Binomial distribution Questions in English

(A) Let $X$ be the number of tails obtained in $100$ tosses of a fair coin. $X$ follows a binomial distribution $B(n, p)$ where $n = 100$ and $p = \frac{1}{2}$.
The probability of getting tails $k$ times is given by $P(X=k) = {}^{100}C_k (\frac{1}{2})^k (\frac{1}{2})^{100-k} = {}^{100}C_k (\frac{1}{2})^{100}$.
We need to find the probability of getting an odd number of tails,which is $P(X=1) + P(X=3) + \dots + P(X=99)$.
This sum is equal to $(\frac{1}{2})^{100} \times ({}^{100}C_1 + {}^{100}C_3 + \dots + {}^{100}C_{99})$.
We know that the sum of odd-indexed binomial coefficients ${}^{n}C_1 + {}^{n}C_3 + \dots = 2^{n-1}$.
For $n=100$,the sum is $2^{100-1} = 2^{99}$.
Therefore,the required probability is $(\frac{1}{2})^{100} \times 2^{99} = \frac{2^{99}}{2^{100}} = \frac{1}{2}$.

(A) Given that $X$ is a binomial variate with range $\{0, 1, 2, 3, 4, 5, 6\}$,so the number of trials $n = 6$.
The probability mass function of a binomial distribution is given by $P(X=k) = {^nC_k} p^k q^{n-k}$,where $q = 1-p$.
According to the problem,$P(X=2) = 4 P(X=4)$.
Substituting the values:
${^6C_2} p^2 q^4 = 4 \cdot {^6C_4} p^4 q^2$.
Since ${^6C_2} = \frac{6 \times 5}{2 \times 1} = 15$ and ${^6C_4} = {^6C_2} = 15$,we have:
$15 p^2 q^4 = 4 \cdot 15 p^4 q^2$.
Dividing both sides by $15 p^2 q^2$ (assuming $p, q \neq 0$):
$q^2 = 4 p^2$.
Substituting $q = 1-p$:
$(1-p)^2 = 4 p^2$.
Taking the square root on both sides:
$1-p = 2p$ or $1-p = -2p$.
Case $1$: $1 = 3p \Rightarrow p = \frac{1}{3}$.
Case $2$: $1 = -p \Rightarrow p = -1$ (which is impossible as $0 \leq p \leq 1$).
Therefore,the parameter $p = \frac{1}{3}$.

(D) Given that the probability of success $p = \frac{1}{4}$.
Then,the probability of failure $q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4}$.
The standard deviation $(SD)$ of a binomial distribution is given by $\sqrt{npq} = 3$.
Squaring both sides,we get $npq = 9$.
Substituting the values of $p$ and $q$:
$n \times \frac{1}{4} \times \frac{3}{4} = 9$
$n \times \frac{3}{16} = 9$
$n = 9 \times \frac{16}{3} = 48$.
The mean of a binomial distribution is given by $np$.
Mean $= 48 \times \frac{1}{4} = 12$.

(B) For a binomial distribution $X \sim B(n, p)$,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $q = 1-p$.
Given $\mu - \sigma^2 = 1$,we have $np - npq = 1$,which simplifies to $np(1-q) = 1$,so $np^2 = 1$.
Given $2 P(X=2) = 3 P(X=1)$,we use the formula $P(X=k) = \binom{n}{k} p^k q^{n-k}$.
$2 \binom{n}{2} p^2 q^{n-2} = 3 \binom{n}{1} p^1 q^{n-1}$.
$2 \cdot \frac{n(n-1)}{2} p^2 q^{n-2} = 3n p q^{n-1}$.
$(n-1) p = 3q = 3(1-p)$.
$np - p = 3 - 3p \implies np + 2p = 3$.
Since $np^2 = 1$,we have $n = \frac{1}{p^2}$.
Substituting $n$: $\frac{1}{p^2} \cdot p + 2p = 3 \implies \frac{1}{p} + 2p = 3$.
$1 + 2p^2 = 3p \implies 2p^2 - 3p + 1 = 0$.
$(2p-1)(p-1) = 0$. Since $p < 1$,$p = \frac{1}{2}$.
Then $n = \frac{1}{(1/2)^2} = 4$.
We need $n^2 P(X>1) = 16(1 - P(X=0) - P(X=1))$.
$P(X=0) = (1/2)^4 = 1/16$.
$P(X=1) = \binom{4}{1} (1/2)^1 (1/2)^3 = 4 \cdot (1/16) = 4/16$.
$P(X>1) = 1 - (1/16 + 4/16) = 11/16$.
$n^2 P(X>1) = 16 \cdot (11/16) = 11$.

(A) For a binomial distribution with parameters $n$ and $p$,the mean is $\mu = np = x$ and the variance is $\sigma^2 = npq = 5$,where $q = 1 - p$.
Since $np = x$ and $npq = 5$,we have $xq = 5$,which implies $q = \frac{5}{x}$.
Since $0 < q < 1$,we must have $0 < \frac{5}{x} < 1$,which implies $x > 5$.
Also,$p = 1 - q = 1 - \frac{5}{x} = \frac{x-5}{x}$.
Since $0 < p < 1$,we have $0 < \frac{x-5}{x} < 1$,which is consistent with $x > 5$.
We know that $n = \frac{x}{p} = \frac{x}{(x-5)/x} = \frac{x^2}{x-5}$.
Since $n$ must be a positive integer,$x-5$ must be a divisor of $x^2$.
We can write $x^2 = (x-5)(x+5) + 25$,so $n = x+5 + \frac{25}{x-5}$.
For $n$ to be an integer,$x-5$ must be a divisor of $25$.
The divisors of $25$ are $1, 5, 25$.
Case $1$: $x-5 = 1 \implies x = 6$. Then $n = 6+5 + 25/1 = 36$.
Case $2$: $x-5 = 5 \implies x = 10$. Then $n = 10+5 + 25/5 = 20$.
Case $3$: $x-5 = 25 \implies x = 30$. Then $n = 30+5 + 25/25 = 36$.
Thus,the possible values for $x$ are $6, 10, 30$.

(C) When $8$ coins are tossed simultaneously,the total number of possible outcomes is $2^8 = 256$.
Let $X$ be the number of heads obtained. $X$ follows a binomial distribution with $n = 8$ and $p = 0.5$.
The probability of getting at least $6$ heads is $P(X \ge 6) = P(X = 6) + P(X = 7) + P(X = 8)$.
Using the formula $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$,we have:
$P(X = 6) = \binom{8}{6} (0.5)^8 = 28 \times \frac{1}{256} = \frac{28}{256}$.
$P(X = 7) = \binom{8}{7} (0.5)^8 = 8 \times \frac{1}{256} = \frac{8}{256}$.
$P(X = 8) = \binom{8}{8} (0.5)^8 = 1 \times \frac{1}{256} = \frac{1}{256}$.
Summing these probabilities: $P(X \ge 6) = \frac{28 + 8 + 1}{256} = \frac{37}{256}$.

(D) Let $n$ be the number of tosses. The probability of getting $k$ heads in $n$ tosses is given by the binomial distribution $P(X=k) = {^nC_k} (\frac{1}{2})^n$.
The probability of getting at least two heads is $P(X \ge 2) = 1 - P(X=0) - P(X=1)$.
$P(X=0) = {^nC_0} (\frac{1}{2})^n = \frac{1}{2^n}$.
$P(X=1) = {^nC_1} (\frac{1}{2})^n = \frac{n}{2^n}$.
So,$P(X \ge 2) = 1 - \frac{1+n}{2^n}$.
We want $1 - \frac{1+n}{2^n} \ge 0.96$,which implies $\frac{1+n}{2^n} \le 0.04 = \frac{1}{25}$.
Testing values for $n$:
For $n=7$: $\frac{1+7}{2^7} = \frac{8}{128} = \frac{1}{16} = 0.0625 > 0.04$.
For $n=8$: $\frac{1+8}{2^8} = \frac{9}{256} \approx 0.03515 < 0.04$.
Thus,the minimum number of tosses required is $8$.

(A) Let $X$ be the binomial variate for which mean $= 4$ and variance $= 3$.
Then $np = 4$ and $npq = 3$.
Dividing the variance by the mean,we get $q = \frac{3}{4}$.
Since $p = 1 - q$,we have $p = 1 - \frac{3}{4} = \frac{1}{4}$.
Substituting $p$ into $np = 4$,we get $n \times \frac{1}{4} = 4$,so $n = 16$.
The probability mass function is $P(X=k) = {}^{n}C_k p^k q^{n-k}$.
We need to calculate $2^{32} P\left(X=\frac{16}{2}\right) = 2^{32} P(X=8)$.
$P(X=8) = {}^{16}C_8 \left(\frac{1}{4}\right)^8 \left(\frac{3}{4}\right)^{16-8} = {}^{16}C_8 \left(\frac{1}{4}\right)^8 \left(\frac{3}{4}\right)^8 = {}^{16}C_8 \frac{3^8}{4^{16}}$.
Since $4^{16} = (2^2)^{16} = 2^{32}$,we have $P(X=8) = {}^{16}C_8 \frac{3^8}{2^{32}}$.
Therefore,$2^{32} P(X=8) = 2^{32} \times {}^{16}C_8 \frac{3^8}{2^{32}} = {}^{16}C_8 (3^8)$.

(D) For a binomial distribution $B(n, p)$,the variance is given by $\operatorname{Var}(X) = npq$,where $q = 1 - p$.
Given $n = 15$ and $\operatorname{Var}(X) = 3.15$.
Substituting the values,we get $15 \times p(1 - p) = 3.15$.
Dividing by $15$,we get $p(1 - p) = \frac{3.15}{15} = 0.21$.
This simplifies to $p - p^2 = 0.21$,or $p^2 - p + 0.21 = 0$.
Solving the quadratic equation using the factorization method:
$p^2 - 0.7p - 0.3p + 0.21 = 0$
$p(p - 0.7) - 0.3(p - 0.7) = 0$
$(p - 0.7)(p - 0.3) = 0$.
Thus,the possible values for $p$ are $0.7$ and $0.3$.

(A) For a binomial distribution $X \sim B(n, p)$,the variance is given by $Var(X) = n \times p \times q$,where $q = 1 - p$.
Given $X \sim B(n_1, 0.5)$,so $p_1 = 0.5$ and $q_1 = 1 - 0.5 = 0.5$.
The variance is $n_1 \times 0.5 \times 0.5 = 6$.
$n_1 \times 0.25 = 6 \Rightarrow n_1 = \frac{6}{0.25} = 24$.
Given $Y \sim B(n_2, 0.4)$,so $p_2 = 0.4$ and $q_2 = 1 - 0.4 = 0.6$.
The variance is $n_2 \times 0.4 \times 0.6 = 6$.
$n_2 \times 0.24 = 6 \Rightarrow n_2 = \frac{6}{0.24} = 25$.
Therefore,$\sqrt{n_1 + n_2} = \sqrt{24 + 25} = \sqrt{49} = 7$.

(C) Let $n$ be the number of days,so $n = 7$.
Let $p$ be the probability that $A$ wakes up before the alarm,so $p = 0.4$.
Let $q$ be the probability that $A$ does not wake up before the alarm,so $q = 1 - p = 1 - 0.4 = 0.6$.
This follows a binomial distribution $B(n, p)$.
The mean of a binomial distribution is given by $E(X) = n \cdot p$.
Mean $= 7 \times 0.4 = 2.8$.
The variance of a binomial distribution is given by $Var(X) = n \cdot p \cdot q$.
Variance $= 7 \times 0.4 \times 0.6 = 2.8 \times 0.6 = 1.68$.
Thus,the mean and variance are $2.8$ and $1.68$ respectively.

(C) Given,mean $np = 2$ $(i)$
And variance $npq = 1$ $(ii)$
Dividing $(ii)$ by $(i)$,we get $q = \frac{1}{2}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $(i)$,we get $n \times \frac{1}{2} = 2$,so $n = 4$.
The binomial distribution is given by $P(X = k) = {}^nC_k p^k q^{n-k} = {}^4C_k (\frac{1}{2})^k (\frac{1}{2})^{4-k} = {}^4C_k (\frac{1}{2})^4$.
We need to find $P(X > 1) = P(X = 2) + P(X = 3) + P(X = 4)$.
$P(X > 1) = {}^4C_2 (\frac{1}{2})^4 + {}^4C_3 (\frac{1}{2})^4 + {}^4C_4 (\frac{1}{2})^4$.
$P(X > 1) = (6 + 4 + 1) \times \frac{1}{16} = \frac{11}{16}$.

(A) Let $n = 4$ be the number of children in each family. The probability of a child being a boy $(B)$ or a girl $(G)$ is $P(B) = P(G) = \frac{1}{2}$.
For a family with $4$ children,the total number of possible outcomes is $2^4 = 16$.
The event of having children of both sexes is the complement of the event where all children are of the same sex (all boys or all girls).
$P(\text{all boys}) = (\frac{1}{2})^4 = \frac{1}{16}$.
$P(\text{all girls}) = (\frac{1}{2})^4 = \frac{1}{16}$.
$P(\text{both sexes}) = 1 - [P(\text{all boys}) + P(\text{all girls})] = 1 - [\frac{1}{16} + \frac{1}{16}] = 1 - \frac{2}{16} = 1 - \frac{1}{8} = \frac{7}{8}$.
For $800$ families,the expected number of families with children of both sexes is $800 \times \frac{7}{8} = 700$.

(B) Given that $X$ is a binomial variate with mean $np = 6$ and variance $npq = 2$.
From these,we have $6q = 2$,which implies $q = \frac{1}{3}$.
Thus,$p = 1 - q = 1 - \frac{1}{3} = \frac{2}{3}$.
Substituting $p$ in $np = 6$,we get $n \times \frac{2}{3} = 6$,so $n = 9$.
We need to calculate $P(5 \leq X \leq 7) = P(X=5) + P(X=6) + P(X=7)$.
Using the formula $P(X=k) = {}^nC_k p^k q^{n-k}$:
$P(X=5) = {}^9C_5 (\frac{2}{3})^5 (\frac{1}{3})^4 = 126 \times \frac{32}{3^9} = \frac{4032}{19683}$.
$P(X=6) = {}^9C_6 (\frac{2}{3})^6 (\frac{1}{3})^3 = 84 \times \frac{64}{3^9} = \frac{5376}{19683}$.
$P(X=7) = {}^9C_7 (\frac{2}{3})^7 (\frac{1}{3})^2 = 36 \times \frac{128}{3^9} = \frac{4608}{19683}$.
Summing these values: $P(5 \leq X \leq 7) = \frac{4032 + 5376 + 4608}{19683} = \frac{14016}{19683}$.
Dividing numerator and denominator by $3$: $\frac{4672}{6561}$.

(A) Let $X$ be the number of defective locks in a box. Since the number of locks $n=100$ is large and the probability of a defect $p=0.02$ is small,we use the Poisson distribution with mean $\lambda = np = 100 \times 0.02 = 2$.
The probability of having $r$ defective locks is given by $P(X=r) = \frac{e^{-\lambda} \lambda^r}{r!} = \frac{e^{-2} 2^r}{r!}$.
The manufacturer guarantees that there are not more than $2$ defective locks,meaning the box meets the quality if $X \le 2$.
The box fails to meet the guarantee if $X > 2$.
The probability of failure is $P(X > 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]$.
Substituting the values:
$P(X > 2) = 1 - [\frac{e^{-2} 2^0}{0!} + \frac{e^{-2} 2^1}{1!} + \frac{e^{-2} 2^2}{2!}] = 1 - [e^{-2} + 2e^{-2} + 2e^{-2}] = 1 - 5e^{-2}$.

(C) For a binomial distribution,the mean is given by $np = \frac{4}{3}$ $(i)$ and the variance is given by $npq = \frac{8}{9}$ $(ii)$.
Dividing equation $(ii)$ by equation $(i)$,we get:
$\frac{npq}{np} = \frac{8/9}{4/3} \implies q = \frac{8}{9} \times \frac{3}{4} = \frac{2}{3}$.
Since $p + q = 1$,we have $p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$.
Substituting $p = \frac{1}{3}$ into equation $(i)$:
$n \times \frac{1}{3} = \frac{4}{3} \implies n = 4$.
The probability mass function for a binomial distribution is $P(X=k) = {}^nC_k p^k q^{n-k}$.
For $k=2$,$P(X=2) = {}^4C_2 \times (\frac{1}{3})^2 \times (\frac{2}{3})^{4-2}$.
$P(X=2) = 6 \times \frac{1}{9} \times \frac{4}{9} = \frac{24}{81} = \frac{8}{27}$.

(C) Given that $n=6$ and $P(X=2)=9 P(X=4)$.
Using the binomial probability formula $P(X=k) = {}^n C_k p^k q^{n-k}$:
${}^6 C_2 p^2 q^4 = 9 \cdot {}^6 C_4 p^4 q^2$
Since ${}^6 C_2 = 15$ and ${}^6 C_4 = 15$,we have:
$15 p^2 q^4 = 9 \cdot 15 p^4 q^2$
Dividing both sides by $15 p^2 q^2$ (assuming $p, q \neq 0$):
$q^2 = 9 p^2$
Taking the square root on both sides:
$q = 3p$
Since $p + q = 1$,we substitute $q = 3p$:
$p + 3p = 1 \Rightarrow 4p = 1 \Rightarrow p = \frac{1}{4}$
Then $q = 1 - \frac{1}{4} = \frac{3}{4}$.
The variance of a binomial distribution is given by $npq$:
$\text{Variance} = 6 \cdot \frac{1}{4} \cdot \frac{3}{4} = \frac{18}{16} = \frac{9}{8}$.

(D) The probability mass function of a binomial distribution is given by $P(X=k) = \binom{n}{k} p^k q^{n-k}$,where $q = 1-p$.
Given $n=4$ and $P(X=0) = \frac{16}{81}$.
Substituting $k=0$ into the formula: $P(X=0) = \binom{4}{0} p^0 q^{4-0} = q^4$.
So,$q^4 = \frac{16}{81} = (\frac{2}{3})^4$.
This implies $q = \frac{2}{3}$.
Since $p+q=1$,we have $p = 1 - \frac{2}{3} = \frac{1}{3}$.
Now,we need to find $P(X=4)$:
$P(X=4) = \binom{4}{4} p^4 q^{4-4} = 1 \times (\frac{1}{3})^4 \times 1 = \frac{1}{81}$.
Thus,the correct option is $D$.

(A) We know that for a binomial distribution,mean $\mu = np$ and variance $\sigma^2 = npq$.
Given $\mu = 2\sigma^2$,we have $np = 2npq$,which implies $q = \frac{1}{2}$.
Since $p + q = 1$,we get $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Given $\mu + \sigma^2 = 3$,we have $np + npq = 3$.
Substituting $p = \frac{1}{2}$ and $q = \frac{1}{2}$,we get $\frac{n}{2} + \frac{n}{4} = 3$.
Multiplying by $4$,we get $2n + n = 12$,so $3n = 12$,which gives $n = 4$.
Now,$P(X \leq 3) = 1 - P(X = 4)$.
Using the binomial probability formula $P(X=k) = {^nC_k} p^k q^{n-k}$,we have $P(X=4) = {^4C_4} (\frac{1}{2})^4 (\frac{1}{2})^0 = 1 \times \frac{1}{16} = \frac{1}{16}$.
Therefore,$P(X \leq 3) = 1 - \frac{1}{16} = \frac{15}{16}$.

(D) Let $p$ be the probability that a student is good at mathematics,so $p = 0.6 = \frac{3}{5}$.
Let $q$ be the probability that a student is not good at mathematics,so $q = 1 - 0.6 = 0.4 = \frac{2}{5}$.
For a group of $n = 8$ students,the probability of having exactly $X = 2$ students good at mathematics follows the binomial distribution formula $P(X=k) = { }^n C_k \times p^k \times q^{n-k}$.
Substituting the values: $P(X=2) = { }^8 C_2 \times (0.6)^2 \times (0.4)^6$.
$P(X=2) = \frac{8 \times 7}{2} \times \left(\frac{3}{5}\right)^2 \times \left(\frac{2}{5}\right)^6$.
$P(X=2) = 28 \times \frac{3^2}{5^2} \times \frac{2^6}{5^6} = (2^2 \times 7) \times \frac{3^2 \times 2^6}{5^8} = \frac{2^8 \times 3^2 \times 7}{5^8}$.

(D) Let $X$ be the number of engineering students in a sample of $n = 8$ people. This follows a binomial distribution with $p = \frac{1}{5}$ and $q = 1 - p = \frac{4}{5}$.
We need to find $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$.
$P(X=0) = \binom{8}{0} (\frac{1}{5})^0 (\frac{4}{5})^8 = (\frac{4}{5})^8 = \frac{4^8}{5^8}$.
$P(X=1) = \binom{8}{1} (\frac{1}{5})^1 (\frac{4}{5})^7 = 8 \times \frac{1}{5} \times \frac{4^7}{5^7} = 2 \times \frac{4 \times 4^7}{5^8} = 2 \times \frac{4^8}{5^8}$.
$P(X=2) = \binom{8}{2} (\frac{1}{5})^2 (\frac{4}{5})^6 = 28 \times \frac{1}{25} \times \frac{4^6}{5^6} = 28 \times \frac{4^6}{5^8} = 28 \times \frac{4^6}{5^8}$.
Summing these: $P(X \le 2) = \frac{4^8 + 2 \times 4^8 + 28 \times 4^6}{5^8} = \frac{4^6(16 + 32 + 28)}{5^8} = \frac{76 \times 4^6}{5^8} = \frac{19 \times 4 \times 4^6}{5^8} = 19 \times \frac{4^7}{5^8}$.

(A) The total number of outcomes when two dice are rolled is $6 \times 6 = 36$.
The outcomes where the sum is at most $7$ are:
Sum $= 2: (1,1)$ ($1$ outcome)
Sum $= 3: (1,2), (2,1)$ ($2$ outcomes)
Sum $= 4: (1,3), (2,2), (3,1)$ ($3$ outcomes)
Sum $= 5: (1,4), (2,3), (3,2), (4,1)$ ($4$ outcomes)
Sum $= 6: (1,5), (2,4), (3,3), (4,2), (5,1)$ ($5$ outcomes)
Sum $= 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ ($6$ outcomes)
Total favorable outcomes $= 1+2+3+4+5+6 = 21$.
Thus,$x = \frac{21}{36} = \frac{7}{12}$.
For $y$,the probability of getting a sum of $7$ in one roll is $p = \frac{6}{36} = \frac{1}{6}$. The probability of not getting a sum of $7$ is $q = 1 - \frac{1}{6} = \frac{5}{6}$.
When rolled $n$ times,the probability of getting a sum of $7$ at least once is $y = 1 - q^n = 1 - (\frac{5}{6})^n$.
We want $y > x$,so $1 - (\frac{5}{6})^n > \frac{7}{12} \Rightarrow (\frac{5}{6})^n < 1 - \frac{7}{12} = \frac{5}{12}$.
For $n=1: \frac{5}{6} \approx 0.833 > 0.416$
For $n=2: \frac{25}{36} \approx 0.694 > 0.416$
For $n=3: \frac{125}{216} \approx 0.578 > 0.416$
For $n=4: \frac{625}{1296} \approx 0.482 > 0.416$
For $n=5: \frac{3125}{7776} \approx 0.401 < 0.416$.
Thus,the minimum $n$ is $5$.

(C) Let the probability of a bomb hitting the target be $p = \frac{3}{4}$.
Thus,the probability of a bomb missing the target is $q = 1 - p = \frac{1}{4}$.
Let $n$ be the number of bombs dropped. 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 $1 - [P(X = 0) + P(X = 1)] \geq 0.99$.
$P(X = 0) + P(X = 1) \leq 0.01 = \frac{1}{100}$.
Using the binomial distribution: $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$.
${}^{n}C_{0} (\frac{3}{4})^{0} (\frac{1}{4})^{n} + {}^{n}C_{1} (\frac{3}{4})^{1} (\frac{1}{4})^{n-1} \leq \frac{1}{100}$.
$\frac{1}{4^{n}} + n \cdot \frac{3}{4} \cdot \frac{4}{4^{n}} \leq \frac{1}{100}$.
$\frac{1 + 3n}{4^{n}} \leq \frac{1}{100} \Rightarrow 4^{n} \geq 100(3n + 1) = 300n + 100$.
For $n = 5$: $4^{5} = 1024$ and $300(5) + 100 = 1600$. ($1024 < 1600$,false).
For $n = 6$: $4^{6} = 4096$ and $300(6) + 100 = 1900$. ($4096 \geq 1900$,true).
Thus,the minimum number of bombs is $6$.

(B) Given that a fair coin is tossed $K$ times,the probability of getting $X$ heads follows a binomial distribution with $p = \frac{1}{2}$ and $q = \frac{1}{2}$.
Given $P(X=4) = P(X=6)$.
Using the binomial probability formula $P(X=r) = {}^K C_r p^r q^{K-r}$,we have:
${}^K C_4 (\frac{1}{2})^K = {}^K C_6 (\frac{1}{2})^K$
${}^K C_4 = {}^K C_6$
Since ${}^n C_x = {}^n C_y$ implies $n = x + y$ (for $x \neq y$),we get $K = 4 + 6 = 10$.
For a binomial distribution with $p = q = \frac{1}{2}$,the probability $P(X=r)$ is maximum at the mean value.
For $n = 10$,the maximum probability occurs at $r = \frac{n}{2} = \frac{10}{2} = 5$.

(C) Let $n = 5$ be the number of dice thrown.
For a specific face (e.g.,$1$),the probability of it appearing on a single die is $p = \frac{1}{6}$ and the probability of it not appearing is $q = \frac{5}{6}$.
Since there are $6$ possible faces,the probability that at least $3$ dice show the same face is $6 \times P(X \geq 3)$,where $X$ follows a binomial distribution $B(5, \frac{1}{6})$.
$P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5)$.
$P(X = 3) = \binom{5}{3} (\frac{1}{6})^3 (\frac{5}{6})^2 = 10 \times \frac{25}{6^5} = \frac{250}{6^5}$.
$P(X = 4) = \binom{5}{4} (\frac{1}{6})^4 (\frac{5}{6})^1 = 5 \times \frac{5}{6^5} = \frac{25}{6^5}$.
$P(X = 5) = \binom{5}{5} (\frac{1}{6})^5 = 1 \times \frac{1}{6^5} = \frac{1}{6^5}$.
Total probability $= 6 \times (\frac{250 + 25 + 1}{6^5}) = 6 \times \frac{276}{6^5} = \frac{276}{6^4}$.

(A) According to the Binomial probability distribution,let $n$ be the number of times the coin is tossed.
Probability of getting $5$ heads is $P(X=5) = {}^{n}C_{5} (\frac{1}{2})^{n-5} (\frac{1}{2})^{5} = {}^{n}C_{5} (\frac{1}{2})^{n}$.
Probability of getting $7$ heads is $P(X=7) = {}^{n}C_{7} (\frac{1}{2})^{n}$.
Given that $P(X=5) = P(X=7)$,we have ${}^{n}C_{5} = {}^{n}C_{7}$.
Using the property ${}^{n}C_{x} = {}^{n}C_{y} \Rightarrow x+y=n$,we get $n = 5+7 = 12$.
Now,the probability of getting $4$ heads is $P(X=4) = {}^{12}C_{4} (\frac{1}{2})^{12}$.
$P(X=4) = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} \times \frac{1}{4096} = 495 \times \frac{1}{4096} = \frac{495}{4096}$.
Thus,option $A$ is correct.

(D) Total number of outcomes $= 2^4 = 16$.
Let $X$ be the number of girls.
We need to find the probability of at least two girls,i.e.,$P(X \ge 2)$.
This can be calculated as $P(X \ge 2) = 1 - \{P(X=0) + P(X=1)\}$.
For $P(X=0)$ (no girls,all boys): The only outcome is $\{BBBB\}$,so $P(X=0) = \frac{1}{16}$.
For $P(X=1)$ (exactly one girl): The outcomes are $\{GBBB, BGBB, BBGB, BBBG\}$,so $P(X=1) = \frac{4}{16}$.
Therefore,$P(X \ge 2) = 1 - \left(\frac{1}{16} + \frac{4}{16}\right) = 1 - \frac{5}{16} = \frac{11}{16}$.

(B) The probability of success in a single trial is $p = \frac{\alpha}{\beta}$.
The probability of failure in a single trial is $q = 1 - p = \frac{\beta - \alpha}{\beta}$.
We need the probability of failing at least $(n-1)$ times in $n$ trials,which means failing $(n-1)$ times or failing $n$ times.
This is equivalent to succeeding at most $1$ time.
Using the binomial distribution formula $P(X = k) = {}^{n}C_{k} p^k q^{n-k}$:
$P(X \le 1) = P(X = 0) + P(X = 1)$
$P(X = 0) = {}^{n}C_{0} p^0 q^n = 1 \cdot 1 \cdot \left(\frac{\beta - \alpha}{\beta}\right)^n = \frac{(\beta - \alpha)^n}{\beta^n}$
$P(X = 1) = {}^{n}C_{1} p^1 q^{n-1} = n \cdot \left(\frac{\alpha}{\beta}\right) \cdot \left(\frac{\beta - \alpha}{\beta}\right)^{n-1} = \frac{n \alpha (\beta - \alpha)^{n-1}}{\beta^n}$
Adding these probabilities:
$P = \frac{(\beta - \alpha)^n + n \alpha (\beta - \alpha)^{n-1}}{\beta^n}$
$P = \frac{(\beta - \alpha)^{n-1} [(\beta - \alpha) + n \alpha]}{\beta^n}$
$P = \frac{(\beta - \alpha)^{n-1} (n \alpha + \beta - \alpha)}{\beta^n}$
Thus,option $B$ is correct.

(C) Let $n$ be the number of tosses. For a fair coin,the probability of heads $p = \frac{1}{2}$ and the probability of tails $q = \frac{1}{2}$.
Using the binomial distribution formula $P(X=k) = {}^nC_k p^k q^{n-k}$,we have:
$P(X=5) = P(X=4)$
${}^nC_5 (\frac{1}{2})^5 (\frac{1}{2})^{n-5} = {}^nC_4 (\frac{1}{2})^4 (\frac{1}{2})^{n-4}$
${}^nC_5 (\frac{1}{2})^n = {}^nC_4 (\frac{1}{2})^n$
${}^nC_5 = {}^nC_4$
Using the property ${}^nC_r = {}^nC_{n-r}$,we get $n = 5 + 4 = 9$.
Now,we need to find the probability of getting $6$ heads:
$P(X=6) = {}^9C_6 (\frac{1}{2})^6 (\frac{1}{2})^3 = {}^9C_3 (\frac{1}{2})^9$
$P(X=6) = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \times \frac{1}{512} = 84 \times \frac{1}{512} = \frac{21}{128}$.

(A) Let $H$ denote a head and $T$ denote a tail. The points awarded are $2$ for $H$ and $1$ for $T$.
When three coins are tossed,let $n_H$ be the number of heads and $n_T$ be the number of tails.
The total points $S = 2n_H + 1n_T$.
Since $n_H + n_T = 3$,we have $n_T = 3 - n_H$.
Substituting this,$S = 2n_H + (3 - n_H) = n_H + 3$.
For $S$ to be an odd number,$n_H + 3$ must be odd,which means $n_H$ must be even.
Possible values for $n_H$ are $0$ and $2$.
Case $1$: $n_H = 0$ (all tails). The probability is $\binom{3}{0} (\frac{1}{2})^0 (\frac{1}{2})^3 = \frac{1}{8}$.
Case $2$: $n_H = 2$ (two heads,one tail). The probability is $\binom{3}{2} (\frac{1}{2})^2 (\frac{1}{2})^1 = 3 \times \frac{1}{8} = \frac{3}{8}$.
The total probability is $\frac{1}{8} + \frac{3}{8} = \frac{4}{8} = \frac{1}{2}$.

(C) Given $X \sim B(n, p)$ with $n=7$. The probability mass function is $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$.
Given $P(X=3) = P(X=5)$,we have:
$\binom{7}{3} p^3 (1-p)^{7-3} = \binom{7}{5} p^5 (1-p)^{7-5}$
$\binom{7}{3} p^3 (1-p)^4 = \binom{7}{5} p^5 (1-p)^2$
Since $\binom{7}{3} = \binom{7}{4} = 35$ and $\binom{7}{5} = \binom{7}{2} = 21$,we get:
$35 p^3 (1-p)^4 = 21 p^5 (1-p)^2$
Dividing both sides by $7 p^3 (1-p)^2$ (assuming $p \neq 0, 1$):
$5 (1-p)^2 = 3 p^2$
$5 (1 - 2p + p^2) = 3 p^2$
$5 - 10p + 5p^2 = 3p^2$
$2p^2 - 10p + 5 = 0$
Using the quadratic formula $p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$p = \frac{10 \pm \sqrt{100 - 4(2)(5)}}{2(2)} = \frac{10 \pm \sqrt{100 - 40}}{4} = \frac{10 \pm \sqrt{60}}{4} = \frac{10 \pm 2\sqrt{15}}{4} = \frac{5 \pm \sqrt{15}}{2}$
Since $0 \le p \le 1$,we take $p = \frac{5 - \sqrt{15}}{2}$.

(B) For a binomial distribution,the mean is given by $np = \frac{4}{3}$ and the variance is given by $npq = \frac{10}{9}$.
Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{10/9}{4/3} = \frac{10}{9} \times \frac{3}{4} = \frac{30}{36} = \frac{5}{6}$.
Since $p + q = 1$,we have $p = 1 - \frac{5}{6} = \frac{1}{6}$.
Substituting $p = \frac{1}{6}$ into $np = \frac{4}{3}$,we get $n(\frac{1}{6}) = \frac{4}{3}$,which implies $n = \frac{4}{3} \times 6 = 8$.
We need to find $P(X \geq 6) = P(X=6) + P(X=7) + P(X=8)$.
The probability mass function is $P(X=k) = \binom{n}{k} p^k q^{n-k} = \binom{8}{k} (\frac{1}{6})^k (\frac{5}{6})^{8-k}$.
$P(X=6) = \binom{8}{6} (\frac{1}{6})^6 (\frac{5}{6})^2 = 28 \times \frac{25}{6^8} = \frac{700}{6^8}$.
$P(X=7) = \binom{8}{7} (\frac{1}{6})^7 (\frac{5}{6})^1 = 8 \times \frac{5}{6^8} = \frac{40}{6^8}$.
$P(X=8) = \binom{8}{8} (\frac{1}{6})^8 (\frac{5}{6})^0 = 1 \times \frac{1}{6^8} = \frac{1}{6^8}$.
Summing these,$P(X \geq 6) = \frac{700 + 40 + 1}{6^8} = \frac{741}{6^8}$.

(A) Let $X$ be the number of defective units in a sample of $n = 5$ units.
The probability of a unit being defective is $p = \frac{1}{8}$,and the probability of a unit being non-defective is $q = 1 - p = \frac{7}{8}$.
Since the trials are independent,$X$ follows a binomial distribution $B(n, p) = B(5, \frac{1}{8})$.
The probability of having at most one defective unit is $P(X \le 1) = P(X = 0) + P(X = 1)$.
Using the binomial probability formula $P(X = k) = \binom{n}{k} p^k q^{n-k}$:
$P(X = 0) = \binom{5}{0} (\frac{1}{8})^0 (\frac{7}{8})^5 = 1 \times 1 \times (\frac{7}{8})^5 = \frac{16807}{32768}$.
$P(X = 1) = \binom{5}{1} (\frac{1}{8})^1 (\frac{7}{8})^4 = 5 \times \frac{1}{8} \times \frac{2401}{4096} = \frac{12005}{32768}$.
$P(X \le 1) = \frac{16807 + 12005}{32768} = \frac{28812}{32768} = \frac{7203}{8192}$.
None of the given options match the calculated result $\frac{7203}{8192}$.

(A) For a binomial distribution $X \sim B(n, p)$,the mean is given by $E(X) = np = 1$.
Thus,$p = \frac{1}{n}$ and $q = 1 - p = 1 - \frac{1}{n} = \frac{n-1}{n}$.
The probability mass function is $P(X=k) = \binom{n}{k} p^k q^{n-k}$.
Given $P(X=2) = \binom{n}{2} p^2 q^{n-2} = \frac{27}{128}$.
Substituting $p$ and $q$:
$\frac{n(n-1)}{2} \times (\frac{1}{n})^2 \times (\frac{n-1}{n})^{n-2} = \frac{27}{128}$
$\frac{n-1}{2n} \times \frac{(n-1)^{n-2}}{n^{n-2}} = \frac{27}{128}$
$\frac{(n-1)^{n-1}}{2n^{n-1}} = \frac{27}{128}$
For $n=4$:
$\frac{(4-1)^{4-1}}{2(4)^{4-1}} = \frac{3^3}{2(4^3)} = \frac{27}{2(64)} = \frac{27}{128}$.
This satisfies the given condition.
The variance is $Var(X) = npq = 1 \times q = q$.
Since $p = \frac{1}{4}$,$q = 1 - \frac{1}{4} = \frac{3}{4}$.
Therefore,the variance is $\frac{3}{4}$.

(D) Given that $X \sim B(6, p)$ is a binomial variate with $n=6$ and probability of success $p$. The probability mass function is $P(X=x) = {}^{6}C_{x} p^{x} q^{6-x}$,where $q = 1-p$.
Given $\frac{P(X=4)}{P(X=2)} = \frac{1}{9}$.
Substituting the formula: $\frac{{}^{6}C_{4} p^{4} q^{2}}{{}^{6}C_{2} p^{2} q^{4}} = \frac{1}{9}$.
Since ${}^{6}C_{4} = {}^{6}C_{2} = 15$,the expression simplifies to: $\frac{p^{2}}{q^{2}} = \frac{1}{9}$.
Taking the square root on both sides: $\frac{p}{q} = \frac{1}{3}$ (since $p, q > 0$).
Substituting $q = 1-p$: $\frac{p}{1-p} = \frac{1}{3}$.
$3p = 1-p \Rightarrow 4p = 1 \Rightarrow p = \frac{1}{4}$.

(C) Let $X$ be the number of correct answers. Since each question has two options (True or False),the probability of a correct answer is $p = \frac{1}{2}$ and the probability of an incorrect answer is $q = \frac{1}{2}$.
This follows a binomial distribution $B(n, p)$ where $n = 6$ and $p = \frac{1}{2}$.
The student passes if he gets $4, 5,$ or $6$ correct answers.
The probability of passing is $P(X \ge 4) = P(X=4) + P(X=5) + P(X=6)$.
Using the formula $P(X=k) = {}^nC_k p^k q^{n-k}$:
$P(X=4) = {}^6C_4 (\frac{1}{2})^4 (\frac{1}{2})^2 = 15 \times (\frac{1}{2})^6 = \frac{15}{64}$.
$P(X=5) = {}^6C_5 (\frac{1}{2})^5 (\frac{1}{2})^1 = 6 \times (\frac{1}{2})^6 = \frac{6}{64}$.
$P(X=6) = {}^6C_6 (\frac{1}{2})^6 (\frac{1}{2})^0 = 1 \times (\frac{1}{2})^6 = \frac{1}{64}$.
Total probability $= \frac{15+6+1}{64} = \frac{22}{64} = \frac{11}{32}$.

(A) We know that for a binomial distribution:
Mean $= np = 4$
Variance $= npq = \frac{4}{3}$
Dividing the variance by the mean,we get:
$\frac{npq}{np} = \frac{4/3}{4} \Rightarrow q = \frac{1}{3}$
Since $p + q = 1$,we have $p = 1 - \frac{1}{3} = \frac{2}{3}$
Substituting $p$ into the mean equation:
$n \times \frac{2}{3} = 4 \Rightarrow n = 6$
The probability mass function is $P(X=k) = {}^nC_k p^k q^{n-k}$
For $X=2$:
$P(X=2) = {}^6C_2 \times (\frac{2}{3})^2 \times (\frac{1}{3})^4$
$P(X=2) = \frac{6 \times 5}{2 \times 1} \times \frac{4}{9} \times \frac{1}{81}$
$P(X=2) = 15 \times \frac{4}{729} = \frac{60}{729} = \frac{20}{243}$

(B) Let $p$ be the probability of success and $q$ be the probability of failure. Given $p = 5q$. Since $p + q = 1$,we have $5q + q = 1$,which implies $6q = 1$,so $q = \frac{1}{6}$ and $p = \frac{5}{6}$.
For $n = 5$ trials,the probability of getting at most one success is given by $P(X \le 1) = P(X = 0) + P(X = 1)$.
Using the binomial distribution formula $P(X = k) = ^nC_k p^k q^{n-k}$:
$P(X = 0) = ^5C_0 \left(\frac{5}{6}\right)^0 \left(\frac{1}{6}\right)^5 = 1 \times 1 \times \frac{1}{6^5} = \frac{1}{6^5}$.
$P(X = 1) = ^5C_1 \left(\frac{5}{6}\right)^1 \left(\frac{1}{6}\right)^4 = 5 \times \frac{5}{6} \times \frac{1}{6^4} = \frac{25}{6^5}$.
Therefore,$P(X \le 1) = \frac{1}{6^5} + \frac{25}{6^5} = \frac{26}{6^5}$.
Thus,option $(b)$ is correct.

(B) The probability of getting $k$ successes in $n$ trials is given by the binomial distribution formula $P(X=r) = {}^{n}C_r p^r q^{n-r}$.
Here,$n=15$,$p=1/2$ (probability of tail),and $q=1/2$ (probability of head).
We need to find the probability of getting at least $3$ tails,i.e.,$P(X \geq 3)$.
This can be calculated as $P(X \geq 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]$.
$P(X=0) = {}^{15}C_0 (1/2)^0 (1/2)^{15} = 1 \times (1/2)^{15} = 1/2^{15}$.
$P(X=1) = {}^{15}C_1 (1/2)^1 (1/2)^{14} = 15 \times (1/2)^{15} = 15/2^{15}$.
$P(X=2) = {}^{15}C_2 (1/2)^2 (1/2)^{13} = \frac{15 \times 14}{2} \times (1/2)^{15} = 105/2^{15}$.
Summing these probabilities: $P(X < 3) = \frac{1 + 15 + 105}{2^{15}} = \frac{121}{2^{15}}$.
Therefore,$P(X \geq 3) = 1 - \frac{121}{2^{15}}$.

(B) When $2n$ unbiased coins are tossed,the total number of outcomes is $2^{2n}$.
The probability of getting $r$ heads in $2n$ tosses is given by the binomial distribution: $P(r) = \frac{1}{2^{2n}} \binom{2n}{r}$.
The number of heads is equal to the number of tails when the number of heads is exactly $n$.
The probability of getting exactly $n$ heads is $P(n) = \frac{1}{2^{2n}} \binom{2n}{n} = \frac{1}{2^{2n}} \cdot \frac{(2n)!}{(n!)^2}$.
The probability that the number of heads is not equal to the number of tails is $1 - P(n)$.
Therefore,the required probability is $1 - \frac{(2n)!}{(n!)^2} \cdot \frac{1}{2^{2n}}$.

(C) Let $X$ be the number of defective bolts in a sample of $n=4$. The probability of a bolt being defective is $p = 20\% = 0.2 = \frac{1}{5}$.
Thus,the probability of a bolt being non-defective is $q = 1 - p = 1 - \frac{1}{5} = \frac{4}{5}$.
Since the selection is random,$X$ follows a binomial distribution $B(n, p)$ with $n=4$ and $p=\frac{1}{5}$.
The probability mass function is given by $P(X=k) = {}^nC_k p^k q^{n-k}$.
We need to find the probability that less than $2$ bolts are defective,i.e.,$P(X < 2) = P(X=0) + P(X=1)$.
$P(X=0) = {}^4C_0 \left(\frac{1}{5}\right)^0 \left(\frac{4}{5}\right)^4 = 1 \times 1 \times \frac{256}{625} = \frac{256}{625}$.
$P(X=1) = {}^4C_1 \left(\frac{1}{5}\right)^1 \left(\frac{4}{5}\right)^3 = 4 \times \frac{1}{5} \times \frac{64}{125} = \frac{256}{625}$.
Therefore,$P(X < 2) = \frac{256}{625} + \frac{256}{625} = \frac{512}{625} = 0.8192$.

(C) This is a binomial distribution problem where the number of trials $n = 3$.
Success is defined as getting $1$ or $6$ on a die.
The probability of success in a single trial is $p = \frac{2}{6} = \frac{1}{3}$.
The probability of failure is $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.
For a binomial distribution,the variance is given by the formula $Var(X) = npq$.
Substituting the values,we get:
$Var(X) = 3 \times \frac{1}{3} \times \frac{2}{3} = \frac{2}{3}$.

(D) Let the probability of getting a head be $p$.
Then the probability of not getting a head is $1-p$.
The person wins in an odd number of tosses if the first head appears on the $1^{st}, 3^{rd}, 5^{th}, \dots$ toss.
This forms a geometric series: $p + (1-p)^2 p + (1-p)^4 p + \dots = 3/4$.
Using the sum formula for an infinite geometric series $S = \frac{a}{1-r}$,where $a = p$ and $r = (1-p)^2$:
$\frac{p}{1-(1-p)^2} = \frac{3}{4}$
$\frac{p}{1-(1-2p+p^2)} = \frac{3}{4}$
$\frac{p}{2p-p^2} = \frac{3}{4}$
$\frac{1}{2-p} = \frac{3}{4}$
$4 = 6 - 3p \implies 3p = 2 \implies p = 2/3$.
If $5$ such coins are tossed at a time,the probability that a head appears on all $5$ coins is $p^5 = (2/3)^5 = 32/243$.

(C) Let $n=10$ be the number of tosses,$p$ be the probability of success (getting a head) $= \frac{1}{2}$,and $q$ be the probability of failure (getting a tail) $= \frac{1}{2}$.
Let $X$ be the number of successes. We want the probability of having more failures than successes. This means $X < 5$,i.e.,$X \in \{0, 1, 2, 3, 4\}$.
However,the question asks for more failures than successes,which means $X < 5$. Since the total number of trials is $10$,the number of failures is $10-X$. We want $10-X > X$,which implies $2X < 10$,or $X < 5$.
Thus,we need to calculate $P(X \leq 4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)$.
Since the distribution is symmetric for $p=q=\frac{1}{2}$,$P(X \leq 4) = P(X \geq 6)$.
$P(X \geq 6) = P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10)$.
$P(X \geq 6) = \sum_{r=6}^{10} {}^{10}C_r (\frac{1}{2})^{10} = \frac{1}{2^{10}} ({}^{10}C_6 + {}^{10}C_7 + {}^{10}C_8 + {}^{10}C_9 + {}^{10}C_{10})$.
$= \frac{1}{2^{10}} (210 + 120 + 45 + 10 + 1) = \frac{386}{2^{10}} = \frac{193}{2^9}$.

(B) Consider the Binomial probability distribution where $n = 5$ and $p = 0.2 = \frac{1}{5}$.
Probability of success $p = \frac{1}{5}$,so probability of failure $q = 1 - \frac{1}{5} = \frac{4}{5}$.
We need to find the probability that at least two subscribers receive the promotion,which is $P(X \geq 2)$.
This can be calculated as $P(X \geq 2) = 1 - [P(X = 0) + P(X = 1)]$.
$P(X = 0) = { }^5 C_0 \left(\frac{1}{5}\right)^0 \left(\frac{4}{5}\right)^5 = 1 \times 1 \times \frac{1024}{3125} = \frac{1024}{3125}$.
$P(X = 1) = { }^5 C_1 \left(\frac{1}{5}\right)^1 \left(\frac{4}{5}\right)^4 = 5 \times \frac{1}{5} \times \frac{256}{625} = \frac{256}{625} = \frac{1280}{3125}$.
Therefore,$P(X \geq 2) = 1 - \left(\frac{1024 + 1280}{3125}\right) = 1 - \frac{2304}{3125} = \frac{3125 - 2304}{3125} = \frac{821}{3125}$.