Binomial distribution Questions in English

(C) Let the coin be tossed $n$ times.
Probability of getting a head in a single toss is $p = \frac{1}{2}$.
Probability of not getting a head (tail) is $q = 1 - p = \frac{1}{2}$.
The probability of getting at least one head in $n$ tosses is $P(X \geq 1) = 1 - P(X = 0)$.
Given that $P(X \geq 1) > \frac{99}{100}$.
Therefore,$1 - P(X = 0) > \frac{99}{100}$.
Since $P(X = 0) = (\frac{1}{2})^n$,we have $1 - (\frac{1}{2})^n > \frac{99}{100}$.
This simplifies to $1 - \frac{99}{100} > (\frac{1}{2})^n$,which means $\frac{1}{100} > (\frac{1}{2})^n$.
Taking the reciprocal,we get $100 < 2^n$.
For $n = 6$,$2^6 = 64$,which is less than $100$.
For $n = 7$,$2^7 = 128$,which is greater than $100$.
Thus,the minimum number of tosses required is $7$.

(A) Let $X$ be the number of patients who recover. This follows a binomial distribution $B(n, p)$ where $n = 6$ and $p = 0.6$.
Here,$q = 1 - p = 1 - 0.6 = 0.4$.
We need to find the probability that half of the $6$ patients recover,which is $P(X = 3)$.
Using the binomial probability formula $P(X = k) = ^nC_k p^k q^{n-k}$:
$P(X = 3) = ^6C_3 \times (0.6)^3 \times (0.4)^{6-3}$
$P(X = 3) = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times (0.6)^3 \times (0.4)^3$
$P(X = 3) = 20 \times 0.216 \times 0.064$
$P(X = 3) = 0.27648 \approx 0.2762$ (rounding to the given option).

(D) Let $X$ be the number of heads,which follows a binomial distribution $B(n, p)$ with $n = 100$.
The probability of getting $k$ heads is given by $P(X=k) = {}^{n}C_{k} p^{k} (1-p)^{n-k}$.
We are given that $P(X=50) = P(X=51)$.
Substituting the values,we get:
${}^{100}C_{50} p^{50} (1-p)^{50} = {}^{100}C_{51} p^{51} (1-p)^{49}$.
Dividing both sides by $p^{50} (1-p)^{49}$,we get:
${}^{100}C_{50} (1-p) = {}^{100}C_{51} p$.
Rearranging to solve for $p$:
$\frac{1-p}{p} = \frac{{}^{100}C_{51}}{{}^{100}C_{50}} = \frac{100!}{51! 49!} \times \frac{50! 50!}{100!} = \frac{50}{51}$.
Thus,$51(1-p) = 50p$.
$51 - 51p = 50p$.
$101p = 51$.
$p = \frac{51}{101}$.

(B) The probability mass function for a Binomial distribution is given by $P(X=k) = { }^n C_k p^k (1-p)^{n-k}$.
Given $n=4$,we have:
$P(X=3) = { }^4 C_3 p^3 (1-p)^1 = 4p^3(1-p)$
$P(X=2) = { }^4 C_2 p^2 (1-p)^2 = 6p^2(1-p)^2$
Given the condition $2 P(X=3) = 3 P(X=2)$:
$2 \times [4p^3(1-p)] = 3 \times [6p^2(1-p)^2]$
$8p^3(1-p) = 18p^2(1-p)^2$
Dividing both sides by $2p^2(1-p)$ (assuming $p \neq 0, 1$):
$4p = 9(1-p)$
$4p = 9 - 9p$
$13p = 9 \implies p = \frac{9}{13}$
Then $q = 1 - p = 1 - \frac{9}{13} = \frac{4}{13}$.
The variance of a Binomial distribution is $npq$:
$\text{Variance} = 4 \times \frac{9}{13} \times \frac{4}{13} = \frac{144}{169}$.

(D) Given,$P(X=4) = \frac{135}{2^{12}}$.
Using the binomial probability formula $P(X=k) = {}^nC_k p^k q^{n-k}$,we have:
${}^6C_4 p^4 q^2 = \frac{135}{2^{12}}$.
Since ${}^6C_4 = 15$,we get $15 p^4 q^2 = \frac{135}{2^{12}}$.
Dividing by $15$,$p^4 q^2 = \frac{9}{2^{12}} = \frac{3^2}{(2^6)^2} = \left(\frac{3}{64}\right)^2$.
Taking the square root,$p^2 q = \frac{3}{64}$.
Substituting $q = 1-p$,we have $p^2(1-p) = \frac{3}{64}$.
By inspection,if $p = \frac{1}{4}$,then $p^2(1-p) = (\frac{1}{16})(\frac{3}{4}) = \frac{3}{64}$.
Thus,$p = \frac{1}{4}$ and $q = \frac{3}{4}$.
The variance of a binomial distribution is given by $npq$.
Variance $= 6 \times \frac{1}{4} \times \frac{3}{4} = \frac{18}{16} = \frac{9}{8}$.

(B) Given the Binomial distribution $B(n, p)$ with $n=7$,the probability mass function is $P(X=k) = {^nC_k} p^k q^{n-k}$,where $q = 1-p$.
Given $P(X=3) = 5 P(X=4)$.
Substituting the formula:
${^7C_3} p^3 q^4 = 5 \times {^7C_4} p^4 q^3$
Since ${^7C_3} = {^7C_4} = 35$,we can divide both sides by $35 p^3 q^3$:
$q = 5p$
Since $q = 1-p$,we have $1-p = 5p$,which implies $6p = 1$,so $p = \frac{1}{6}$.
Then $q = 1 - \frac{1}{6} = \frac{5}{6}$.
The variance of a Binomial distribution is given by $Var(X) = npq$.
$Var(X) = 7 \times \frac{1}{6} \times \frac{5}{6} = \frac{35}{36}$.

(D) For a Binomial distribution with $n$ trials,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $q = 1 - p$.
Given $n = 5$ and $\mu + \sigma^2 = 1.8$.
Substituting the values,we get:
$np + npq = 1.8$
$5p + 5p(1 - p) = 1.8$
$5p + 5p - 5p^2 = 1.8$
$10p - 5p^2 = 1.8$
$5p^2 - 10p + 1.8 = 0$
Multiplying by $10$ to clear the decimal:
$50p^2 - 100p + 18 = 0$
Dividing by $2$:
$25p^2 - 50p + 9 = 0$
Using the quadratic formula $p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$p = \frac{50 \pm \sqrt{2500 - 4(25)(9)}}{50} = \frac{50 \pm \sqrt{2500 - 900}}{50} = \frac{50 \pm \sqrt{1600}}{50} = \frac{50 \pm 40}{50}$
$p_1 = \frac{90}{50} = 1.8$ (Not possible since $0 \le p \le 1$)
$p_2 = \frac{10}{50} = 0.2$
Thus,$p = 0.2$.

(D) Total number of balls $= 4 + 3 = 7$.
Probability of drawing a black ball,$p = \frac{3}{7}$.
Probability of drawing a red ball (not black),$q = 1 - p = 1 - \frac{3}{7} = \frac{4}{7}$.
Since the ball is replaced,the trials are independent,and we use the binomial distribution $P(X = x) = {}^nC_x p^x q^{n-x}$,where $n = 3$.
For $X = 0$: $P(X = 0) = {}^3C_0 (\frac{3}{7})^0 (\frac{4}{7})^3 = (\frac{4}{7})^3$.
For $X = 1$: $P(X = 1) = {}^3C_1 (\frac{3}{7})^1 (\frac{4}{7})^2 = 3 \times \frac{3}{7} \times (\frac{4}{7})^2 = \frac{9}{7} (\frac{4}{7})^2$.
For $X = 2$: $P(X = 2) = {}^3C_2 (\frac{3}{7})^2 (\frac{4}{7})^1 = 3 \times (\frac{3}{7})^2 \times \frac{4}{7} = \frac{12}{7} (\frac{3}{7})^2$.
For $X = 3$: $P(X = 3) = {}^3C_3 (\frac{3}{7})^3 (\frac{4}{7})^0 = (\frac{3}{7})^3$.
Comparing this with the given options,option $D$ matches the calculated distribution.

(A) This problem follows a binomial distribution with $n = 5$ trials and probability of success $p = 0.3$ (science student). The probability of failure is $q = 1 - p = 0.7$.
The probability of having $x = 2$ science students is given by the formula $P(X = x) = {}^nC_x \cdot p^x \cdot q^{n-x}$.
Substituting the values: $P(X = 2) = {}^5C_2 \cdot (0.3)^2 \cdot (0.7)^3$.
Calculating the values:
${}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$.
$(0.3)^2 = 0.09$.
$(0.7)^3 = 0.343$.
Therefore,$P(X = 2) = 10 \times 0.09 \times 0.343 = 0.9 \times 0.343 = 0.3087$.

(D) Here,$n=5$,$p=\frac{1}{2}$,$q=\frac{1}{2}$.
Probability of getting at least $4$ successes is given by $P(X \ge 4) = P(X=4) + P(X=5)$.
Using the binomial distribution formula $P(X=r) = {}^nC_r p^r q^{n-r}$:
$P(X=4) = {}^5C_4 \left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^1 = 5 \times \frac{1}{16} \times \frac{1}{2} = \frac{5}{32}$.
$P(X=5) = {}^5C_5 \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^0 = 1 \times \frac{1}{32} \times 1 = \frac{1}{32}$.
Therefore,$P(X \ge 4) = \frac{5}{32} + \frac{1}{32} = \frac{6}{32} = \frac{3}{16}$.

(C) For a binomial distribution,the standard deviation is given by $\sigma = \sqrt{npq}$.
Here,the number of trials $n = 500$.
The probability of getting a six in a single toss is $p = \frac{1}{6}$.
The probability of not getting a six is $q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.
Substituting these values into the formula:
$\sigma = \sqrt{500 \times \frac{1}{6} \times \frac{5}{6}}$
$\sigma = \sqrt{\frac{2500}{36}}$
$\sigma = \frac{50}{6} = \frac{25}{3}$.

(C) The probability mass function for a Binomial distribution is given by $P(X=k) = {}^nC_k p^k q^{n-k}$,where $p+q=1$.
Given $n=6$,we have $P(X=4) = {}^6C_4 p^4 q^2$ and $P(X=2) = {}^6C_2 p^2 q^4$.
According to the problem,$9 P(X=4) = P(X=2)$.
Substituting the values: $9 \times {}^6C_4 p^4 q^2 = {}^6C_2 p^2 q^4$.
Since ${}^6C_4 = \frac{6 \times 5}{2 \times 1} = 15$ and ${}^6C_2 = \frac{6 \times 5}{2 \times 1} = 15$,the equation becomes:
$9 \times 15 p^4 q^2 = 15 p^2 q^4$.
Dividing both sides by $15 p^2 q^2$ (assuming $p, q \neq 0$):
$9 p^2 = q^2$.
Taking the square root of both sides: $3p = q$.
Since $p+q=1$,we substitute $p = 1-q$:
$3(1-q) = q \Rightarrow 3 - 3q = q \Rightarrow 4q = 3 \Rightarrow q = \frac{3}{4}$.

(A) Given $X \sim B\left(8, \frac{1}{2}\right)$,we have $n=8$,$p=\frac{1}{2}$,and $q=1-p=\frac{1}{2}$.
We need to find $P(|X-4| \leq 2)$.
This inequality is equivalent to $-2 \leq X-4 \leq 2$,which simplifies to $2 \leq X \leq 6$.
Thus,$P(2 \leq X \leq 6) = P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6)$.
Alternatively,$P(2 \leq X \leq 6) = 1 - [P(X=0) + P(X=1) + P(X=7) + P(X=8)]$.
Using the binomial probability formula $P(X=k) = \binom{n}{k} p^k q^{n-k} = \binom{8}{k} \left(\frac{1}{2}\right)^8$:
$P(X=0) = \binom{8}{0} \left(\frac{1}{2}\right)^8 = 1 \cdot \frac{1}{256} = \frac{1}{256}$.
$P(X=1) = \binom{8}{1} \left(\frac{1}{2}\right)^8 = 8 \cdot \frac{1}{256} = \frac{8}{256}$.
$P(X=7) = \binom{8}{7} \left(\frac{1}{2}\right)^8 = 8 \cdot \frac{1}{256} = \frac{8}{256}$.
$P(X=8) = \binom{8}{8} \left(\frac{1}{2}\right)^8 = 1 \cdot \frac{1}{256} = \frac{1}{256}$.
Sum $= \frac{1+8+8+1}{256} = \frac{18}{256} = \frac{9}{128}$.
Therefore,$P(2 \leq X \leq 6) = 1 - \frac{9}{128} = \frac{119}{128}$.

(D) The probability mass function for a Binomial distribution is given by $P(X=k) = {n \choose k} p^k q^{n-k}$,where $p+q=1$.
Given $n=4$,the equation $2 P(X=3) = 3 P(X=2)$ becomes:
$2 \times {4 \choose 3} p^3 q^1 = 3 \times {4 \choose 2} p^2 q^2$
Since ${4 \choose 3} = 4$ and ${4 \choose 2} = 6$,we have:
$2 \times 4 \times p^3 q = 3 \times 6 \times p^2 q^2$
$8 p^3 q = 18 p^2 q^2$
Dividing both sides by $2 p^2 q$ (assuming $p, q \neq 0$):
$4 p = 9 q$
Since $p = 1-q$,substitute this into the equation:
$4(1-q) = 9q$
$4 - 4q = 9q$
$4 = 13q$
$q = \frac{4}{13}$

(C) Given that for a Binomial distribution,the mean is $np = 2$ and the variance is $npq = 1$.
Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{1}{2}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $np = 2$,we get $n(\frac{1}{2}) = 2$,so $n = 4$.
The probability $P(X=0)$ is given by the formula $P(X=k) = { }^n C_k p^k q^{n-k}$.
For $X=0$,$P(X=0) = { }^4 C_0 (\frac{1}{2})^0 (\frac{1}{2})^4 = 1 \times 1 \times \frac{1}{16} = \frac{1}{16}$.
The odds in favor of an event $E$ are given by $P(E) : (1 - P(E))$.
Therefore,the odds in favor of $X=0$ are $\frac{1}{16} : (1 - \frac{1}{16}) = \frac{1}{16} : \frac{15}{16} = 1 : 15$.

(C) Let $n = 100$ be the number of trials,$p = \frac{1}{2}$ be the probability of getting a head,and $q = \frac{1}{2}$ be the probability of getting a tail.
The probability of getting a head an odd number of times is given by the sum of probabilities for $r = 1, 3, 5, \dots, 99$.
$P(r \text{ is odd}) = \sum_{r \in \{1, 3, \dots, 99\}} \binom{100}{r} p^r q^{100-r}$
$P(r \text{ is odd}) = \sum_{r \in \{1, 3, \dots, 99\}} \binom{100}{r} \left(\frac{1}{2}\right)^r \left(\frac{1}{2}\right)^{100-r}$
$P(r \text{ is odd}) = \left(\frac{1}{2}\right)^{100} \sum_{r \in \{1, 3, \dots, 99\}} \binom{100}{r}$
We know that the sum of binomial coefficients for odd $r$ is $2^{n-1}$. Thus,$\sum_{r \in \{1, 3, \dots, 99\}} \binom{100}{r} = 2^{100-1} = 2^{99}$.
Therefore,$P(r \text{ is odd}) = \left(\frac{1}{2}\right)^{100} \times 2^{99} = \frac{2^{99}}{2^{100}} = \frac{1}{2}$.

(D) For a Binomial distribution,$P(X=k) = {}^nC_k p^k q^{n-k}$,where $q = 1-p$.
Given $n=6$,the relation is $4 P(X=4) = P(X=2)$.
Substituting the formula: $4 \cdot {}^6C_4 p^4 q^{6-4} = {}^6C_2 p^2 q^{6-2}$.
$4 \cdot {}^6C_4 p^4 q^2 = {}^6C_2 p^2 q^4$.
Since ${}^6C_4 = {}^6C_2 = 15$,we have:
$4 \cdot 15 \cdot p^4 q^2 = 15 \cdot p^2 q^4$.
Dividing both sides by $15 p^2 q^2$ (assuming $p, q \neq 0$):
$4 p^2 = q^2$.
Taking the square root on both sides: $2p = q$.
Since $p + q = 1$,we substitute $q = 2p$:
$p + 2p = 1 \Rightarrow 3p = 1 \Rightarrow p = \frac{1}{3}$.

(B) The probability of a case being solved is $p = 25 \% = \frac{1}{4}$.
Thus,the probability of a case not being solved is $q = 1 - p = \frac{3}{4}$.
Given $n = 6$ trials,we need to find the probability that at least $5$ cases are solved,which is $P(X \ge 5) = P(X = 5) + P(X = 6)$.
Using the binomial distribution formula $P(X = x) = {}^nC_x p^x q^{n-x}$:
$P(X = 5) = {}^6C_5 \left(\frac{1}{4}\right)^5 \left(\frac{3}{4}\right)^1 = 6 \times \frac{1}{4^5} \times \frac{3}{4} = \frac{18}{4^6}$.
$P(X = 6) = {}^6C_6 \left(\frac{1}{4}\right)^6 \left(\frac{3}{4}\right)^0 = 1 \times \frac{1}{4^6} \times 1 = \frac{1}{4^6}$.
Therefore,$P(X \ge 5) = \frac{18}{4^6} + \frac{1}{4^6} = \frac{19}{4^6} = \frac{19}{4096}$.

(A) For a binomial distribution,the mean is given by $\mu = np = 4$ and the variance is given by $\sigma^2 = npq = 2$.
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 getting $x$ successes is given by $P(X=x) = { }^n C_x p^x q^{n-x}$.
For $x = 2$,$P(X=2) = { }^8 C_2 \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^6 = \frac{8 \times 7}{2 \times 1} \times \left(\frac{1}{2}\right)^8$.
$P(X=2) = 28 \times \frac{1}{256} = \frac{28}{256}$.

(D) Let $p$ be the probability of getting a doublet in a single throw of a pair of dice. $A$ pair of dice has $36$ total outcomes,and there are $6$ possible doublets: $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$.
Thus,$p = \frac{6}{36} = \frac{1}{6}$ and $q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.
The dice are thrown $n = 4$ times. Let $X$ be the random variable representing the number of doublets obtained. This follows a binomial distribution $B(n, p) = B(4, \frac{1}{6})$.
The probability $P(X=k)$ is given by $\binom{n}{k} p^k q^{n-k}$.
For $X=0: P(0) = \binom{4}{0} (\frac{1}{6})^0 (\frac{5}{6})^4 = 1 \times 1 \times \frac{625}{1296} = \frac{625}{1296}$.
For $X=1: P(1) = \binom{4}{1} (\frac{1}{6})^1 (\frac{5}{6})^3 = 4 \times \frac{1}{6} \times \frac{125}{216} = \frac{500}{1296} = \frac{125}{324}$.
For $X=2: P(2) = \binom{4}{2} (\frac{1}{6})^2 (\frac{5}{6})^2 = 6 \times \frac{1}{36} \times \frac{25}{36} = \frac{150}{1296} = \frac{25}{216}$.
For $X=3: P(3) = \binom{4}{3} (\frac{1}{6})^3 (\frac{5}{6})^1 = 4 \times \frac{1}{216} \times \frac{5}{6} = \frac{20}{1296} = \frac{5}{324}$.
For $X=4: P(4) = \binom{4}{4} (\frac{1}{6})^4 (\frac{5}{6})^0 = 1 \times \frac{1}{1296} \times 1 = \frac{1}{1296}$.

(A) Let $X$ be the number of heads in $n=100$ tosses. Since the coin is fair,the probability of getting a head is $p = \frac{1}{2}$ and the probability of getting a tail is $q = \frac{1}{2}$.
$X$ follows a binomial distribution $B(n, p) = B(100, \frac{1}{2})$.
The probability of getting an even number of heads is $P(X \in \{0, 2, 4, \dots, 100\}) = \sum_{k \in \{0, 2, \dots, 100\}} \binom{100}{k} p^k q^{100-k}$.
Since $p=q=\frac{1}{2}$,this becomes $\left(\frac{1}{2}\right)^{100} \sum_{k \in \{0, 2, \dots, 100\}} \binom{100}{k}$.
We know that the sum of even binomial coefficients is $\sum_{k \text{ even}} \binom{n}{k} = 2^{n-1}$.
Thus,the probability is $\left(\frac{1}{2}\right)^{100} \times 2^{100-1} = \frac{2^{99}}{2^{100}} = \frac{1}{2}$.

(B) Let $p$ be the probability of getting a head,so $p = \frac{1}{2}$.
Let $q$ be the probability of not getting a head,so $q = 1 - p = \frac{1}{2}$.
The probability of getting $x$ heads in $n$ tosses is given by the binomial distribution formula: $P(X=x) = {}^{n}C_{x} p^x q^{n-x} = {}^{n}C_{x} (\frac{1}{2})^n$.
Given that $P(X=7) = P(X=9)$,we have:
${}^{n}C_{7} (\frac{1}{2})^n = {}^{n}C_{9} (\frac{1}{2})^n$.
This implies ${}^{n}C_{7} = {}^{n}C_{9}$.
Using the property ${}^{n}C_{r} = {}^{n}C_{n-r}$,we get $n = 7 + 9 = 16$.
Now,we need to find the probability of getting $2$ heads,which is $P(X=2)$:
$P(X=2) = {}^{16}C_{2} (\frac{1}{2})^{16} = \frac{16 \times 15}{2} \times \frac{1}{2^{16}} = 120 \times \frac{1}{2^{16}} = \frac{15 \times 8}{2^{16}} = \frac{15}{2^{13}}$.

(A) Given $X \sim B(4, p)$,where $n=4$. The probability mass function is $P(X=k) = {}^{n}C_{k} p^{k} q^{n-k}$.
Given the condition $2 P(X=3) = 3 P(X=2)$.
Substituting the values: $2 \times ({}^{4}C_{3} p^{3} q^{1}) = 3 \times ({}^{4}C_{2} p^{2} q^{2})$.
Calculating combinations: $2 \times (4 p^{3} q) = 3 \times (6 p^{2} q^{2})$.
Simplifying: $8 p^{3} q = 18 p^{2} q^{2}$.
Dividing both sides by $2 p^{2} q$ (assuming $p, q \neq 0$): $4 p = 9 q$.
Since $q = 1 - p$,we have $4 p = 9(1 - p)$.
$4 p = 9 - 9 p$.
$13 p = 9$.
Therefore,$p = \frac{9}{13}$.

(D) Let $X$ denote the number of defective bulbs out of $20$ bulbs selected.
Given that the total number of bulbs is $100$ and $10$ are defective,the probability $p$ of selecting a defective bulb is $p = \frac{10}{100} = \frac{1}{10}$.
Consequently,the probability $q$ of selecting a non-defective bulb is $q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10}$.
We are selecting $n = 20$ bulbs. The probability of getting no defective bulb is given by the binomial distribution formula $P(X = k) = {}^{n}C_{k} \cdot p^{k} \cdot q^{n-k}$.
For $k = 0$,$P(X = 0) = {}^{20}C_{0} \cdot \left(\frac{1}{10}\right)^{0} \cdot \left(\frac{9}{10}\right)^{20-0}$.
$P(X = 0) = 1 \cdot 1 \cdot \left(\frac{9}{10}\right)^{20} = \left(\frac{9}{10}\right)^{20}$.

(C) Let $X$ be the number of heads in $8$ tosses of a fair coin. Here,$n=8$,$p=\frac{1}{2}$,and $q=\frac{1}{2}$.
We want to find the probability that heads show more than tails,which means $X > 4$.
Since the total number of outcomes is $2^8 = 256$,and the distribution is symmetric,we have $P(X < 4) = P(X > 4)$.
We know that $\sum_{k=0}^{8} P(X=k) = 1$,so $P(X < 4) + P(X=4) + P(X > 4) = 1$.
Thus,$2P(X > 4) + P(X=4) = 1$,which implies $P(X > 4) = \frac{1 - P(X=4)}{2}$.
Calculating $P(X=4) = {}^{8}C_{4} \left(\frac{1}{2}\right)^{8} = \frac{70}{256}$.
Therefore,$P(X > 4) = \frac{1 - \frac{70}{256}}{2} = \frac{\frac{186}{256}}{2} = \frac{93}{256}$.

(B) The probability of missing the target is $q = 0.2 = \frac{1}{5}$.
So,the probability of hitting the target is $p = 1 - q = 1 - 0.2 = 0.8 = \frac{4}{5}$.
Given $n = 10$ bombs are dropped,and we want exactly $r = 2$ hits.
Using the binomial distribution formula $P(X = r) = {}^{n}C_{r} p^{r} q^{n-r}$:
$P(X = 2) = {}^{10}C_{2} \times (0.8)^{2} \times (0.2)^{8}$
$P(X = 2) = \frac{10 \times 9}{2 \times 1} \times \left(\frac{4}{5}\right)^{2} \times \left(\frac{1}{5}\right)^{8}$
$P(X = 2) = 45 \times \frac{16}{25} \times \frac{1}{5^{8}}$
$P(X = 2) = 45 \times \frac{16}{5^{2} \times 5^{8}} = 45 \times \frac{16}{5^{10}}$
$P(X = 2) = (9 \times 5) \times \frac{16}{5^{10}} = \frac{9 \times 16}{5^{9}} = \frac{144}{5^{9}}$
Thus,the correct option is $B$.

(B) Let the probability of a person having a common cold be $p = \frac{10}{100} = \frac{1}{10}$.
Then,the probability of a person not having a common cold is $q = 1 - p = \frac{9}{10}$.
We are selecting $n = 5$ persons. Let $X$ be the number of persons having a common cold. $X$ follows a binomial distribution $B(n, p) = B(5, 0.1)$.
We need to find the probability that at most one person has a common cold,which is $P(X \le 1) = P(X = 0) + P(X = 1)$.
Using the binomial probability formula $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$:
$P(X = 0) = {}^{5}C_{0} \left(\frac{1}{10}\right)^{0} \left(\frac{9}{10}\right)^{5} = 1 \times 1 \times \frac{59049}{100000} = 0.59049$.
$P(X = 1) = {}^{5}C_{1} \left(\frac{1}{10}\right)^{1} \left(\frac{9}{10}\right)^{4} = 5 \times \frac{1}{10} \times \frac{6561}{10000} = \frac{32805}{100000} = 0.32805$.
Therefore,$P(X \le 1) = 0.59049 + 0.32805 = 0.91854$.
Rounding to four decimal places,the probability is $0.9185$.

(D) For a binomial distribution $X \sim B(n, p)$,the mean is given by $E(X) = np$ and the variance is given by $\operatorname{Var}(X) = npq$,where $q = 1 - p$.
Given $E(X) = np = 4$.
Given $\operatorname{Var}(X) = npq = 2.4$.
Substituting $np = 4$ into the variance equation: $4q = 2.4$.
Solving for $q$: $q = \frac{2.4}{4} = 0.6 = \frac{3}{5}$.
Since $p = 1 - q$,we have $p = 1 - 0.6 = 0.4 = \frac{2}{5}$.
Now,substitute $p$ back into the mean equation: $n \times \frac{2}{5} = 4$.
$n = 4 \times \frac{5}{2} = 10$.
Thus,the value of $n$ is $10$.

(B) For a binomial distribution,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $q = 1-p$.
Given $n = 5$ and $np + npq = 1 \cdot 8$.
Substituting the values,we get $5p + 5pq = 1 \cdot 8$.
$5p(1 + q) = 1 \cdot 8$.
Since $q = 1 - p$,we have $5p(1 + 1 - p) = 1 \cdot 8$.
$5p(2 - p) = 1 \cdot 8$.
$10p - 5p^2 = 1 \cdot 8$.
$5p^2 - 10p + 1 \cdot 8 = 0$.
Multiplying by $5$ to simplify: $25p^2 - 50p + 9 = 0$.
Using the quadratic formula $p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$p = \frac{50 \pm \sqrt{2500 - 4(25)(9)}}{50} = \frac{50 \pm \sqrt{2500 - 900}}{50} = \frac{50 \pm \sqrt{1600}}{50} = \frac{50 \pm 40}{50}$.
$p = \frac{90}{50} = 1 \cdot 8$ (not possible as $0 \le p \le 1$) or $p = \frac{10}{50} = 0 \cdot 2$.
Thus,$p = 0 \cdot 2$.

(B) Let $X$ be the number of successes in $n = 100$ trials. This follows a binomial distribution $B(n, p)$.
Here,the probability of getting an even number in a single throw of a die is $p = \frac{3}{6} = \frac{1}{2}$.
The probability of failure is $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.
The variance of a binomial distribution is given by the formula $\text{Var}(X) = npq$.
Substituting the values,we get $\text{Var}(X) = 100 \times \frac{1}{2} \times \frac{1}{2} = 100 \times \frac{1}{4} = 25$.
Therefore,the variance is $25$.

(C) The probability of getting a head in a single toss is $p = \frac{1}{2}$ and the probability of getting a tail is $q = \frac{1}{2}$.
For $n = 15$ trials,the probability of getting $r = 10$ heads is given by the binomial distribution formula: $P(X = r) = {}^{n}C_{r} p^{r} q^{n-r}$.
Substituting the values,we get: $P(X = 10) = {}^{15}C_{10} \left(\frac{1}{2}\right)^{10} \left(\frac{1}{2}\right)^{5}$.
Since ${}^{15}C_{10} = {}^{15}C_{5}$,we have: $P(X = 10) = {}^{15}C_{5} \times \left(\frac{1}{2}\right)^{15}$.
Calculating the combination: ${}^{15}C_{5} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} = 3003$.
Thus,the probability is: $P(X = 10) = \frac{3003}{2^{15}} = \frac{3003}{32768}$.

(A) The probability mass function of a Binomial distribution is given by $P(X=k) = {}^{n}C_{k} p^{k} (1-p)^{n-k}$.
Given $n=6$,we have $P(X=4) = {}^{6}C_{4} p^{4} (1-p)^{2}$ and $P(X=2) = {}^{6}C_{2} p^{2} (1-p)^{4}$.
According to the problem,$9 P(X=4) = P(X=2)$.
Substituting the values,we get $9 \cdot {}^{6}C_{4} p^{4} (1-p)^{2} = {}^{6}C_{2} p^{2} (1-p)^{4}$.
Since ${}^{6}C_{4} = {}^{6}C_{2} = 15$,the equation simplifies to $9 p^{4} (1-p)^{2} = p^{2} (1-p)^{4}$.
Dividing both sides by $p^{2} (1-p)^{2}$ (assuming $p \neq 0, 1$),we get $9 p^{2} = (1-p)^{2}$.
Taking the square root of both sides,$3p = 1-p$ (since $p$ must be positive).
$4p = 1$,which gives $p = 1/4$.

(C) For a binomial distribution $B(n, p)$,the probability mass function is given by $P(X=k) = \binom{n}{k} p^k q^{n-k}$,where $q = 1-p$ and $n = 33$.
Given $3 P(X=0) = P(X=1)$,we substitute the values:
$3 \binom{33}{0} p^0 q^{33} = \binom{33}{1} p^1 q^{32}$
$3 \times 1 \times q^{33} = 33 \times p \times q^{32}$
Since $q \neq 0$,we divide both sides by $q^{32}$:
$3q = 33p$
$q = 11p$
Since $q = 1-p$,we have $1-p = 11p$,which implies $1 = 12p$,so $p = \frac{1}{12}$.
Then $q = 1 - \frac{1}{12} = \frac{11}{12}$.
The variance of a binomial distribution is given by $Var(X) = npq$.
$Var(X) = 33 \times \frac{1}{12} \times \frac{11}{12} = \frac{33 \times 11}{144} = \frac{363}{144}$.
Simplifying by dividing by $3$: $\frac{121}{48}$.

(B) The given probability mass function is $P(X=x) = \frac{{}^4C_x}{2^4} = {}^4C_x \left(\frac{1}{2}\right)^x \left(\frac{1}{2}\right)^{4-x}$.
This is a binomial distribution $B(n, p)$ with $n = 4$ and $p = \frac{1}{2}$.
The mean $\mu$ of a binomial distribution is given by $\mu = np = 4 \times \frac{1}{2} = 2$.
The variance $\sigma^2$ of a binomial distribution is given by $\sigma^2 = npq$,where $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.
Thus,$\sigma^2 = 4 \times \frac{1}{2} \times \frac{1}{2} = 1$.
Therefore,$\mu = 2$ and $\sigma^2 = 1$.

(B) Given $X \sim B(n, p)$ with $n=35$. The probability mass function is $P(X=k) = \binom{n}{k} p^k q^{n-k}$,where $q = 1-p$.
Given $7 P(X=0) = P(X=1)$.
Substituting the values: $7 \binom{35}{0} p^0 q^{35} = \binom{35}{1} p^1 q^{34}$.
$7 q = 35 p \implies q = 5p$.
Since $p+q=1$,we have $p+5p=1 \implies 6p=1 \implies p = \frac{1}{6}$ and $q = \frac{5}{6}$.
Now,we need to find $\frac{P(X=15)}{P(X=20)} = \frac{\binom{35}{15} p^{15} q^{20}}{\binom{35}{20} p^{20} q^{15}} = \frac{\binom{35}{15}}{\binom{35}{20}} \cdot \frac{q^5}{p^5} = \frac{\binom{35}{15}}{\binom{35}{15}} \cdot (\frac{q}{p})^5$.
Since $\binom{35}{15} = \binom{35}{20}$,the ratio is $(q/p)^5 = (5)^5 = 3125$.

(D) The total number of cards is $52$. The number of kings in the deck is $4$.
Since the cards are drawn with replacement,the probability of drawing a king in a single trial is $p = \frac{4}{52} = \frac{1}{13}$.
The probability of not drawing a king is $q = 1 - p = 1 - \frac{1}{13} = \frac{12}{13}$.
Let $X$ be the random variable representing the number of kings in $n = 2$ trials. $X$ follows a binomial distribution $B(n, p) = B(2, \frac{1}{13})$.
The probability distribution of $X$ is:
$P(X=0) = \binom{2}{0} p^0 q^2 = 1 \times 1 \times (\frac{12}{13})^2 = \frac{144}{169}$
$P(X=1) = \binom{2}{1} p^1 q^1 = 2 \times \frac{1}{13} \times \frac{12}{13} = \frac{24}{169}$
$P(X=2) = \binom{2}{2} p^2 q^0 = 1 \times (\frac{1}{13})^2 \times 1 = \frac{1}{169}$
We need to find $E(X^2) = \sum x^2 P(X=x)$.
$E(X^2) = (0^2 \times \frac{144}{169}) + (1^2 \times \frac{24}{169}) + (2^2 \times \frac{1}{169})$
$E(X^2) = 0 + \frac{24}{169} + \frac{4}{169} = \frac{28}{169}$.

(C) The total number of cards is $52$. The number of jacks in the pack is $4$.
Since the cards are drawn with replacement,the probability of drawing a jack in a single draw is $p = \frac{4}{52} = \frac{1}{13}$,and the probability of not drawing a jack is $q = 1 - \frac{1}{13} = \frac{12}{13}$.
This follows a binomial distribution $B(n, p)$ where $n=2$ and $p=\frac{1}{13}$.
$P(X=1) = \binom{2}{1} \times p^1 \times q^1 = 2 \times \frac{1}{13} \times \frac{12}{13} = \frac{24}{169}$.
$P(X=2) = \binom{2}{2} \times p^2 \times q^0 = 1 \times \left(\frac{1}{13}\right)^2 \times 1 = \frac{1}{169}$.
Therefore,$P(X=1) + P(X=2) = \frac{24}{169} + \frac{1}{169} = \frac{25}{169}$.

(A) Let $X$ be the number of times a black ball is drawn in $3$ independent trials. This follows a binomial distribution $B(n, p)$ where $n=3$.
The probability of drawing a black ball in a single draw is $p = \frac{3}{4+3} = \frac{3}{7}$.
The probability of not drawing a black ball is $q = 1 - p = 1 - \frac{3}{7} = \frac{4}{7}$.
The probability distribution is given by $P(X=k) = \binom{n}{k} p^k q^{n-k}$ for $k = 0, 1, 2, 3$.
For $k=0$: $P(X=0) = \binom{3}{0} (\frac{3}{7})^0 (\frac{4}{7})^3 = 1 \times 1 \times (\frac{4}{7})^3 = (\frac{4}{7})^3$.
For $k=1$: $P(X=1) = \binom{3}{1} (\frac{3}{7})^1 (\frac{4}{7})^2 = 3 \times \frac{3}{7} \times (\frac{4}{7})^2 = \frac{9}{7} \times (\frac{4}{7})^2$.
For $k=2$: $P(X=2) = \binom{3}{2} (\frac{3}{7})^2 (\frac{4}{7})^1 = 3 \times (\frac{3}{7})^2 \times \frac{4}{7} = \frac{12}{7} \times (\frac{3}{7})^2$.
For $k=3$: $P(X=3) = \binom{3}{3} (\frac{3}{7})^3 (\frac{4}{7})^0 = 1 \times (\frac{3}{7})^3 \times 1 = (\frac{3}{7})^3$.
Comparing these values with the given options,option $(A)$ is correct.

(B) For a Binomial Distribution,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $n$ is the number of trials,$p$ is the probability of success,and $q = 1-p$ is the probability of failure.
Given $n = 10$ and $\text{mean} + \text{variance} = \frac{15}{2}$.
Substituting the formulas: $np + npq = \frac{15}{2}$.
Since $q = 1-p$,we have $np + np(1-p) = \frac{15}{2}$.
$10p + 10p(1-p) = 7.5$.
$10p + 10p - 10p^2 = 7.5$.
$20p - 10p^2 = 7.5$.
Divide by $2.5$: $8p - 4p^2 = 3$.
$4p^2 - 8p + 3 = 0$.
Solving the quadratic equation: $(2p-1)(2p-3) = 0$.
This gives $p = \frac{1}{2}$ or $p = \frac{3}{2}$.
Since $0 < p < 1$,we must have $p = \frac{1}{2}$.
Then $q = 1 - \frac{1}{2} = \frac{1}{2}$.
Therefore,the variance is $\sigma^2 = npq = 10 \times \frac{1}{2} \times \frac{1}{2} = 2.5$.

(A) The probability mass function for a binomial distribution is given by $P(X=k) = {^nC_k} p^k q^{n-k}$,where $q = 1-p$.
Given $n=6$,we have $P(X=4) = {^6C_4} p^4 q^2$ and $P(X=2) = {^6C_2} p^2 q^4$.
The condition is $9 \cdot P(X=4) = P(X=2)$.
Substituting the values: $9 \cdot {^6C_4} p^4 q^2 = {^6C_2} p^2 q^4$.
Since ${^6C_4} = \frac{6 \times 5}{2 \times 1} = 15$ and ${^6C_2} = \frac{6 \times 5}{2 \times 1} = 15$,the equation becomes:
$9 \cdot 15 \cdot p^4 q^2 = 15 \cdot p^2 q^4$.
Dividing both sides by $15 p^2 q^2$ (assuming $p, q \neq 0$):
$9 p^2 = q^2$.
Taking the square root of both sides: $3p = q$.
Since $q = 1-p$,we have $3p = 1-p$.
$4p = 1$,which gives $p = \frac{1}{4}$.

(D) Let $X \sim B(n, p)$.
Given that the mean $np = 8$ and variance $npq = 4$.
Since $q = 1 - p$,we have $8q = 4$,which implies $q = \frac{1}{2}$ and $p = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $np = 8$,we get $n = 16$.
We need to find $P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$.
Using the formula $P(X=r) = {}^{n}C_{r} p^{r} q^{n-r}$,we have:
$P(X \leq 2) = {}^{16}C_{0} (\frac{1}{2})^{0} (\frac{1}{2})^{16} + {}^{16}C_{1} (\frac{1}{2})^{1} (\frac{1}{2})^{15} + {}^{16}C_{2} (\frac{1}{2})^{2} (\frac{1}{2})^{14}$.
$P(X \leq 2) = \frac{{}^{16}C_{0} + {}^{16}C_{1} + {}^{16}C_{2}}{2^{16}}$.
Calculating the combinations: ${}^{16}C_{0} = 1$,${}^{16}C_{1} = 16$,and ${}^{16}C_{2} = \frac{16 \times 15}{2} = 120$.
Thus,$P(X \leq 2) = \frac{1 + 16 + 120}{2^{16}} = \frac{137}{2^{16}}$.
Comparing this with $\frac{K}{2^{16}}$,we get $K = 137$.

(A) Let $X$ be the random variable representing the number of jacks drawn in $2$ trials with replacement.
The total number of cards is $52$,and the number of jacks is $4$.
The probability of drawing a jack in a single trial is $p = \frac{4}{52} = \frac{1}{13}$.
The probability of not drawing a jack is $q = 1 - p = 1 - \frac{1}{13} = \frac{12}{13}$.
Since the trials are independent (with replacement),$X$ follows a binomial distribution $B(n, p)$ where $n = 2$.
$P(X = x) = ^nC_x p^x q^{n-x}$
$P(X = 0) = ^2C_0 (\frac{1}{13})^0 (\frac{12}{13})^2 = 1 \times 1 \times \frac{144}{169} = \frac{144}{169}$
$P(X = 1) = ^2C_1 (\frac{1}{13})^1 (\frac{12}{13})^1 = 2 \times \frac{1}{13} \times \frac{12}{13} = \frac{24}{169}$
$P(X = 2) = ^2C_2 (\frac{1}{13})^2 (\frac{12}{13})^0 = 1 \times \frac{1}{169} \times 1 = \frac{1}{169}$
Thus,the probability distribution is given by option $(A)$.

(C) Let $X$ be the random variable representing the number of kings drawn.
Since the cards are drawn with replacement,the trials are independent.
Total number of cards = $52$.
Number of kings = $4$.
Probability of drawing a king in a single trial,$p = \frac{4}{52} = \frac{1}{13}$.
Probability of not drawing a king in a single trial,$q = 1 - p = 1 - \frac{1}{13} = \frac{12}{13}$.
Since there are $n = 2$ trials,$X$ follows a binomial distribution $B(n, p) = B(2, \frac{1}{13})$.
The probability distribution is given by $P(X = k) = {}^nC_k p^k q^{n-k}$.
For $X = 0$: $P(X = 0) = {}^2C_0 (\frac{1}{13})^0 (\frac{12}{13})^2 = 1 \cdot 1 \cdot \frac{144}{169} = \frac{144}{169}$.
For $X = 1$: $P(X = 1) = {}^2C_1 (\frac{1}{13})^1 (\frac{12}{13})^1 = 2 \cdot \frac{1}{13} \cdot \frac{12}{13} = \frac{24}{169}$.
For $X = 2$: $P(X = 2) = {}^2C_2 (\frac{1}{13})^2 (\frac{12}{13})^0 = 1 \cdot \frac{1}{169} \cdot 1 = \frac{1}{169}$.
Thus,the probability distribution is:
$X=0, P(X)=\frac{144}{169}$
$X=1, P(X)=\frac{24}{169}$
$X=2, P(X)=\frac{1}{169}$
Therefore,option $(C)$ is correct.

(A) Given $X \sim B(n, p)$ with $n=4$. The probability mass function is $P(X=k) = {}^nC_k p^k q^{n-k}$,where $q = 1-p$.
Given $P(X=0) = \frac{16}{81}$.
Using the formula,$P(X=0) = {}^4C_0 p^0 q^4 = q^4$.
So,$q^4 = \frac{16}{81} = \left(\frac{2}{3}\right)^4$.
Thus,$q = \frac{2}{3}$,which implies $p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$.
Now,we need to find $P(X=4) = {}^4C_4 p^4 q^0 = (1) \left(\frac{1}{3}\right)^4 (1) = \frac{1}{81}$.

(C) Let $p$ be the probability of getting a $6$ on a single die,so $p = \frac{1}{6}$.
Let $q$ be the probability of not getting a $6$,so $q = 1 - p = \frac{5}{6}$.
Since two dice are thrown,$X$ follows a binomial distribution $B(n, p)$ where $n = 2$ and $p = \frac{1}{6}$.
The expectation $E(X)$ of a binomial distribution is given by $E(X) = np$.
$E(X) = 2 \times \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.

(C) The given probability mass function is of a binomial distribution $B(n, p)$ where $n = 6$ and $p = \frac{1}{3}$.
Since $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$,the distribution is $P(X=x) = \binom{6}{x} \left(\frac{1}{3}\right)^{x} \left(\frac{2}{3}\right)^{6-x}$.
The mean $\mu = np = 6 \times \frac{1}{3} = 2$.
The variance $\sigma^{2} = npq = 6 \times \frac{1}{3} \times \frac{2}{3} = \frac{12}{9} = \frac{4}{3}$.
Now,we calculate $2\mu + 12\sigma^{2} = 2(2) + 12\left(\frac{4}{3}\right) = 4 + 16 = 20$.

(A) Given $X \sim B\left(8, \frac{1}{2}\right)$,we have $n=8, p=\frac{1}{2}, q=\frac{1}{2}$.
We need to find $P(|X-4| \leq 2)$.
This inequality $|X-4| \leq 2$ implies $-2 \leq X-4 \leq 2$,which simplifies to $2 \leq X \leq 6$.
Thus,we need to calculate $P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6)$.
The probability mass function is $P(X=k) = {}^{8}C_{k} \left(\frac{1}{2}\right)^{k} \left(\frac{1}{2}\right)^{8-k} = {}^{8}C_{k} \left(\frac{1}{2}\right)^{8}$.
Summing these probabilities:
$P(2 \leq X \leq 6) = \left(\frac{1}{2}\right)^{8} \left[ {}^{8}C_{2} + {}^{8}C_{3} + {}^{8}C_{4} + {}^{8}C_{5} + {}^{8}C_{6} \right]$.
Calculating the combinations:
${}^{8}C_{2} = 28, {}^{8}C_{3} = 56, {}^{8}C_{4} = 70, {}^{8}C_{5} = 56, {}^{8}C_{6} = 28$.
Sum $= 28 + 56 + 70 + 56 + 28 = 238$.
Therefore,$P(2 \leq X \leq 6) = \frac{238}{256} = \frac{119}{128}$.

(A) We have $p = \frac{3}{4}$,so $q = 1 - p = \frac{1}{4}$.
It is given that $P(X \ge 1) \ge 0.9375$.
Using the complement rule,$P(X \ge 1) = 1 - P(X = 0)$.
Thus,$1 - P(X = 0) \ge 0.9375$.
$1 - ^nC_0 (p^0) (q)^n \ge 0.9375$.
$1 - (\frac{1}{4})^n \ge 0.9375$.
$1 - 0.9375 \ge (\frac{1}{4})^n$.
$0.0625 \ge (\frac{1}{4})^n$.
Since $0.0625 = \frac{625}{10000} = \frac{1}{16}$,we have $\frac{1}{16} \ge (\frac{1}{4})^n$.
$(\frac{1}{4})^2 \ge (\frac{1}{4})^n$.
Since the base is less than $1$,the inequality reverses for the exponents: $n \ge 2$.
The least value of $n$ is $2$.

(B) Key Idea: Use $p + q = 1$ and $Var(X) = E(X^2) - (E(X))^2$.
Given that the standard deviation of $X$ is $\sqrt{3pq}$,the variance is $Var(X) = (\sqrt{3pq})^2 = 3pq$.
Given that the mean $E(X) = 3p$.
Using the formula $Var(X) = E(X^2) - (E(X))^2$,we have:
$3pq = E(X^2) - (3p)^2$
$E(X^2) = 3pq + 9p^2$.
Since $p + q = 1$,we can substitute $q = 1 - p$:
$E(X^2) = 3p(1 - p) + 9p^2$
$E(X^2) = 3p - 3p^2 + 9p^2$
$E(X^2) = 3p + 6p^2$
$E(X^2) = 3p(1 + 2p)$.