q^{n-k}$:
$P(X = 4) = {}^5C_4 (0.9)^4 (0.1)^1 = 5 \times (0.9)^4 \times 0.1 = 0.5 \times (0.9)^4$.
$P(X = 5) = {}^5C_5 (0.9)^5 (0.1)^0 = 1 \times (0.9)^5 \times 1 = 0.9 \times (0.9)^4$.
Adding these probabilities:
$P(X \geq 4) = 0.5(0.9)^4 + 0.9(0.9)^4 = (0.5 + 0.9)(0.9)^4 = 1.4(0.9)^4$.

(C) Let $X$ be the number of heads in $100$ tosses. $X$ follows a binomial distribution $B(n, p)$ where $n = 100$.
The probability of getting $k$ heads is given by $P(X=k) = {}^{n}C_{k} p^{k} (1-p)^{n-k}$.
Given that $P(X=50) = P(X=51)$,we have:
${}^{100}C_{50} p^{50} (1-p)^{50} = {}^{100}C_{51} p^{51} (1-p)^{49}$
Dividing both sides by $p^{50} (1-p)^{49}$,we get:
${}^{100}C_{50} (1-p) = {}^{100}C_{51} p$
Using the formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$,we have:
$\frac{100!}{50! 50!} (1-p) = \frac{100!}{51! 49!} p$
$\frac{1-p}{50!} = \frac{p}{51 \times 50! \times \frac{49!}{50!}} \Rightarrow \frac{1-p}{50} = \frac{p}{51}$
$51(1-p) = 50p$
$51 - 51p = 50p$
$101p = 51$
$p = \frac{51}{101}$

(B) For a binomial distribution,the mean is given by $\mu = np = 4$ and the variance is $\sigma^2 = npq = (\sqrt{3})^2 = 3$.
Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{3}{4}$,which implies $q = \frac{3}{4}$.
Since $p + q = 1$,we have $p = 1 - \frac{3}{4} = \frac{1}{4}$.
Substituting $p = \frac{1}{4}$ into $np = 4$,we get $n(\frac{1}{4}) = 4$,so $n = 16$.
We need to find $P(X \geq 1)$. Using the complement rule,$P(X \geq 1) = 1 - P(X = 0)$.
For a binomial distribution,$P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$.
Thus,$P(X = 0) = {}^{16}C_{0} (\frac{1}{4})^{0} (\frac{3}{4})^{16} = 1 \times 1 \times (\frac{3}{4})^{16} = (\frac{3}{4})^{16}$.
Therefore,$P(X \geq 1) = 1 - (\frac{3}{4})^{16}$.

(C) The probability mass function for a binomial distribution is given by $P(X=k) = { }^n C_k p^k q^{n-k}$,where $q = 1-p$.
Given $n=4$,the equation is $P(X=4) = 6 P(X=2)$.
Substituting the formula: ${ }^4 C_4 p^4 q^0 = 6 \cdot { }^4 C_2 p^2 q^2$.
Since ${ }^4 C_4 = 1$ and ${ }^4 C_2 = \frac{4 \times 3}{2 \times 1} = 6$,we have:
$1 \cdot p^4 = 6 \cdot 6 \cdot p^2 q^2$.
$p^4 = 36 p^2 q^2$.
Dividing both sides by $p^2$ (assuming $p \neq 0$):
$p^2 = 36 q^2$.
Taking the square root of both sides:
$p = 6q$ (since $p$ and $q$ are probabilities,they must be positive).
Substitute $q = 1-p$:
$p = 6(1-p)$.
$p = 6 - 6p$.
$7p = 6$.
$p = \frac{6}{7}$.

(A) For a binomial distribution,the mean is given by $E(X) = np = 3$ and the variance is given by $Var(X) = npq = 2$.
Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{2}{3}$,which implies $q = \frac{2}{3}$.
Since $p + q = 1$,we have $p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$.
Substituting $p = \frac{1}{3}$ into $np = 3$,we get $n \times \frac{1}{3} = 3$,which implies $n = 9$.
The binomial distribution is given by $(q + p)^n = \left(\frac{2}{3} + \frac{1}{3}\right)^9$.

(C) Let $X$ be the random variable representing the number of mathematics books selected in $3$ trials.
Total number of books = $3 + 2 = 5$.
The probability of selecting a mathematics book in a single trial is $p = \frac{3}{5}$.
The probability of selecting a physics book is $q = 1 - p = 1 - \frac{3}{5} = \frac{2}{5}$.
Since the books are replaced after each selection,the trials are independent,and the random variable $X$ follows a binomial distribution $B(n, p)$ with $n = 3$ and $p = \frac{3}{5}$.
The mean of a binomial distribution is given by $E[X] = np$.
Substituting the values,we get $E[X] = 3 \times \frac{3}{5} = \frac{9}{5}$.

(B) The total number of cards is $52$. The number of ace cards is $4$. Since the cards are drawn with replacement,the probability of drawing an ace in a single trial is $p = \frac{4}{52} = \frac{1}{13}$. The probability of not drawing an ace is $q = 1 - p = \frac{12}{13}$.
This is a binomial distribution $B(n, p)$ where $n = 2$ and $p = \frac{1}{13}$.
The mean of a binomial distribution is given by $E(X) = np$.
Mean $= 2 \times \frac{1}{13} = \frac{2}{13}$.

(A) For a binomial distribution,the mean is $np = 16/5$ and the variance is $npq = 48/25$.
Dividing the variance by the mean,we get $q = (48/25) / (16/5) = (48/25) \times (5/16) = 3/5$.
Since $p + q = 1$,we have $p = 1 - 3/5 = 2/5$.
Substituting $p$ into the mean equation: $n(2/5) = 16/5 \Rightarrow n = 8$.
We need to find $P(X > 1) = 1 - P(X = 0) - P(X = 1)$.
$P(X = 0) = ^8C_0 (2/5)^0 (3/5)^8 = (3/5)^8$.
$P(X = 1) = ^8C_1 (2/5)^1 (3/5)^7 = 8 \times (2/5) \times (3/5)^7 = (16/5) \times (3/5)^7$.
Thus,$P(X > 1) = 1 - (3/5)(3/5)^7 - (16/5)(3/5)^7 = 1 - (3/5 + 16/5)(3/5)^7 = 1 - (19/5)(3/5)^7$.
Comparing this with $1 - K(3/5)^7$,we get $K = 19/5$.
Therefore,$5K = 5 \times (19/5) = 19$.

(A) Let $X$ be the random variable representing the number of spade cards drawn. Since the cards are drawn with replacement,this follows a binomial distribution $B(n, p)$ where $n = 2$ and $p = \frac{13}{52} = \frac{1}{4}$.
The probability of success (getting a spade) is $p = \frac{1}{4}$ and the probability of failure is $q = 1 - p = \frac{3}{4}$.
The variance of a binomial distribution is given by $Var(X) = npq$.
Substituting the values: $Var(X) = 2 \times \frac{1}{4} \times \frac{3}{4} = \frac{6}{16} = \frac{3}{8}$.

(D) In a random experiment of throwing $n = 5$ coins,the number of heads $X$ follows a binomial distribution $B(n, p)$,where $n = 5$ and $p = \frac{1}{2}$ (probability of getting a head).
The mean of a binomial distribution is given by the formula $E(X) = np$.
Substituting the values,we get $E(X) = 5 \times \frac{1}{2} = \frac{5}{2}$.
Alternatively,using the probability distribution table:

$X$	$0$	$1$	$2$	$3$	$4$	$5$
$P(X)$	$\frac{1}{32}$	$\frac{5}{32}$	$\frac{10}{32}$	$\frac{10}{32}$	$\frac{5}{32}$	$\frac{1}{32}$

The mean is calculated as $\sum X P(X) = (0 \times \frac{1}{32}) + (1 \times \frac{5}{32}) + (2 \times \frac{10}{32}) + (3 \times \frac{10}{32}) + (4 \times \frac{5}{32}) + (5 \times \frac{1}{32})$
$= 0 + \frac{5}{32} + \frac{20}{32} + \frac{30}{32} + \frac{20}{32} + \frac{5}{32} = \frac{80}{32} = \frac{5}{2}$.

(D) The distribution of successes follows a binomial distribution with parameters $n$ and $p$.
The variance of a binomial distribution is given by the formula $Var(X) = npq$,where $n$ is the number of trials,$p$ is the probability of success,and $q = 1 - p$ is the probability of failure.
In a single roll of a die,the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$. The odd numbers are $\{1, 3, 5\}$.
Therefore,the probability of success $p = \frac{3}{6} = \frac{1}{2}$.
The probability of failure $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.
Given that the die is rolled $n = 5$ times.
Substituting these values into the variance formula:
$Var(X) = n \times p \times q = 5 \times \frac{1}{2} \times \frac{1}{2} = \frac{5}{4}$.

(A) Given that the experiment is conducted $n = 5$ times. Let $p$ be the probability of success and $q = 1 - p$ be the probability of failure. The mean of a binomial distribution is $np$ and the variance is $npq$.
Given $np - npq = \frac{5}{9}$.
Substituting $n = 5$: $5p - 5pq = \frac{5}{9} \implies p - pq = \frac{1}{9}$.
Since $1 - q = p$,we have $p(1 - q) = p^2 = \frac{1}{9}$,which gives $p = \frac{1}{3}$.
Thus,$q = 1 - \frac{1}{3} = \frac{2}{3}$.
The probability of getting at most two successes is $P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$.
Using the formula $P(X = k) = {}^nC_k p^k q^{n-k}$:
$P(X = 0) = {}^5C_0 (\frac{1}{3})^0 (\frac{2}{3})^5 = 1 \times 1 \times \frac{32}{243} = \frac{32}{243}$.
$P(X = 1) = {}^5C_1 (\frac{1}{3})^1 (\frac{2}{3})^4 = 5 \times \frac{1}{3} \times \frac{16}{81} = \frac{80}{243}$.
$P(X = 2) = {}^5C_2 (\frac{1}{3})^2 (\frac{2}{3})^3 = 10 \times \frac{1}{9} \times \frac{8}{27} = \frac{80}{243}$.
Summing these probabilities: $P(X \leq 2) = \frac{32}{243} + \frac{80}{243} + \frac{80}{243} = \frac{192}{243} = \frac{64}{81}$.

(B) Given that $X$ is a binomial variate with $n=6$ and probability of success $p$. Let $q = 1-p$ be the probability of failure.
The probability mass function is given by $P(X=r) = {}^{n}C_{r} p^{r} q^{n-r}$.
We are given the condition $4 P(X=4) = P(X=2)$.
Substituting the formula:
$4 \cdot {}^{6}C_{4} p^{4} q^{6-4} = {}^{6}C_{2} p^{2} q^{6-2}$
$4 \cdot {}^{6}C_{4} p^{4} q^{2} = {}^{6}C_{2} p^{2} q^{4}$
Since ${}^{6}C_{4} = {}^{6}C_{2} = 15$,we can cancel them:
$4 p^{4} q^{2} = p^{2} q^{4}$
Dividing both sides by $p^{2} q^{2}$ (assuming $p, q \neq 0$):
$4 p^{2} = q^{2}$
Taking the square root of both sides:
$2p = q$
Since $q = 1-p$,we have:
$2p = 1-p$
$3p = 1$
$p = 1/3$

(B) For a binomial distribution,the mean is given by $np = 2$ and the variance is given by $npq = 1$.
Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{1}{2}$,which implies $q = \frac{1}{2}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $np = 2$,we get $n \times \frac{1}{2} = 2$,so $n = 4$.
The probability mass function is $P(X = k) = \binom{n}{k} p^k q^{n-k} = \binom{4}{k} (\frac{1}{2})^k (\frac{1}{2})^{4-k} = \binom{4}{k} (\frac{1}{2})^4 = \binom{4}{k} \frac{1}{16}$.
We need to find $P(X \geq 1) = 1 - P(X = 0)$.
$P(X = 0) = \binom{4}{0} (\frac{1}{2})^4 = 1 \times \frac{1}{16} = \frac{1}{16}$.
Therefore,$P(X \geq 1) = 1 - \frac{1}{16} = \frac{15}{16}$.

(B) The probability of getting a head in a single toss is $p = \frac{1}{2}$,and the probability of getting a tail is $q = 1 - p = \frac{1}{2}$.
The probability of getting no heads in $n$ tosses is $P(\text{no heads}) = q^n = (\frac{1}{2})^n$.
The probability of getting at least one head is $P(\text{at least one head}) = 1 - P(\text{no heads}) = 1 - (\frac{1}{2})^n$.
Given that $1 - (\frac{1}{2})^n > 0.8$,we have:
$1 - 0.8 > (\frac{1}{2})^n$
$0.2 > (\frac{1}{2})^n$
$\frac{1}{5} > \frac{1}{2^n}$
$2^n > 5$.
For $n = 2$,$2^2 = 4 < 5$.
For $n = 3$,$2^3 = 8 > 5$.
Thus,the least value of $n$ is $3$.

(D) Let $X$ be the number of correct answers. The $3$ questions have probability of success $p_1 = \frac{1}{4}$ and failure $q_1 = \frac{3}{4}$. The $2$ questions have probability of success $p_2 = \frac{1}{2}$ and failure $q_2 = \frac{1}{2}$.
We want the probability of at least $4$ correct answers,i.e.,$P(X=4) + P(X=5)$.
For $X=5$: All $5$ are correct. $P(X=5) = (\frac{1}{4})^3 \cdot (\frac{1}{2})^2 = \frac{1}{64} \cdot \frac{1}{4} = \frac{1}{256}$.
For $X=4$: Either one of the $3$ questions is wrong,or one of the $2$ questions is wrong.
Case $1$: One of the $3$ questions is wrong. Probability $= {^3C_2} \cdot (\frac{1}{4})^2 \cdot (\frac{3}{4})^1 \cdot (\frac{1}{2})^2 = 3 \cdot \frac{1}{16} \cdot \frac{3}{4} \cdot \frac{1}{4} = \frac{9}{256}$.
Case $2$: One of the $2$ questions is wrong. Probability $= {^3C_3} \cdot (\frac{1}{4})^3 \cdot {^2C_1} \cdot (\frac{1}{2})^1 \cdot (\frac{1}{2})^1 = 1 \cdot \frac{1}{64} \cdot 2 \cdot \frac{1}{4} = \frac{2}{256}$.
Total probability $= P(X=5) + P(X=4) = \frac{1}{256} + \frac{9}{256} + \frac{2}{256} = \frac{12}{256} = \frac{3}{64}$.

(A) For a binomial distribution,the mean is given by $np = 4$ and the variance is given by $npq = 2$.
Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{2}{4} = \frac{1}{2}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $np = 4$,we get $n \times \frac{1}{2} = 4$,which implies $n = 8$.
The probability of exactly $x$ successes is given by $P(X=x) = {}^{n}C_{x} p^{x} q^{n-x}$.
For $x = 2$,$P(X=2) = {}^{8}C_{2} \left(\frac{1}{2}\right)^{2} \left(\frac{1}{2}\right)^{8-2} = {}^{8}C_{2} \left(\frac{1}{2}\right)^{8}$.
Calculating the value,${}^{8}C_{2} = \frac{8 \times 7}{2 \times 1} = 28$.
Thus,$P(X=2) = 28 \times \frac{1}{256} = \frac{28}{256} = \frac{7}{64}$.

(B) Let $n$ be the number of rounds fired.
Probability of hitting the target in a single trial,$p = 10 \% = 0.1$.
Probability of missing the target in a single trial,$q = 1 - p = 0.9$.
The probability of hitting the target at least once in $n$ trials is given by $P(X \geq 1) = 1 - P(X = 0) = 1 - q^n$.
We want this probability to be greater than $50 \%$,so:
$1 - (0.9)^n > 0.5$
$(0.9)^n < 0.5$
Now,we test values for $n$:
For $n = 6$,$(0.9)^6 = 0.531441 > 0.5$.
For $n = 7$,$(0.9)^7 = 0.4782969 < 0.5$.
Thus,the least number of rounds required is $n = 7$.

(B) Let $n$ be the number of times the coin is tossed.
For an unbiased coin,the probability of getting a head is $p = \frac{1}{2}$ and the probability of getting a tail is $q = 1 - p = \frac{1}{2}$.
The probability of getting at least one head is given by $P(X \geq 1) = 1 - P(X = 0)$.
Here,$P(X = 0)$ is the probability of getting no heads (i.e.,all tails),which is $q^n = (\frac{1}{2})^n$.
We are given that $P(X \geq 1) \geq 0.9$.
So,$1 - (\frac{1}{2})^n \geq 0.9$.
$1 - 0.9 \geq (\frac{1}{2})^n$.
$0.1 \geq \frac{1}{2^n}$.
$\frac{1}{10} \geq \frac{1}{2^n}$.
$2^n \geq 10$.
For $n = 3$,$2^3 = 8 < 10$.
For $n = 4$,$2^4 = 16 \geq 10$.
Thus,the minimum number of tosses required is $4$.

(B) Let the coin be tossed $n$ times.
Let getting a head be considered a success. $\therefore p = \frac{1}{2}, q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.
It is given that $P(X = 3) = P(X = 5)$.
Using the binomial distribution formula $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$:
${}^{n}C_{3} (\frac{1}{2})^{3} (\frac{1}{2})^{n-3} = {}^{n}C_{5} (\frac{1}{2})^{5} (\frac{1}{2})^{n-5}$.
Since the powers of $(\frac{1}{2})$ sum to $n$ on both sides,we have ${}^{n}C_{3} = {}^{n}C_{5}$.
Using the property ${}^{n}C_{x} = {}^{n}C_{y} \Rightarrow x + y = n$ (where $x \neq y$),we get $n = 3 + 5 = 8$.
Now,we need to find the probability of getting exactly one head,$P(X = 1)$:
$P(X = 1) = {}^{8}C_{1} (\frac{1}{2})^{1} (\frac{1}{2})^{8-1} = 8 \times (\frac{1}{2})^{8} = \frac{8}{256} = \frac{1}{32}$.

(D) Let $n$ be the number of parts produced. The probability of a part being defective is $p = 0.05 = \frac{1}{20}$.
The probability of a part being non-defective is $q = 1 - 0.05 = 0.95 = \frac{19}{20}$.
We want the probability of at least one defective part to be at least $1/2$,i.e.,$P(X \geq 1) \geq 1/2$.
This is equivalent to $1 - P(X = 0) \geq 1/2$,where $P(X = 0)$ is the probability that no parts are defective.
$1 - (0.95)^n \geq 0.5 \implies 0.5 \geq (0.95)^n$.
Taking $\log_{10}$ on both sides: $\log_{10}(0.5) \geq n \log_{10}(0.95)$.
$-\log_{10}(2) \geq n(\log_{10}(95) - \log_{10}(100))$.
$-0.3 \geq n(1.977 - 2)$.
$-0.3 \geq n(-0.023)$.
Since we are dividing by a negative number,the inequality sign reverses: $n \geq \frac{0.3}{0.023} = \frac{300}{23} \approx 13.04$.
Since $n$ must be an integer,the smallest integer $n$ is $14$.

(D) Total bulbs = $100$. Number of defective bulbs = $10$. Number of non-defective bulbs = $90$.
Probability of selecting a defective bulb $p = \frac{10}{100} = \frac{1}{10}$.
Probability of selecting a non-defective bulb $q = \frac{90}{100} = \frac{9}{10}$.
Since bulbs are selected with replacement,this follows a Binomial Distribution $B(n, p)$ with $n = 8$ and $p = 0.1$.
The probability of getting at least $7$ defective bulbs is $P(X \ge 7) = P(X = 7) + P(X = 8)$.
$P(X = 7) = \binom{8}{7} \times (0.1)^7 \times (0.9)^1 = 8 \times \frac{1}{10^7} \times \frac{9}{10} = \frac{72}{10^8}$.
$P(X = 8) = \binom{8}{8} \times (0.1)^8 \times (0.9)^0 = 1 \times \frac{1}{10^8} \times 1 = \frac{1}{10^8}$.
Total probability = $\frac{72}{10^8} + \frac{1}{10^8} = \frac{73}{10^8}$.

(A) The mean $\bar{X}$ is given by $\frac{\sum f_i x_i}{\sum f_i}$.
Here,$\sum f_i = \sum_{k=0}^n {}^nC_k = 2^n$.
The values are $x_k = k(k-1)$ for $k=0, 1, ..., n$.
The sum $\sum_{k=0}^n k(k-1) {}^nC_k = n(n-1) 2^{n-2}$.
Thus,$\bar{X} = \frac{n(n-1) 2^{n-2}}{2^n} = \frac{n(n-1)}{4} = 60$.
$n^2 - n - 240 = 0 \implies (n-16)(n+15) = 0$. Since $n > 0$,$n = 16$.
The total frequency is $2^{16} = 65536$. The median is the value at the $\frac{65536+1}{2} \approx 32768$th position.
The distribution of frequencies follows the binomial expansion of $(1+1)^n$. The median of a binomial distribution $B(n, p)$ with $p=0.5$ is approximately $np = 16 \times 0.5 = 8$. The value at $k=8$ is $8(8-1) = 56$.