Binomial distribution Questions in English

(B) We know that for a binomial distribution,the mean is given by $\mu = np = 18$ and the variance is given by $\sigma^2 = npq = 12$.
Dividing the variance by the mean,we get: $\frac{npq}{np} = \frac{12}{18}$.
This simplifies to $q = \frac{2}{3}$.
Since $p + q = 1$,we have $p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$.

(D) Given $X \sim B(n, p)$ with $n = 10$ and $p = 0.4$.
Since $q = 1 - p$,we have $q = 1 - 0.4 = 0.6$.
The mean of a binomial distribution is $E(X) = np = 10 \times 0.4 = 4$.
The variance of a binomial distribution is $V(X) = npq = 10 \times 0.4 \times 0.6 = 2.4$.
We know the relationship $V(X) = E(X^2) - (E(X))^2$.
Substituting the values,we get $2.4 = E(X^2) - (4)^2$.
$2.4 = E(X^2) - 16$.
Therefore,$E(X^2) = 16 + 2.4 = 18.4$.

(B) For a binomial distribution $X \sim B(n, p)$,the mean is given by $E(X) = np = 18$ and the variance is given by $Var(X) = npq = 12$.
Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{12}{18}$.
This simplifies to $q = \frac{2}{3}$.
Since $p + q = 1$,we have $p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$.
Substituting $p = \frac{1}{3}$ into the mean equation $np = 18$,we get $n \times \frac{1}{3} = 18$,which implies $n = 54$.
The random variable $X$ can take values from $0, 1, 2, \dots, n$.
Therefore,the possible values of $X$ are $0, 1, 2, \dots, 54$.
The total number of such values is $n + 1 = 54 + 1 = 55$.

(B) Given that $X$ follows a binomial distribution $B(n, p)$ with $n=5$ and $p=\frac{1}{3}$.
Since $q = 1 - p$,we have $q = 1 - \frac{1}{3} = \frac{2}{3}$.
The probability mass function is given by $P(X=k) = { }^n C_k p^k q^{n-k}$.
We need to find $P(2 < X < 4)$,which is equivalent to $P(X=3)$.
Substituting the values into the formula:
$P(X=3) = { }^5 C_3 \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^{5-3}$
$P(X=3) = \frac{5 \times 4}{2 \times 1} \times \left(\frac{1}{27}\right) \times \left(\frac{2}{3}\right)^2$
$P(X=3) = 10 \times \frac{1}{27} \times \frac{4}{9}$
$P(X=3) = \frac{40}{243}$

(B) Given that $X$ follows a binomial distribution $B(n, p)$.
For a binomial distribution,the mean $E(X) = np = 5$ and the variance $\operatorname{Var}(X) = npq = 2.5$.
Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{2.5}{5} = 0.5 = \frac{1}{2}$.
Since $p + q = 1$,we have $p = 1 - 0.5 = 0.5 = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $np = 5$,we get $n \times \frac{1}{2} = 5$,which implies $n = 10$.
We need to find $P(X < 1)$,which is equivalent to $P(X = 0)$.
The probability mass function is given by $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$.
For $k = 0$,$P(X = 0) = {}^{10}C_{0} \left(\frac{1}{2}\right)^{0} \left(\frac{1}{2}\right)^{10-0} = 1 \times 1 \times \left(\frac{1}{2}\right)^{10} = \left(\frac{1}{2}\right)^{10}$.

(B) Let $X$ denote the number of successes (getting an even number) in $100$ trials.
Then $X$ follows a binomial distribution with $n = 100$,$p = \frac{1}{2}$,and $q = \frac{1}{2}$.
Variance of $X = npq = 100 \times \frac{1}{2} \times \frac{1}{2} = 25$.
Standard Deviation = $\sqrt{\text{Variance}} = \sqrt{25} = 5$.

(C) Total number of cards $= 52$. Total number of queens $= 4$.
Probability of getting a queen in one draw,$p = \frac{4}{52} = \frac{1}{13}$.
Probability of not getting a queen,$q = 1 - p = \frac{12}{13}$.
Since the cards are drawn with replacement,this follows a binomial distribution $B(n, p)$ with $n = 2$ and $p = \frac{1}{13}$.
The mean of a binomial distribution is given by $E(X) = np$.
$E(X) = 2 \times \frac{1}{13} = \frac{2}{13}$.

(C) Given $X \sim B(n, p)$ with $p=0.6$ and $E(X)=6$.
Since $E(X) = np$,we have $n(0.6) = 6$,which implies $n = \frac{6}{0.6} = 10$.
We know that $q = 1 - p = 1 - 0.6 = 0.4$.
The variance is given by $\operatorname{Var}(X) = npq$.
Substituting the values,$\operatorname{Var}(X) = (10)(0.6)(0.4) = 2.4$.

(A) For a binomial distribution $X \sim B(n, p)$,the mean is given by $E(X) = np = 18$ and the variance is given by $Var(X) = npq = 12$.
We know that $q = 1 - p$.
Dividing the variance by the mean,we get:
$\frac{npq}{np} = \frac{12}{18} \implies q = \frac{2}{3}$.
Since $p + q = 1$,we have $p = 1 - \frac{2}{3} = \frac{1}{3}$.
Substituting $p = \frac{1}{3}$ into $np = 18$:
$n \times \frac{1}{3} = 18 \implies n = 18 \times 3 = 54$.

(D) The number of green balls drawn follows a binomial distribution $B(n, p)$.
Here,the total number of balls is $15 + 10 = 25$.
The probability of drawing a green ball in a single trial is $p = \frac{15}{25} = \frac{3}{5}$.
The probability of drawing a yellow ball is $q = 1 - p = \frac{10}{25} = \frac{2}{5}$.
The number of trials is $n = 10$.
The variance of a binomial distribution is given by $\sigma^2 = npq$.
Substituting the values,we get $\sigma^2 = 10 \times \frac{3}{5} \times \frac{2}{5} = 10 \times \frac{6}{25} = \frac{60}{25} = \frac{12}{5}$.

(A) For a binomial distribution,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $q = 1 - p$.
Given $n = 5$ and $np + npq = 1.8$.
Substituting $n = 5$ and $q = 1 - p$:
$5p + 5p(1 - p) = 1.8$
$5p + 5p - 5p^2 = 1.8$
$10p - 5p^2 = 1.8$
$5p^2 - 10p + 1.8 = 0$
Multiply by $5$ to clear decimals: $25p^2 - 50p + 9 = 0$.
Using the quadratic formula $p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$p = \frac{50 \pm \sqrt{2500 - 900}}{50} = \frac{50 \pm \sqrt{1600}}{50} = \frac{50 \pm 40}{50}$.
$p = \frac{90}{50} = 1.8$ (not possible as $p \leq 1$) or $p = \frac{10}{50} = 0.2$.
Thus,the probability of success is $0.2$.

(B) For a fair coin,the probability of getting a head in a single toss is $p = \frac{1}{2}$ and the probability of getting a tail is $q = 1 - p = \frac{1}{2}$.
Since the coin is tossed $n = 5$ times,we use the binomial distribution formula: $P(X = k) = \binom{n}{k} p^k q^{n-k}$.
We want the probability of getting exactly $k = 3$ heads.
Substituting the values: $P(X = 3) = \binom{5}{3} (\frac{1}{2})^3 (\frac{1}{2})^{5-3}$.
Calculating the binomial coefficient: $\binom{5}{3} = \frac{5 \times 4}{2 \times 1} = 10$.
Thus,$P(X = 3) = 10 \times (\frac{1}{2})^3 \times (\frac{1}{2})^2 = 10 \times (\frac{1}{2})^5$.
$P(X = 3) = 10 \times \frac{1}{32} = \frac{10}{32} = \frac{5}{16}$.
Therefore,the correct option is $B$.

(A) For a Binomial distribution $B(n, p)$,the probability mass function is given by $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$.
Here,$n = 10$ and $p = \frac{1}{2}$,so $1-p = \frac{1}{2}$.
Thus,$P(X = k) = \binom{10}{k} \left( \frac{1}{2} \right)^k \left( \frac{1}{2} \right)^{10-k} = \binom{10}{k} \left( \frac{1}{2} \right)^{10}$.
We need to find $P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$.
$P(X=0) = \binom{10}{0} \left( \frac{1}{2} \right)^{10} = 1 \times \left( \frac{1}{2} \right)^{10}$.
$P(X=1) = \binom{10}{1} \left( \frac{1}{2} \right)^{10} = 10 \times \left( \frac{1}{2} \right)^{10}$.
$P(X=2) = \binom{10}{2} \left( \frac{1}{2} \right)^{10} = \frac{10 \times 9}{2} \times \left( \frac{1}{2} \right)^{10} = 45 \times \left( \frac{1}{2} \right)^{10}$.
Adding these probabilities: $P(X \leq 2) = (1 + 10 + 45) \left( \frac{1}{2} \right)^{10} = 56 \left( \frac{1}{2} \right)^{10}$.
Comparing this with $m \left( \frac{1}{2} \right)^{10}$,we get $m = 56$.

(B) For a binomial distribution,the mean is given by $np = 6$ and the variance is given by $npq = 3$.
Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{3}{6} = \frac{1}{2}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $np = 6$,we get $n(\frac{1}{2}) = 6$,which implies $n = 12$.
The probability mass function is $P(X = k) = \binom{n}{k} p^k q^{n-k} = \binom{12}{k} (\frac{1}{2})^k (\frac{1}{2})^{12-k} = \binom{12}{k} (\frac{1}{2})^{12}$.
We need to find $P(X < 2) = P(X = 0) + P(X = 1)$.
$P(X = 0) = \binom{12}{0} (\frac{1}{2})^{12} = 1 \times \frac{1}{4096} = \frac{1}{4096}$.
$P(X = 1) = \binom{12}{1} (\frac{1}{2})^{12} = 12 \times \frac{1}{4096} = \frac{12}{4096}$.
Therefore,$P(X < 2) = \frac{1}{4096} + \frac{12}{4096} = \frac{13}{4096}$.

(B) Total number of bulbs = $100$.
Number of defective bulbs = $10$.
Number of non-defective bulbs = $100 - 10 = 90$.
The probability of picking a non-defective bulb in a single draw is $p = \frac{90}{100} = \frac{9}{10}$.
Since we are selecting $5$ bulbs,and assuming the selection is done with replacement (or the sample size is small relative to the population),the probability that none of the $5$ bulbs are defective is given by the binomial probability $P(X=0) = \left(\frac{9}{10}\right)^5$.
Thus,the correct option is $B$.

(D) Let $p$ be the probability that a student is a swimmer and $q$ be the probability that a student is not a swimmer.
Given $q = \frac{1}{5}$,so $p = 1 - q = 1 - \frac{1}{5} = \frac{4}{5}$.
We need to find the probability that out of $n = 5$ students,$x = 4$ are swimmers.
Using the Binomial Distribution formula: $P(X = x) = ^nC_x \cdot p^x \cdot q^{n-x}$.
Substituting the values: $P(X = 4) = ^5C_4 \cdot \left(\frac{4}{5}\right)^4 \cdot \left(\frac{1}{5}\right)^{5-4}$.
$P(X = 4) = 5 \cdot \left(\frac{4}{5}\right)^4 \cdot \frac{1}{5} = \left(\frac{4}{5}\right)^4$.
Since this value is not explicitly listed in the options $A, B, C$,the correct choice is $D$.

(B) This is a binomial distribution problem where $n = 5$ and $p = \frac{1}{5}$.
The probability of success (student is a singer) is $p = \frac{1}{5}$.
The probability of failure (student is not a singer) is $q = 1 - p = 1 - \frac{1}{5} = \frac{4}{5}$.
We want to find the probability that $x = 4$ students are singers.
The formula for binomial probability is $P(X = x) = \binom{n}{x} p^x q^{n-x}$.
Substituting the values: $P(X = 4) = \binom{5}{4} \left(\frac{1}{5}\right)^4 \left(\frac{4}{5}\right)^{5-4}$.
$P(X = 4) = 5 \times \left(\frac{1}{5}\right)^4 \times \frac{4}{5}$.
$P(X = 4) = 5 \times \frac{1}{5^4} \times \frac{4}{5} = \frac{4}{5^4} = 4 \left(\frac{1}{5}\right)^4$.
Thus,the correct option is $B$.

(C) For a binomial distribution,the variance is given by the formula $Var(X) = n \times p \times q$,where $q = 1 - p$.
Given $n = 5$ and $p = 0.30$.
Then $q = 1 - 0.30 = 0.70$.
Substituting these values into the formula:
$Var(X) = 5 \times 0.30 \times 0.70$
$Var(X) = 1.5 \times 0.70 = 1.05$.
Therefore,the correct option is $C$.

(C) Given that,the probability of selecting a defective pen is $p = \frac{10}{100} = \frac{1}{10}$.
So,the probability of selecting a non-defective pen is $q = 1 - p = \frac{9}{10}$.
Since the pens are drawn with replacement,this follows a binomial distribution with $n = 5$ trials.
We need to find the probability that at most one pen is defective,i.e.,$P(X \leq 1) = P(X = 0) + P(X = 1)$.
Using the binomial formula $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$:
$P(X = 0) = {}^{5}C_{0} \left(\frac{1}{10}\right)^{0} \left(\frac{9}{10}\right)^{5} = \left(\frac{9}{10}\right)^{5}$.
$P(X = 1) = {}^{5}C_{1} \left(\frac{1}{10}\right)^{1} \left(\frac{9}{10}\right)^{4} = 5 \times \frac{1}{10} \times \left(\frac{9}{10}\right)^{4} = \frac{1}{2} \left(\frac{9}{10}\right)^{4}$.
Therefore,$P(X \leq 1) = \left(\frac{9}{10}\right)^{5} + \frac{1}{2} \left(\frac{9}{10}\right)^{4}$.

(D) Total number of bulbs $N = 100$.
Number of defective bulbs $D = 10$.
Number of non-defective bulbs $G = 100 - 10 = 90$.
We are drawing a sample of $n = 5$ bulbs without replacement.
The probability that none is defective is given by the hypergeometric distribution formula:
$P(X = 0) = \frac{\binom{90}{5} \binom{10}{0}}{\binom{100}{5}}$
$P(X = 0) = \frac{\frac{90 \times 89 \times 88 \times 87 \times 86}{5 \times 4 \times 3 \times 2 \times 1}}{\frac{100 \times 99 \times 98 \times 97 \times 96}{5 \times 4 \times 3 \times 2 \times 1}}$
$P(X = 0) = \frac{90 \times 89 \times 88 \times 87 \times 86}{100 \times 99 \times 98 \times 97 \times 96} \approx (0.9)^{5}$.
Given the options provided,the intended model is binomial approximation (sampling with replacement),where $p = \frac{90}{100} = 0.9$ is the probability of picking a non-defective bulb.
Thus,$P(X = 0) = (0.9)^{5} = (\frac{9}{10})^{5}$.

(B) Let $p$ be the probability of taking a step forward,so $p = 0.4$.
Let $q$ be the probability of taking a step backward,so $q = 0.6$.
Total steps $n = 11$.
Let $x$ be the number of forward steps and $y$ be the number of backward steps.
We have $x + y = 11$.
For the man to be one step away from the starting point,the net displacement must be $x - y = 1$ or $x - y = -1$.
Case $1$: $x - y = 1$. Adding the equations,$2x = 12 \implies x = 6$,so $y = 5$.
The probability for this case is $^{11}C_{6} \times p^{6} \times q^{5} = ^{11}C_{6} \times (0.4)^{6} \times (0.6)^{5}$.
Case $2$: $x - y = -1$. Adding the equations,$2x = 10 \implies x = 5$,so $y = 6$.
The probability for this case is $^{11}C_{5} \times p^{5} \times q^{6} = ^{11}C_{5} \times (0.4)^{5} \times (0.6)^{6}$.
Since $^{11}C_{6} = ^{11}C_{5}$,the total probability is $^{11}C_{6} \times (0.4)^{5} \times (0.6)^{5} \times (0.4 + 0.6) = ^{11}C_{6} \times (0.24)^{5} \times 1 = ^{11}C_{6} \times (0.24)^{5}$.

(C) Given,$n=10$.
Probability of getting an odd number in a single throw,$p = \frac{3}{6} = \frac{1}{2}$.
Probability of not getting an odd number,$q = 1 - p = \frac{1}{2}$.
We need to find the probability that an odd number appears at least once,which is $P(X \geq 1)$.
Using the complement rule,$P(X \geq 1) = 1 - P(X=0)$.
Using the binomial distribution formula $P(X=k) = {}^{n}C_{k} p^{k} q^{n-k}$,we have:
$P(X=0) = {}^{10}C_{0} (\frac{1}{2})^{0} (\frac{1}{2})^{10} = 1 \times 1 \times \frac{1}{1024} = \frac{1}{1024}$.
Therefore,$P(X \geq 1) = 1 - \frac{1}{1024} = \frac{1023}{1024}$.

(D) The probability mass function for a binomial distribution is given by $P(X=r) = {}^{n}C_{r} q^{n-r} p^{r}$,where $q = 1-p$.
Given $n=6$,we have:
$P(X=2) = {}^{6}C_{2} q^{4} p^{2} = 15 q^{4} p^{2} = 12$ (Equation $1$)
$P(X=3) = {}^{6}C_{3} q^{3} p^{3} = 20 q^{3} p^{3} = 5$ (Equation $2$)
Dividing Equation $1$ by Equation $2$:
$\frac{15 q^{4} p^{2}}{20 q^{3} p^{3}} = \frac{12}{5}$
$\frac{3q}{4p} = \frac{12}{5}$
$15q = 48p$
$5q = 16p$
Since $q = 1-p$,we substitute:
$5(1-p) = 16p$
$5 - 5p = 16p$
$5 = 21p$
$p = \frac{5}{21}$

(B) Given that the random variable $X$ follows a binomial distribution with parameters $n=5$ and $p$. The probability mass function is given by $P(X=k) = { }^n C_k p^k (1-p)^{n-k}$.
Given $P(X=2) = 9 P(X=3)$.
Substituting the values: ${ }^5 C_2 p^2 (1-p)^{5-2} = 9 \times { }^5 C_3 p^3 (1-p)^{5-3}$.
Since ${ }^5 C_2 = \frac{5 \times 4}{2 \times 1} = 10$ and ${ }^5 C_3 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$,we have:
$10 p^2 (1-p)^3 = 9 \times 10 p^3 (1-p)^2$.
Dividing both sides by $10 p^2 (1-p)^2$ (assuming $p \neq 0, 1$):
$(1-p) = 9p$.
$1 = 10p$.
$p = 1/10$.

(D) Given the expression: $\sum_{x=0}^{\infty} \frac{k^x}{x !} \lim _{n}$ ${\rightarrow \infty} \frac{n !}{(n-x) !}\left(1-\frac{k}{n}\right)^{n-x}\left(\frac{1}{n}\right)^x$
$= \lim _{n}$ ${\rightarrow \infty} \sum_{x=0}^n \frac{n !}{x !(n-x) !}\left(1-\frac{k}{n}\right)^{n-x}\left(\frac{k}{n}\right)^x$
$= \lim _{n \rightarrow \infty} \sum_{x=0}^n { }^n C_x \left(1-\frac{k}{n}\right)^{n-x}\left(\frac{k}{n}\right)^x$
Using the Binomial Theorem,$\sum_{x=0}^n { }^n C_x a^{n-x} b^x = (a+b)^n$:
$= \lim _{n \rightarrow \infty} \left(1-\frac{k}{n} + \frac{k}{n}\right)^n$
$= \lim _{n \rightarrow \infty} (1)^n = 1$
Thus,the correct option is $D$.

(A) This is a binomial distribution problem where $n = 6$ (number of coins) and $p = 1/2$ (probability of getting a head).
The probability of getting $5$ heads in one trial is $P(X=5) = \binom{6}{5} (1/2)^5 (1/2)^1 = 6 \times (1/2)^6 = 6/64 = 3/32$.
We are performing $N = 320$ trials. Let $Y$ be the number of times we get $5$ heads in $320$ trials.
Then $Y$ follows a binomial distribution with $N = 320$ and $p' = 3/32$.
The mean $\lambda = Np' = 320 \times (3/32) = 30$.
Using the Poisson approximation for the binomial distribution,$P(Y=k) = \frac{\lambda^k e^{-\lambda}}{k!}$.
For $k = 2$,$P(Y=2) = \frac{30^2 \times e^{-30}}{2!} = \frac{900 \times e^{-30}}{2} = 450 \times e^{-30}$.
Since the options are provided in a specific format,the closest match based on the Poisson formula is $30^2 \times \frac{e^{-30}}{2}$.

(B) We know that the mean of a binomial distribution is given by $\mu = np = 9$.
The standard deviation is given by $\sigma = \sqrt{npq} = \frac{3}{2}$.
Squaring the standard deviation,we get $npq = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$.
Substituting $np = 9$ into the equation $npq = \frac{9}{4}$,we get $9q = \frac{9}{4}$,which implies $q = \frac{1}{4}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{4} = \frac{3}{4}$.
Now,$np = 9 \implies n \times \frac{3}{4} = 9 \implies n = 9 \times \frac{4}{3} = 12$.
The binomial distribution is given by $(p + q)^n = \left(\frac{3}{4} + \frac{1}{4}\right)^{12}$.

(B) Given,$p = 0.3, q = 1 - p = 0.7$.
Mean $(\mu) = np = 0.3n$.
Standard deviation $(\sigma) = \sqrt{npq} = \sqrt{n(0.3)(0.7)} = \sqrt{0.21n}$.
Given that $\mu = 3\sigma$.
Substituting the values,we get $0.3n = 3\sqrt{0.21n}$.
Dividing both sides by $3$,we get $0.1n = \sqrt{0.21n}$.
Squaring both sides,we get $(0.1n)^2 = 0.21n$.
$0.01n^2 = 0.21n$.
$n^2 = 21n$.
Since $n \neq 0$,we have $n = 21$.

(C) The numbers greater than $4$ on a die are $5$ and $6$.
The probability of success $p$ is $\frac{2}{6} = \frac{1}{3}$.
The probability of failure $q$ is $1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.
Since the die is thrown $n = 2$ times,the variance of the binomial distribution is given by $\sigma^2 = npq$.
$\sigma^2 = 2 \times \frac{1}{3} \times \frac{2}{3} = \frac{4}{9}$.

(B) The total number of outcomes when a coin is tossed $8$ times is $2^8 = 256$.
Let $H$ denote head and $T$ denote tail.
We want the number of outcomes where $H$ appears consecutively at least $5$ times.
Case $1$: Exactly $5$ consecutive heads.
- $HHHHHTTT$ (and shifts: $THHHHHT T, TTHHHHHT, TTTHHHHH$): $4$ ways.
- $HHHHHTHT$ (and shifts: $THHHHHTH, TTHHHHHT$): $3$ ways.
- $HHHHHTTH$ (and shifts: $THHHHHTT, TTHHHHHT$): $3$ ways.
- $THHHHHTH$ is already counted.
- $HHHHHTHH$ is not possible as it would be $6$ heads.
Using the inclusion-exclusion principle or manual counting for sequences of length $8$ with at least $5$ consecutive $H$:
- Sequences with $5$ consecutive $H$: $HHHHHTXX$ ($4$ ways),$THHHHHTX$ ($3$ ways),$XXTHHHHH$ ($4$ ways).
- Total favorable outcomes are $10$.
Probability $= \frac{10}{256} = \frac{5}{128}$.

(B) The total number of ways to distribute $7$ different balls into $4$ different boxes is $4^7$.
To find the number of ways such that the first box contains exactly $3$ balls,we first choose $3$ balls out of $7$ in ${}^7C_3$ ways.
The remaining $4$ balls can be distributed among the other $3$ boxes in $3^4$ ways.
Thus,the number of favorable ways is ${}^7C_3 \times 3^4$.
The required probability is $\frac{{}^7C_3 \times 3^4}{4^7} = \frac{35 \times 3^4}{4^7} = \frac{35}{4^3} \times \left(\frac{3}{4}\right)^4 = \frac{35}{64} \left(\frac{3}{4}\right)^4$.

(D) When two dice are thrown,the total number of outcomes is $6 \times 6 = 36$.
For a sum of $6$,the favorable outcomes are $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$,so there are $5$ outcomes. The probability of getting a sum of $6$ in one trial is $p_1 = \frac{5}{36}$.
The probability of not getting a sum of $6$ in one trial is $q_1 = 1 - \frac{5}{36} = \frac{31}{36}$.
The probability of getting a sum of $6$ at least once in $9$ trials is $P_1 = 1 - (q_1)^9 = 1 - \left(\frac{31}{36}\right)^9$.
For a sum of $8$,the favorable outcomes are $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$,so there are $5$ outcomes. The probability of getting a sum of $8$ in one trial is $p_2 = \frac{5}{36}$.
The probability of not getting a sum of $8$ in one trial is $q_2 = 1 - \frac{5}{36} = \frac{31}{36}$.
The probability of getting a sum of $8$ at least once in $9$ trials is $P_2 = 1 - (q_2)^9 = 1 - \left(\frac{31}{36}\right)^9$.
Since $P_1 = P_2$,the ratio $P_1 : P_2 = 1 : 1$.

(C) For a single die,the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$.
Multiples of $3$ on a die are $\{3, 6\}$.
The probability of getting a multiple of $3$ on one die is $P(M) = \frac{2}{6} = \frac{1}{3}$.
The probability of $NOT$ getting a multiple of $3$ on one die is $P(M') = 1 - \frac{1}{3} = \frac{2}{3}$.
Since $12$ dice are thrown independently,the probability that a multiple of $3$ does not appear on any of the $12$ dice is $\left(\frac{2}{3}\right)^{12}$.

(A) For a single die,the total number of outcomes is $6$. The probability of getting the number $1$ on the face is $P(1) = \frac{1}{6}$.
Therefore,the probability of not getting the number $1$ on the face of a single die is $P(\text{not } 1) = 1 - \frac{1}{6} = \frac{5}{6}$.
Since $4$ dice are thrown simultaneously,the events are independent.
Thus,the probability that none of the $4$ dice shows the number $1$ is $\left(\frac{5}{6}\right) \times \left(\frac{5}{6}\right) \times \left(\frac{5}{6}\right) \times \left(\frac{5}{6}\right) = \left(\frac{5}{6}\right)^4 = \frac{625}{1296}$.

(B) Let $p$ be the probability that a ship sinks,so $p = \frac{1}{9}$.
Let $q$ be the probability that a ship arrives safely,so $q = 1 - \frac{1}{9} = \frac{8}{9}$.
We are looking for the probability that exactly $3$ ships sink out of $6$ ships.
Using the binomial distribution formula $P(X=k) = {}^n C_k \cdot p^k \cdot q^{n-k}$,where $n=6$,$k=3$,$p=\frac{1}{9}$,and $q=\frac{8}{9}$:
$P(X=3) = {}^6 C_3 \cdot \left(\frac{1}{9}\right)^3 \cdot \left(\frac{8}{9}\right)^3$
$P(X=3) = {}^6 C_3 \cdot \frac{1}{9^3} \cdot \frac{8^3}{9^3}$
$P(X=3) = {}^6 C_3 \cdot \frac{8^3}{9^6}$

(A) The probability of selecting a normal specimen from a pool of $100$ blood samples is $p = \frac{25}{100} = \frac{1}{4}$.
Thus,$q = 1 - p = \frac{3}{4}$.
Number of trials $n = 10$.
We need to find the probability of having at least two normal specimens,$P(X \geq 2)$.
$P(X \geq 2) = 1 - P(X < 2) = 1 - [P(X = 0) + P(X = 1)]$.
Using the binomial distribution formula $P(X = k) = {}^{n}C_{k} p^k q^{n-k}$:
$P(X = 0) = {}^{10}C_{0} (\frac{1}{4})^0 (\frac{3}{4})^{10} = (\frac{3}{4})^{10}$.
$P(X = 1) = {}^{10}C_{1} (\frac{1}{4})^1 (\frac{3}{4})^9 = 10 \times \frac{1}{4} \times (\frac{3}{4})^9 = \frac{10}{4} (\frac{3}{4})^9$.
$P(X < 2) = (\frac{3}{4})^{10} + \frac{10}{4} (\frac{3}{4})^9 = (\frac{3}{4})^9 [\frac{3}{4} + \frac{10}{4}] = (\frac{3}{4})^9 [\frac{13}{4}]$.
Therefore,$P(X \geq 2) = 1 - \frac{13}{4} (\frac{3}{4})^9$.

(A) Let $p$ be the probability of severe fog on a day,so $p = 0.6 = \frac{3}{5}$.
Let $q$ be the probability of no severe fog on a day,so $q = 1 - 0.6 = 0.4 = \frac{2}{5}$.
The number of days in a week is $n = 7$.
Using the binomial distribution formula $P(X = k) = {n \choose k} p^k q^{n-k}$,for $k = 2$:
$P(X = 2) = {7 \choose 2} \times (\frac{3}{5})^2 \times (\frac{2}{5})^5$
$P(X = 2) = 21 \times \frac{9}{25} \times \frac{32}{3125}$
$P(X = 2) = \frac{21 \times 9 \times 32}{5^7} = \frac{6048}{5^7}$.

(D) Total number of outcomes when a coin is tossed $6$ times is $2^6 = 64$.
Let $X$ be the number of heads. We want the probability $P(X > 3)$,which means $X$ can be $4, 5,$ or $6$.
The number of ways to get $r$ heads in $n$ tosses is given by $\binom{n}{r}$.
Number of ways to get $4$ heads $= \binom{6}{4} = \frac{6!}{4!2!} = 15$.
Number of ways to get $5$ heads $= \binom{6}{5} = \frac{6!}{5!1!} = 6$.
Number of ways to get $6$ heads $= \binom{6}{6} = \frac{6!}{6!0!} = 1$.
Total favorable outcomes $= 15 + 6 + 1 = 22$.
Required Probability $= \frac{22}{64} = \frac{11}{32}$.
Hence,option $D$ is correct.

(D) Let $X$ be the number of heads obtained by $P$ and $Y$ be the number of heads obtained by $Q$. Both $X$ and $Y$ follow a binomial distribution $B(n=3, p=0.5)$.
The probability of getting $r$ heads in $3$ tosses is given by $P(X=r) = ^{3}C_{r} (0.5)^3$.
We want to find $P(X=Y) = \sum_{r=0}^{3} P(X=r) \times P(Y=r)$.
Since $P(X=r) = P(Y=r)$,this becomes $\sum_{r=0}^{3} [P(X=r)]^2$.
$P(X=0) = ^{3}C_{0} (0.5)^3 = \frac{1}{8}$,$P(X=1) = ^{3}C_{1} (0.5)^3 = \frac{3}{8}$,$P(X=2) = ^{3}C_{2} (0.5)^3 = \frac{3}{8}$,$P(X=3) = ^{3}C_{3} (0.5)^3 = \frac{1}{8}$.
$P(X=Y) = (\frac{1}{8})^2 + (\frac{3}{8})^2 + (\frac{3}{8})^2 + (\frac{1}{8})^2 = \frac{1+9+9+1}{64} = \frac{20}{64} = \frac{5}{16}$.
Thus,option $D$ is correct.

(D) Let $X_1, X_2, X_3$ be the numbers on the three books drawn with replacement. Each $X_i \in \{1, 2, \dots, 20\}$.
Total number of outcomes $= 20^3$.
We want the largest number to be $7$. This means all three numbers must be $\le 7$,and at least one number must be $7$.
Let $E$ be the event that the largest number is $\le 7$. Then $P(E) = \left(\frac{7}{20}\right)^3$.
Let $F$ be the event that the largest number is $\le 6$. Then $P(F) = \left(\frac{6}{20}\right)^3$.
The probability that the largest number is exactly $7$ is $P(E) - P(F) = \left(\frac{7}{20}\right)^3 - \left(\frac{6}{20}\right)^3$.
Therefore,option $D$ is correct.

(C) The total number of possible outcomes when tossing $6$ coins is $2^6 = 64$.
We need to find the probability of getting at least $4$ heads,which is $P(X \ge 4) = P(X=4) + P(X=5) + P(X=6)$.
Using the binomial distribution formula $P(X=r) = \binom{n}{r} p^r q^{n-r}$,where $n=6$ and $p=q=1/2$:
$P(X=4) = \binom{6}{4} (1/2)^6 = \frac{15}{64}$
$P(X=5) = \binom{6}{5} (1/2)^6 = \frac{6}{64}$
$P(X=6) = \binom{6}{6} (1/2)^6 = \frac{1}{64}$
Summing these probabilities: $P(X \ge 4) = \frac{15+6+1}{64} = \frac{22}{64} = \frac{11}{32}$.
Thus,the correct option is $C$.

(B) Let $p$ be the probability of a climber returning safely,so $p = \frac{9}{10}$.
Let $q$ be the probability of a climber not returning safely,so $q = 1 - p = \frac{1}{10}$.
We use the binomial distribution formula $P(X = k) = {}^nC_k \cdot p^k \cdot q^{n-k}$,where $n = 5$.
We need to find the probability that at least $4$ climbers return safely,which is $P(X \ge 4) = P(X = 4) + P(X = 5)$.
$P(X = 4) = {}^5C_4 \cdot (\frac{9}{10})^4 \cdot (\frac{1}{10})^1 = 5 \cdot \frac{9^4}{10^5}$.
$P(X = 5) = {}^5C_5 \cdot (\frac{9}{10})^5 = 1 \cdot \frac{9^5}{10^5}$.
$P(X \ge 4) = \frac{5 \cdot 9^4}{10^5} + \frac{9^5}{10^5} = \frac{9^4(5 + 9)}{10^5} = \frac{9^4 \cdot 14}{100000} = \frac{9^4 \cdot 7}{50000}$.

(B) Let $S_A$ be the sum of the two dice thrown by $A$ and $S_B$ be the sum of the two dice thrown by $B$.
For the sum of two dice to be even,both dice must show either both even numbers or both odd numbers.
The probability of getting an even sum for one person is $P(\text{even}) = \frac{18}{36} = \frac{1}{2}$.
Since $A$ and $B$ throw simultaneously and independently,the probability that both get an even sum in a single trial is $P(A_{\text{even}} \cap B_{\text{even}}) = P(A_{\text{even}}) \times P(B_{\text{even}}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
Since they perform this experiment $100$ times,the probability that both get an even sum in all $100$ throws is $\left(\frac{1}{4}\right)^{100}$.

(D) Let $X$ be the number of times we get $1, 3,$ or $4$ in $(2n+1)$ throws. The probability of success $p$ in a single throw is $p = \frac{3}{6} = \frac{1}{2}$.
The probability of failure $q$ is $q = 1 - p = \frac{1}{2}$.
Since the die is thrown $(2n+1)$ times,$X$ follows a binomial distribution $B(2n+1, \frac{1}{2})$.
We want to find $P(X \le n) = \sum_{k=0}^{n} \binom{2n+1}{k} (\frac{1}{2})^{2n+1}$.
Using the property of binomial coefficients,$\sum_{k=0}^{n} \binom{2n+1}{k} = \sum_{k=n+1}^{2n+1} \binom{2n+1}{k} = \frac{1}{2} \times 2^{2n+1} = 2^{2n}$.
Thus,$P(X \le n) = 2^{2n} \times (\frac{1}{2})^{2n+1} = \frac{2^{2n}}{2^{2n+1}} = \frac{1}{2}$.

(D) The total number of questions is $n = 8$. Since each question is true or false,the probability of answering correctly is $p = \frac{1}{2}$ and incorrectly is $q = \frac{1}{2}$.
Let $X$ be the number of correct answers. $X$ follows a binomial distribution $B(8, \frac{1}{2})$.
The student passes if $X \ge 6$.
$P(\text{Pass}) = P(X=6) + P(X=7) + P(X=8)$
$P(\text{Pass}) = \binom{8}{6}(\frac{1}{2})^8 + \binom{8}{7}(\frac{1}{2})^8 + \binom{8}{8}(\frac{1}{2})^8$
$P(\text{Pass}) = \frac{28 + 8 + 1}{256} = \frac{37}{256}$.
The probability that the student fails is $P(\text{Fail}) = 1 - P(\text{Pass})$.
$P(\text{Fail}) = 1 - \frac{37}{256} = \frac{219}{256}$.

(A) When two coins are tossed,the sample space is $S = \{(H, H), (H, T), (T, H), (T, T)\}$.
There are $4$ possible outcomes.
The event of getting a tail on both coins is $(T, T)$,which has a probability of $\frac{1}{4}$.
The probability of $NOT$ getting a tail on both coins in a single toss is $1 - \frac{1}{4} = \frac{3}{4}$.
Since the coins are tossed $50$ times independently,the probability of not getting a tail on both coins in all $50$ tosses is $\left(\frac{3}{4}\right)^{50}$.

(A) Let the probability of getting a head be $x$ and the probability of getting a tail be $(1-x)$.
Tom starts the game,so he wins if he gets a head on his turn (1st,3rd,5th,... toss).
The probability of Tom winning is $x + (1-x)^2 x + (1-x)^4 x + \dots = \frac{5}{8}$.
This is an infinite geometric series with first term $a = x$ and common ratio $r = (1-x)^2$.
The sum is $\frac{x}{1-(1-x)^2} = \frac{5}{8}$.
$\frac{x}{1-(1-2x+x^2)} = \frac{5}{8} \Rightarrow \frac{x}{2x-x^2} = \frac{5}{8}$.
Since $x \neq 0$,we have $\frac{1}{2-x} = \frac{5}{8} \Rightarrow 8 = 10 - 5x \Rightarrow 5x = 2 \Rightarrow x = \frac{2}{5}$.
Now,for $n=5$ tosses,the probability of getting exactly $r=3$ heads is given by the binomial distribution $P(X=r) = { }^n C_r p^r q^{n-r}$.
Here $p = \frac{2}{5}$,$q = 1 - \frac{2}{5} = \frac{3}{5}$,$n=5$,$r=3$.
$P(X=3) = { }^5 C_3 \left(\frac{2}{5}\right)^3 \left(\frac{3}{5}\right)^2 = 10 \times \frac{8}{125} \times \frac{9}{25} = \frac{720}{3125} = \frac{144}{625}$.

(B) The probability of getting at least one head is given by $1 - P(\text{no head})$.
Since the coin is unbiased,the probability of getting no head in $n$ tosses is $\left(\frac{1}{2}\right)^n$.
Therefore,the probability of getting at least one head is $1 - \left(\frac{1}{2}\right)^n$.
According to the problem,$1 - \left(\frac{1}{2}\right)^n > 0.8$.
This simplifies to $1 - 0.8 > \left(\frac{1}{2}\right)^n$,which means $0.2 > \left(\frac{1}{2}\right)^n$.
Writing $0.2$ as $\frac{1}{5}$,we get $\frac{1}{5} > \frac{1}{2^n}$,which implies $2^n > 5$.
For $n=1$,$2^1 = 2 < 5$.
For $n=2$,$2^2 = 4 < 5$.
For $n=3$,$2^3 = 8 > 5$.
Thus,the least value of $n$ is $3$.

(A) When a pair of dice is rolled,the total number of possible outcomes is $6 \times 6 = 36$.
The outcome where $6$ appears on both dice is $(6, 6)$,which is only $1$ outcome.
The probability of getting $6$ on both dice in a single roll is $P(E) = \frac{1}{36}$.
The probability of not getting $6$ on both dice in a single roll is $P(E') = 1 - \frac{1}{36} = \frac{35}{36}$.
Since the dice are rolled $24$ times independently,the probability of not getting $6$ on both dice in any of the $24$ rolls is the product of the probabilities for each roll.
Therefore,the required probability is $\left(\frac{35}{36}\right)^{24}$.