Probability distribution Questions in English

(C) Given the probability distribution:
Mean $m = \sum p_i x_i = (0 \times \frac{1}{3}) + (1 \times \frac{1}{2}) + (2 \times 0) + (3 \times \frac{1}{6})$
$m = 0 + \frac{1}{2} + 0 + \frac{1}{2} = 1$
Variance $\sigma^2 = \sum p_i (x_i - m)^2$
$\sigma^2 = \frac{1}{3}(0 - 1)^2 + \frac{1}{2}(1 - 1)^2 + 0(2 - 1)^2 + \frac{1}{6}(3 - 1)^2$
$\sigma^2 = \frac{1}{3}(1) + \frac{1}{2}(0) + 0 + \frac{1}{6}(4)$
$\sigma^2 = \frac{1}{3} + 0 + 0 + \frac{4}{6} = \frac{1}{3} + \frac{2}{3} = 1$
Thus,$m = 1$ and $\sigma^2 = 1$,so $m = \sigma^2 = 1$.

(D) Let $X$ be the sum of the numbers on the two dice. The possible values for $X$ are $2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$.
The probability distribution is as follows:
$P(X=2) = \frac{1}{36}, P(X=3) = \frac{2}{36}, P(X=4) = \frac{3}{36}, P(X=5) = \frac{4}{36}, P(X=6) = \frac{5}{36}, P(X=7) = \frac{6}{36}, P(X=8) = \frac{5}{36}, P(X=9) = \frac{4}{36}, P(X=10) = \frac{3}{36}, P(X=11) = \frac{2}{36}, P(X=12) = \frac{1}{36}$.
The mean $E(X)$ is given by $\sum X_i P(X_i)$:
$E(X) = 2(\frac{1}{36}) + 3(\frac{2}{36}) + 4(\frac{3}{36}) + 5(\frac{4}{36}) + 6(\frac{5}{36}) + 7(\frac{6}{36}) + 8(\frac{5}{36}) + 9(\frac{4}{36}) + 10(\frac{3}{36}) + 11(\frac{2}{36}) + 12(\frac{1}{36})$
$E(X) = \frac{2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12}{36} = \frac{252}{36} = 7$.

(B) Given $n = 2000$ and $p = 0.001$.
Using the Poisson distribution,the parameter $\lambda = np = 2000 \times 0.001 = 2$.
The probability of $X$ individuals suffering a reaction is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.
We need to find $P(X > 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]$.
Calculating the individual probabilities:
$P(X=0) = \frac{e^{-2} 2^0}{0!} = e^{-2}$
$P(X=1) = \frac{e^{-2} 2^1}{1!} = 2e^{-2}$
$P(X=2) = \frac{e^{-2} 2^2}{2!} = 2e^{-2}$
Summing these: $P(X \le 2) = e^{-2} + 2e^{-2} + 2e^{-2} = 5e^{-2}$.
Therefore,$P(X > 2) = 1 - 5e^{-2} = 1 - \frac{5}{e^2}$.

(B) The number of floors available to leave,excluding the ground floor,is $6$.
Each of the $5$ persons can choose any of the $6$ floors independently.
Therefore,the total number of ways the $5$ persons can leave is $6^5$.
The number of ways the $5$ persons can leave at $5$ different floors is given by the permutation formula $^6P_5$.
$^6P_5 = \frac{6!}{(6-5)!} = 6 \times 5 \times 4 \times 3 \times 2 = 720$.
The total number of outcomes is $6^5 = 7776$.
The probability is $\frac{^6P_5}{6^5} = \frac{720}{7776}$.
Simplifying the fraction: $\frac{720}{7776} = \frac{5}{54}$.

(C) The average number of errors per page is given by $\lambda_{page} = \frac{250}{500} = 0.5$.
For a sample of $n = 2$ pages,the parameter for the Poisson distribution becomes $\lambda = n \times \lambda_{page} = 2 \times 0.5 = 1$.
The probability of having $X$ errors in the sample is given by the Poisson formula $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.
For no errors,we set $k = 0$:
$P(X=0) = \frac{e^{-1} \times 1^0}{0!} = \frac{e^{-1} \times 1}{1} = e^{-1}$.

(C) Let $H$ denote a head and $T$ denote a tail. For each $H$,the gain is $+2$ points,and for each $T$,the loss is $-1$ point.
When three coins are tossed,the possible outcomes for the number of heads $(n_H)$ and tails $(n_T)$ are as follows:
$1$. Three tails $(0H, 3T)$: Score $X = 0(2) + 3(-1) = -3$.
$2$. Two tails and one head $(1H, 2T)$: Score $X = 1(2) + 2(-1) = 2 - 2 = 0$.
$3$. One tail and two heads $(2H, 1T)$: Score $X = 2(2) + 1(-1) = 4 - 1 = 3$.
$4$. Three heads $(3H, 0T)$: Score $X = 3(2) + 0(-1) = 6$.
Thus,the possible values for the total score $X$ are $\{-3, 0, 3, 6\}$.
Therefore,the range of $X$ is $\{-3, 0, 3, 6\}$.

(A) The number of customers visiting the shop follows a Poisson distribution with parameter $\lambda = 4$.
The probability mass function is given by $P(X=x) = \frac{\lambda^x e^{-\lambda}}{x!}$.
We need to find the probability that more than $2$ customers visit,which is $P(X > 2)$.
$P(X > 2) = 1 - P(X \leq 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]$.
Substituting the values:
$P(X=0) = \frac{4^0 e^{-4}}{0!} = e^{-4}$.
$P(X=1) = \frac{4^1 e^{-4}}{1!} = 4e^{-4}$.
$P(X=2) = \frac{4^2 e^{-4}}{2!} = \frac{16e^{-4}}{2} = 8e^{-4}$.
Therefore,$P(X > 2) = 1 - [e^{-4} + 4e^{-4} + 8e^{-4}] = 1 - 13e^{-4}$.
This can be written as $1 - \frac{13}{e^4} = \frac{e^4 - 13}{e^4}$.

(C) For a probability distribution,the sum of all probabilities must be equal to $1$,i.e.,$\sum_{x=1}^{\infty} P(X=x) = 1$.
Given $P(X=x) = c\left(\frac{2}{3}\right)^x$.
Substituting the values of $x$: $c\left(\frac{2}{3}\right) + c\left(\frac{2}{3}\right)^2 + c\left(\frac{2}{3}\right)^3 + \ldots = 1$.
Taking $c$ as a common factor: $c \left[ \frac{2}{3} + \left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^3 + \ldots \right] = 1$.
The expression inside the bracket is an infinite geometric series with first term $a = \frac{2}{3}$ and common ratio $r = \frac{2}{3}$.
The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$.
So,$c \left[ \frac{2/3}{1 - 2/3} \right] = 1$.
$c \left[ \frac{2/3}{1/3} \right] = 1$.
$c(2) = 1$.
Therefore,$c = \frac{1}{2}$.

(A) The average arrival rate,$\lambda$,is $240 \text{ vehicles/hour} = \frac{240}{3600} \text{ vehicles/second} = \frac{1}{15} \text{ vehicles/second}$.
For a time interval $t = 30 \text{ seconds}$,the expected number of arrivals is $\mu = \lambda t = \frac{1}{15} \times 30 = 2$.
According to the Poisson distribution,the probability of $n$ arrivals is $P(n) = \frac{\mu^n e^{-\mu}}{n!}$.
We need to find the probability of more than two vehicles arriving,i.e.,$P(n > 2) = 1 - [P(0) + P(1) + P(2)]$.
Calculating individual probabilities:
$P(0) = \frac{2^0 e^{-2}}{0!} = e^{-2}$
$P(1) = \frac{2^1 e^{-2}}{1!} = 2e^{-2}$
$P(2) = \frac{2^2 e^{-2}}{2!} = \frac{4e^{-2}}{2} = 2e^{-2}$
Summing these probabilities: $P(n \leq 2) = e^{-2} + 2e^{-2} + 2e^{-2} = 5e^{-2}$.
Therefore,$P(n > 2) = 1 - 5e^{-2} = 1 - \frac{5}{e^2} = \frac{e^2 - 5}{e^2}$.

(A) Given the probability mass function $P(X=n) = \frac{k(n+1)}{3^n}$ for $n \in \{0, 1, 2, \dots\}$.
Since the sum of all probabilities must be $1$,we have $\sum_{n=0}^{\infty} P(X=n) = 1$.
$k \sum_{n=0}^{\infty} \frac{n+1}{3^n} = k \left( \frac{1}{3^0} + \frac{2}{3^1} + \frac{3}{3^2} + \frac{4}{3^3} + \dots \right) = 1$.
This is an arithmetico-geometric series with $a=1$,$d=1$,and $r=\frac{1}{3}$.
The sum of the series is $S = \frac{a}{1-r} + \frac{dr}{(1-r)^2} = \frac{1}{1-1/3} + \frac{1 \cdot (1/3)}{(1-1/3)^2} = \frac{3}{2} + \frac{1/3}{4/9} = \frac{3}{2} + \frac{3}{4} = \frac{9}{4}$.
Thus,$k \cdot \frac{9}{4} = 1 \implies k = \frac{4}{9}$.
We need to find $P(X < 2) = P(X=0) + P(X=1)$.
$P(X=0) = k \cdot \frac{0+1}{3^0} = k \cdot 1 = \frac{4}{9}$.
$P(X=1) = k \cdot \frac{1+1}{3^1} = k \cdot \frac{2}{3} = \frac{4}{9} \cdot \frac{2}{3} = \frac{8}{27}$.
$P(X < 2) = \frac{4}{9} + \frac{8}{27} = \frac{12+8}{27} = \frac{20}{27}$.

(A) The probability mass function of a Poisson distribution is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.
Given $P(X=3) = P(X=5)$,we have $\frac{e^{-\lambda} \lambda^3}{3!} = \frac{e^{-\lambda} \lambda^5}{5!}$.
Canceling $e^{-\lambda}$ and $\lambda^3$ from both sides,we get $\frac{1}{6} = \frac{\lambda^2}{120}$.
Thus,$\lambda^2 = \frac{120}{6} = 20$,which implies $\lambda = \sqrt{20}$.
Now,we need to find $P(X=4) = \frac{e^{-\lambda} \lambda^4}{4!}$.
Substituting $\lambda = \sqrt{20}$ and $\lambda^2 = 20$,we get $P(X=4) = \frac{e^{-\sqrt{20}} (20)^2}{24} = \frac{400}{24 e^{\sqrt{20}}} = \frac{50}{3 e^{\sqrt{20}}}$.

(B) For a Poisson distribution with parameter $\lambda$,the probability mass function is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.
Given $\frac{P(X=5)}{P(X=2)} = \frac{1}{7500}$,we have $\frac{\frac{e^{-\lambda} \lambda^5}{5!}}{\frac{e^{-\lambda} \lambda^2}{2!}} = \frac{\lambda^3}{5 \times 4 \times 3} = \frac{\lambda^3}{60} = \frac{1}{7500}$.
Thus,$\lambda^3 = \frac{60}{7500} = \frac{6}{750} = \frac{1}{125}$.
Taking the cube root,$\lambda = \frac{1}{5}$.
Checking with the second condition: $\frac{P(X=5)}{P(X=3)} = \frac{\frac{e^{-\lambda} \lambda^5}{5!}}{\frac{e^{-\lambda} \lambda^3}{3!}} = \frac{\lambda^2}{5 \times 4} = \frac{\lambda^2}{20}$.
Substituting $\lambda = \frac{1}{5}$,we get $\frac{(1/5)^2}{20} = \frac{1/25}{20} = \frac{1}{500}$.
This matches the given condition. Therefore,the mean of the distribution is $\lambda = \frac{1}{5}$.

(C) The number of errors follows a Poisson distribution with parameter $\lambda = n \times p_{error}$.
Given $n = 40$ pages and the rate of errors is $1$ per $10$ pages,the average number of errors is $\lambda = 40 \times \frac{1}{10} = 4$.
The probability of having $X$ errors is given by $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$.
We need the probability that the errors are at most $2$,which is $p = P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$.
$p = e^{-4} \left( \frac{4^0}{0!} + \frac{4^1}{1!} + \frac{4^2}{2!} \right) = e^{-4} (1 + 4 + 8) = 13 e^{-4}$.
We are asked to find $e^2 p$.
$e^2 p = e^2 \times 13 e^{-4} = 13 e^{-2}$.

(B) Let $X$ be the random variable representing the number of bad eggs drawn. The total number of eggs is $12$,and we draw $3$ eggs without replacement. The possible values for $X$ are $0, 1, 2$.
The probability distribution is calculated as follows:
$P(X=0) = \frac{{}^{10}C_3}{{}^{12}C_3} = \frac{120}{220} = \frac{12}{22}$
$P(X=1) = \frac{{}^{10}C_2 \times {}^{2}C_1}{{}^{12}C_3} = \frac{45 \times 2}{220} = \frac{90}{220} = \frac{9}{22}$
$P(X=2) = \frac{{}^{10}C_1 \times {}^{2}C_2}{{}^{12}C_3} = \frac{10 \times 1}{220} = \frac{10}{220} = \frac{1}{22}$
Now,we calculate the mean $E(X) = \sum x_i P_i$:
$E(X) = (0 \times \frac{12}{22}) + (1 \times \frac{9}{22}) + (2 \times \frac{1}{22}) = \frac{9+2}{22} = \frac{11}{22} = \frac{1}{2}$
Next,we calculate $E(X^2) = \sum x_i^2 P_i$:
$E(X^2) = (0^2 \times \frac{12}{22}) + (1^2 \times \frac{9}{22}) + (2^2 \times \frac{1}{22}) = \frac{9+4}{22} = \frac{13}{22}$
Finally,the variance is given by $Var(X) = E(X^2) - [E(X)]^2$:
$Var(X) = \frac{13}{22} - (\frac{1}{2})^2 = \frac{13}{22} - \frac{1}{4} = \frac{26-11}{44} = \frac{15}{44}$

(A) Given that the total number of messages $n = 500$ and the probability of a message being transmitted correctly is $0.98$.
Therefore,the probability of a message being transmitted incorrectly is $p = 1 - 0.98 = 0.02$.
Since $n$ is large and $p$ is small,we use the Poisson distribution with parameter $\lambda = np = 500 \times 0.02 = 10$.
The probability of $X$ incorrectly transmitted messages is given by $P(X=r) = \frac{\lambda^r e^{-\lambda}}{r!}$.
We need to find the probability that at most one message is transmitted incorrectly,i.e.,$P(X \leq 1) = P(X=0) + P(X=1)$.
$P(X=0) = \frac{10^0 e^{-10}}{0!} = e^{-10}$.
$P(X=1) = \frac{10^1 e^{-10}}{1!} = 10 e^{-10}$.
Thus,$P(X \leq 1) = e^{-10} + 10 e^{-10} = 11 e^{-10} = \frac{11}{e^{10}}$.

(B) Let $X$ be the number of calls made per hour. Since the average number of calls is $5$,$X$ follows a Poisson distribution with parameter $\lambda = 5$.
The cost of $X$ calls is $2X$. We want to find the probability that the cost exceeds $Rs. 4$,i.e.,$P(2X > 4) = P(X > 2)$.
$P(X > 2) = 1 - P(X \leq 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]$.
Using the Poisson formula $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$:
$P(X=0) = e^{-5} \frac{5^0}{0!} = e^{-5}$.
$P(X=1) = e^{-5} \frac{5^1}{1!} = 5e^{-5}$.
$P(X=2) = e^{-5} \frac{5^2}{2!} = \frac{25}{2} e^{-5}$.
Summing these: $P(X \leq 2) = e^{-5} (1 + 5 + 12.5) = 18.5 e^{-5} = \frac{37}{2} e^{-5}$.
Therefore,$P(X > 2) = 1 - \frac{37}{2 e^5} = \frac{2 e^5 - 37}{2 e^5}$.

(B) The probability of a bad reaction is $p = 0.01$ and the number of people is $n = 300$.
Since $n$ is large and $p$ is small,we use the Poisson distribution as an approximation to the binomial distribution.
The mean $\mu$ is given by $\mu = n \times p = 300 \times 0.01 = 3$.
The Poisson probability formula is $P(X = x) = \frac{e^{-\mu} \cdot \mu^x}{x!}$.
For $x = 2$,we have $P(X = 2) = \frac{e^{-3} \cdot 3^2}{2!} = \frac{9}{2 e^3}$.

(A) When $3$ coins are tossed,the sample space $S$ contains $2^3 = 8$ outcomes: $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.
Let $H$ be the number of heads and $T$ be the number of tails. The net gain $X$ is given by $X = 10T - 5H$. Since $H + T = 3$,we have $T = 3 - H$.
Thus,$X = 10(3 - H) - 5H = 30 - 15H$.
Calculating $X$ for each outcome:
- $HHH (H=3): X = 30 - 15(3) = -15$
- $HHT, HTH, THH (H=2): X = 30 - 15(2) = 0$
- $HTT, THT, TTH (H=1): X = 30 - 15(1) = 15$
- $TTT (H=0): X = 30 - 15(0) = 30$
The probability distribution is:

$x$	$-15$	$0$	$15$	$30$
$P(x)$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$

The mean $E(X) = \Sigma x P(x) = (-15 \times \frac{1}{8}) + (0 \times \frac{3}{8}) + (15 \times \frac{3}{8}) + (30 \times \frac{1}{8})$
$E(X) = \frac{-15 + 0 + 45 + 30}{8} = \frac{60}{8} = \frac{15}{2}$.

(D) Let $X_1, X_2, X_3$ be the random variables representing the numbers appearing on the three dice respectively.
Each $X_i$ can take values from the set $\{1, 2, 3, 4, 5, 6\}$ with equal probability $P(X_i = k) = \frac{1}{6}$ for $k \in \{1, 2, 3, 4, 5, 6\}$.
The mean (expected value) of a single die is $E[X_i] = \sum_{k=1}^{6} k \cdot P(X_i = k) = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5$.
Let $S$ be the sum of the numbers appearing on the three dice,so $S = X_1 + X_2 + X_3$.
By the linearity of expectation,the mean of the sum is $E[S] = E[X_1 + X_2 + X_3] = E[X_1] + E[X_2] + E[X_3]$.
Substituting the mean of a single die,we get $E[S] = 3.5 + 3.5 + 3.5 = 10.5$.
Therefore,the mean of the sum of the numbers is $10.5$.

(D) When two dice are thrown,the total number of outcomes is $6 \times 6 = 36$.
The sum of the numbers on the two dice can be even or odd.
The number of outcomes where the sum is even is $18$,and the number of outcomes where the sum is odd is $18$.
Thus,$P(\text{sum is even}) = \frac{18}{36} = \frac{1}{2}$ and $P(\text{sum is odd}) = \frac{18}{36} = \frac{1}{2}$.
Let $X$ be the random variable representing the points obtained by $A$.
If the sum is even,$X = \frac{1}{2}$ with probability $\frac{1}{2}$.
If the sum is odd,$X = 1$ with probability $\frac{1}{2}$.
The arithmetic mean (expected value) of $X$ is $E(X) = \sum x_i p_i = (\frac{1}{2} \times \frac{1}{2}) + (1 \times \frac{1}{2}) = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$.

(B) For a probability distribution,the sum of all probabilities must be $1$: $\sum_{k=0}^{\infty} P(X=k) = 1$.
Substituting the given expression: $c \sum_{k=0}^{\infty} \frac{2k+3}{3^k} = 1$.
Let $S = \sum_{k=0}^{\infty} \frac{2k+3}{3^k} = 3 \sum_{k=0}^{\infty} \frac{1}{3^k} + 2 \sum_{k=0}^{\infty} \frac{k}{3^k}$.
The first part is a geometric series: $3 \times \frac{1}{1 - 1/3} = 3 \times \frac{3}{2} = \frac{9}{2}$.
The second part is an arithmetico-geometric series: $\sum_{k=0}^{\infty} k x^k = \frac{x}{(1-x)^2}$. For $x = 1/3$,this is $\frac{1/3}{(1-1/3)^2} = \frac{1/3}{4/9} = \frac{3}{4}$.
So,$S = \frac{9}{2} + 2 \times \frac{3}{4} = \frac{9}{2} + \frac{3}{2} = 6$.
Since $c \times S = 1$,we have $c = 1/6$.
Now,$P(X=3) = \frac{(2(3)+3)c}{3^3} = \frac{9c}{27} = \frac{c}{3}$.
Substituting $c = 1/6$,we get $P(X=3) = \frac{1/6}{3} = \frac{1}{18}$.

(C) Step $1$: Find the value of $k$. The sum of probabilities must be $1$.
$3k + k + k + 2k + k = 1 \implies 8k = 1 \implies k = \frac{1}{8}$.
Step $2$: Calculate the mean $E(X) = \sum x_i P(X=x_i)$.
$E(X) = 2(3k) + 3(k) + 5(k) + 7(2k) + 12(k) = 6k + 3k + 5k + 14k + 12k = 40k$.
Since $k = \frac{1}{8}$,$E(X) = 40 \times \frac{1}{8} = 5$.
Step $3$: Calculate $E(X^2) = \sum x_i^2 P(X=x_i)$.
$E(X^2) = 2^2(3k) + 3^2(k) + 5^2(k) + 7^2(2k) + 12^2(k) = 12k + 9k + 25k + 98k + 144k = 288k$.
Since $k = \frac{1}{8}$,$E(X^2) = 288 \times \frac{1}{8} = 36$.
Step $4$: Calculate the variance $Var(X) = E(X^2) - [E(X)]^2$.
$Var(X) = 36 - (5)^2 = 36 - 25 = 11$.
Step $5$: Calculate the standard deviation $\sigma = \sqrt{Var(X)} = \sqrt{11}$.

(A) For a discrete random variable,the sum of all probabilities must be equal to $1$.
Therefore,$P(X=1) + P(X=2) + P(X=3) = 1$.
Substituting the given values: $3k^3 + 2k^2 + (7 - 19k) = 1$.
$3k^3 + 2k^2 - 19k + 6 = 0$.
By testing values,if $k = 3$,$3(27) + 2(9) - 19(3) + 6 = 81 + 18 - 57 + 6 = 48 \neq 0$.
If $k = \frac{1}{3}$,$3(\frac{1}{27}) + 2(\frac{1}{9}) - 19(\frac{1}{3}) + 6 = \frac{1}{9} + \frac{2}{9} - \frac{57}{9} + \frac{54}{9} = 0$.
So,$k = \frac{1}{3}$ is a root.
Now,$P(X=3) = 7 - 19k = 7 - 19(\frac{1}{3}) = 7 - \frac{19}{3} = \frac{21 - 19}{3} = \frac{2}{3}$.

(B) We know that the sum of all probabilities in a distribution is $1$,i.e.,$\Sigma P(X = x_i) = 1$.
$\therefore 2k^2 + k + k + k^2 = 1$
$\Rightarrow 3k^2 + 2k - 1 = 0$
Factoring the quadratic equation: $(3k - 1)(k + 1) = 0$.
This gives $k = \frac{1}{3}$ or $k = -1$.
Since the probability $P(X = x_i)$ must be non-negative,$k = \frac{1}{3}$ is the only valid solution.
The mean of $X$ is given by $E(X) = \Sigma x_i P(X = x_i)$.
$E(X) = (1 \times 2k^2) + (2 \times k) + (3 \times k) + (5 \times k^2)$
$E(X) = 2k^2 + 2k + 3k + 5k^2 = 7k^2 + 5k$
Substituting $k = \frac{1}{3}$:
$E(X) = 7(\frac{1}{3})^2 + 5(\frac{1}{3}) = 7(\frac{1}{9}) + \frac{5}{3} = \frac{7}{9} + \frac{15}{9} = \frac{22}{9}$.

(D) We know that the sum of probabilities in a distribution is $1$.
$\Sigma P(X=x) = K + 2K + 3K^2 + K^2 = 1$
$4K^2 + 3K - 1 = 0$
$(4K - 1)(K + 1) = 0$
Since $P(X=x) \geq 0$,we reject $K = -1$. Thus,$K = \frac{1}{4}$.
The distribution is:
$\begin{array}{|c|c|c|c|c|}\hline X=x & 2 & 3 & 5 & 9 \\\hline P(X=x) & \frac{1}{4} & \frac{1}{2} & \frac{3}{16} & \frac{1}{16} \\\hline\end{array}$
$E(X) = \Sigma x P(x) = (2 \times \frac{1}{4}) + (3 \times \frac{1}{2}) + (5 \times \frac{3}{16}) + (9 \times \frac{1}{16}) = \frac{8}{16} + \frac{24}{16} + \frac{15}{16} + \frac{9}{16} = \frac{56}{16} = \frac{7}{2}$
$E(X^2) = \Sigma x^2 P(x) = (4 \times \frac{1}{4}) + (9 \times \frac{1}{2}) + (25 \times \frac{3}{16}) + (81 \times \frac{1}{16}) = 1 + \frac{9}{2} + \frac{75}{16} + \frac{81}{16} = \frac{16 + 72 + 75 + 81}{16} = \frac{244}{16} = \frac{61}{4}$
$\text{Variance} = E(X^2) - [E(X)]^2 = \frac{61}{4} - (\frac{7}{2})^2 = \frac{61}{4} - \frac{49}{4} = \frac{12}{4} = 3$.

(D) The sum of probabilities in a probability distribution is $1$.
$3K^2 + K + K^2 + 2K = 1$
$4K^2 + 3K - 1 = 0$
$(4K - 1)(K + 1) = 0$
Since $P(X=x) \geq 0$,$K$ must be positive,so $K = \frac{1}{4}$.
The distribution is:

$X=x$	$1$	$3$	$5$	$2$
$P(X=x)$	$\frac{3}{16}$	$\frac{1}{4}$	$\frac{1}{16}$	$\frac{1}{2}$

The mean $\mu = E(X) = \sum x_i P(x_i) = 1(\frac{3}{16}) + 3(\frac{1}{4}) + 5(\frac{1}{16}) + 2(\frac{1}{2}) = \frac{3}{16} + \frac{12}{16} + \frac{5}{16} + \frac{16}{16} = \frac{36}{16} = \frac{9}{4}$.
$E(X^2) = \sum x_i^2 P(x_i) = 1^2(\frac{3}{16}) + 3^2(\frac{1}{4}) + 5^2(\frac{1}{16}) + 2^2(\frac{1}{2}) = \frac{3}{16} + \frac{36}{16} + \frac{25}{16} + \frac{32}{16} = \frac{96}{16} = 6$.
Variance $\sigma^2 = E(X^2) - [E(X)]^2 = 6 - (\frac{9}{4})^2 = 6 - \frac{81}{16} = \frac{96 - 81}{16} = \frac{15}{16}$.

(C) The probability mass function of a Poisson distribution is given by $P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$.
Given the condition $3 P(X=2) = P(X=4)$.
Substituting the formula,we get $3 \cdot \frac{e^{-\lambda} \lambda^2}{2!} = \frac{e^{-\lambda} \lambda^4}{4!}$.
Canceling $e^{-\lambda}$ from both sides and simplifying the factorials: $\frac{3 \lambda^2}{2} = \frac{\lambda^4}{24}$.
Multiplying both sides by $24$,we get $36 \lambda^2 = \lambda^4$.
Since $\lambda > 0$,we have $\lambda^2 = 36$,which implies $\lambda = 6$.
Now,we need to find $P(X=6) = \frac{e^{-6} \cdot 6^6}{6!}$.
Calculating the value: $P(X=6) = \frac{e^{-6} \cdot 46656}{720} = \frac{46656}{720 e^6} = \frac{324}{5 e^6}$.

(C) The sum of all probabilities in a distribution must be $1$:
$\Sigma P(X = x) = 0.15 + 0.23 + k + 0.10 + 0.20 + 0.08 + 0.07 + 0.05 = 1$
$0.88 + k = 1$
$k = 0.12$
The event $E$ consists of prime numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$,so $E = \{2, 3, 5, 7\}$.
The event $F$ consists of values less than $4$,so $F = \{1, 2, 3\}$.
The union $E \cup F = \{1, 2, 3, 5, 7\}$.
The probability $P(E \cup F) = P(X=1) + P(X=2) + P(X=3) + P(X=5) + P(X=7)$
$P(E \cup F) = 0.15 + 0.23 + 0.12 + 0.20 + 0.07 = 0.77$.

(C) For a probability distribution,the sum of all probabilities must be $1$.
$\sum P(X = x_{i}) = 0.1 + k + 0.2 + 2k + 3k + k = 1$
$7k + 0.3 = 1 \Rightarrow 7k = 0.7 \Rightarrow k = 0.1$.
Now,we calculate the mean $\mu = E(X) = \sum x_{i} P(x_{i})$:
$\mu = (-2)(0.1) + (-1)(0.1) + (0)(0.2) + (1)(0.2) + (2)(0.3) + (3)(0.1)$
$\mu = -0.2 - 0.1 + 0 + 0.2 + 0.6 + 0.3 = 0.8$.
Next,we calculate $E(X^2) = \sum x_{i}^2 P(x_{i})$:
$E(X^2) = (-2)^2(0.1) + (-1)^2(0.1) + (0)^2(0.2) + (1)^2(0.2) + (2)^2(0.3) + (3)^2(0.1)$
$E(X^2) = 4(0.1) + 1(0.1) + 0 + 1(0.2) + 4(0.3) + 9(0.1)$
$E(X^2) = 0.4 + 0.1 + 0 + 0.2 + 1.2 + 0.9 = 2.8$.
The variance is given by $Var(X) = E(X^2) - [E(X)]^2$.
$Var(X) = 2.8 - (0.8)^2 = 2.8 - 0.64 = 2.16$.

(C) The total number of balls is $5 + 3 = 8$. Two balls are drawn at random. The total number of ways to draw $2$ balls is ${}^8C_2 = \frac{8 \times 7}{2} = 28$.
Let $X$ be the random variable representing the number of white balls drawn. The possible values for $X$ are $0, 1, 2$.
$P(X = 0) = \frac{{}^5C_2}{{}^8C_2} = \frac{10}{28}$.
$P(X = 1) = \frac{{}^5C_1 \times {}^3C_1}{{}^8C_2} = \frac{5 \times 3}{28} = \frac{15}{28}$.
$P(X = 2) = \frac{{}^3C_2}{{}^8C_2} = \frac{3}{28}$.
The mean of $X$ is given by $E(X) = \sum x_i P(x_i) = 0 \times P(X = 0) + 1 \times P(X = 1) + 2 \times P(X = 2)$.
$E(X) = 0 \times \frac{10}{28} + 1 \times \frac{15}{28} + 2 \times \frac{3}{28} = 0 + \frac{15}{28} + \frac{6}{28} = \frac{21}{28} = \frac{3}{4}$.

(B) The sum of all probabilities in a probability distribution is $1$. Therefore,$\sum_{r=0}^{\infty} P(X=r) = 1$.
Given $P(X=r) = k(1+r) 3^{-r}$,we have $k \sum_{r=0}^{\infty} (1+r) \left(\frac{1}{3}\right)^r = 1$.
Let $S = \sum_{r=0}^{\infty} (1+r) x^r$ where $x = \frac{1}{3}$.
This is an arithmetico-geometric series: $S = 1 + 2x + 3x^2 + 4x^3 + \ldots$.
Multiplying by $x$: $xS = x + 2x^2 + 3x^3 + \ldots$.
Subtracting the two: $S(1-x) = 1 + x + x^2 + x^3 + \ldots = \frac{1}{1-x}$.
Thus,$S = \frac{1}{(1-x)^2}$.
For $x = \frac{1}{3}$,$S = \frac{1}{(1 - 1/3)^2} = \frac{1}{(2/3)^2} = \frac{9}{4}$.
So,$k \times \frac{9}{4} = 1 \implies k = \frac{4}{9}$.
Now,$P(X=0) + P(X=1) + P(X=2) = k \left[ (1+0)3^0 + (1+1)3^{-1} + (1+2)3^{-2} \right]$.
$= \frac{4}{9} \left[ 1 + \frac{2}{3} + \frac{3}{9} \right] = \frac{4}{9} \left[ 1 + \frac{2}{3} + \frac{1}{3} \right] = \frac{4}{9} \times 2 = \frac{8}{9}$.

(D) For a probability distribution,the sum of all probabilities must be equal to $1$:
$\sum P(X=x) = 0 + k + 2k + 5k^2 + 2k^2 + 3k = 1$
$7k^2 + 6k - 1 = 0$
$7k^2 + 7k - k - 1 = 0$
$7k(k + 1) - 1(k + 1) = 0$
$(k + 1)(7k - 1) = 0$
Since $k$ must be positive for probabilities,$k = \frac{1}{7}$.
The mean $E(X)$ is given by $\sum x_i P(X=x_i)$:
$E(X) = (0 \times 0) + (2 \times k) + (4 \times 2k) + (6 \times 5k^2) + (8 \times 2k^2) + (10 \times 3k)$
$E(X) = 0 + 2k + 8k + 30k^2 + 16k^2 + 30k$
$E(X) = 46k^2 + 40k$
Substituting $k = \frac{1}{7}$:
$E(X) = 46(\frac{1}{7})^2 + 40(\frac{1}{7})$
$E(X) = \frac{46}{49} + \frac{40}{7} = \frac{46 + 280}{49} = \frac{326}{49}$
Thus,the correct option is $(d)$.

(D) The probability mass function of a Poisson distribution is given by $P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$.
Given $P(X=2) = P(X=3)$,we have:
$\frac{e^{-\lambda} \lambda^2}{2!} = \frac{e^{-\lambda} \lambda^3}{3!}$
$\frac{1}{2} = \frac{\lambda}{6} \implies \lambda = 3$.
Now,we are given $\frac{5}{3} k = P(X=2) = \frac{e^{-3} 3^2}{2!} = \frac{9 e^{-3}}{2}$.
So,$\frac{5}{3} k = \frac{9 e^{-3}}{2} \implies e^{-3} = \frac{10}{27} k$.
We need to find $P(X=5) = \frac{e^{-3} 3^5}{5!} = \frac{e^{-3} \times 243}{120} = \frac{81}{40} e^{-3}$.
Substituting $e^{-3} = \frac{10}{27} k$:
$P(X=5) = \frac{81}{40} \times \frac{10}{27} k = \frac{3}{4} k$.

(A) For any probability distribution,the sum of all probabilities must be $1$.
$\sum P(X = x_i) = 10k + 9k + 8k + 8k + 6k + 5k + 4k + 3k + k = 54k = 1$.
Therefore,$k = \frac{1}{54}$.
The event $A$ consists of prime numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$,which are $\{2, 3, 5, 7\}$.
$P(A) = P(2) + P(3) + P(5) + P(7) = 9k + 8k + 6k + 4k = 27k$.
The event $B$ consists of values $x_i > 5$,which are $\{6, 7, 8, 9\}$.
$P(B) = P(6) + P(7) + P(8) + P(9) = 5k + 4k + 3k + k = 13k$.
The intersection $A \cap B$ consists of values that are both prime and greater than $5$,which is $\{7\}$.
$P(A \cap B) = P(7) = 4k$.
Using the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = 27k + 13k - 4k = 36k$.
Substituting $k = \frac{1}{54}$,we get $P(A \cup B) = 36 \times \frac{1}{54} = \frac{36}{54} = \frac{2}{3}$.

(A) The problem follows a Poisson distribution because the number of trials $n = 2000$ is large and the probability of success $p = 0.001$ is very small.
For a Poisson distribution,the parameter $\lambda$ is given by $\lambda = n \times p$.
$\lambda = 2000 \times 0.001 = 2$.
The probability mass function for a Poisson distribution is $P(X = x) = \frac{e^{-\lambda} \lambda^{x}}{x!}$.
We need to find the probability for exactly $x = 3$ individuals.
Substituting the values,we get $P(X = 3) = \frac{e^{-2} \times 2^{3}}{3!}$.
$P(X = 3) = \frac{e^{-2} \times 8}{6} = \frac{4}{3 e^{2}}$.

(C) For a Poisson distribution with mean '$m$',the probability mass function is given by $P(x = k) = \frac{m^k \cdot e^{-m}}{k!}$ for $k = 0, 1, 2, \dots$.
We need to find $P(x > 0)$.
Using the complement rule,$P(x > 0) = 1 - P(x = 0)$.
Substituting $k = 0$ into the formula,we get $P(x = 0) = \frac{m^0 \cdot e^{-m}}{0!} = \frac{1 \cdot e^{-m}}{1} = e^{-m}$.
Therefore,$P(x > 0) = 1 - e^{-m} = 1 - \frac{1}{e^m}$.
Simplifying this expression,we get $P(x > 0) = \frac{e^m - 1}{e^m}$.

(C) The sum of all probabilities for a random variable must be $1$.
$P(X = 0) + P(X = 1) + P(X = 2) = 1$
$3c^3 + (4c - 10c^2) + (5c - 1) = 1$
$3c^3 - 10c^2 + 9c - 2 = 0$
Factoring the cubic equation,we get $(c - 1)(c - 2)(3c - 1) = 0$.
Since $0 \le P(X) \le 1$,we test the values. If $c = 1$,$P(X = 2) = 5(1) - 1 = 4$,which is impossible. If $c = 2$,$P(X = 2) = 5(2) - 1 = 9$,which is impossible. Thus,$c = \frac{1}{3}$.
Now,$P(X = 0) = 3(\frac{1}{3})^3 = \frac{1}{9}$,$P(X = 1) = 4(\frac{1}{3}) - 10(\frac{1}{3})^2 = \frac{4}{3} - \frac{10}{9} = \frac{2}{9}$,and $P(X = 2) = 5(\frac{1}{3}) - 1 = \frac{2}{3}$.
We need to find $P(0 < X \le 2) = P(X = 1) + P(X = 2)$.
$P(0 < X \le 2) = \frac{2}{9} + \frac{2}{3} = \frac{2}{9} + \frac{6}{9} = \frac{8}{9}$.

(A) For a discrete probability distribution,the sum of all probabilities must be equal to $1$.
$\sum_{x=1}^{\infty} P(X = x) = 1$
$\sum_{x=1}^{\infty} 5r^x = 1$
$5(r + r^2 + r^3 + \dots) = 1$
This is an infinite geometric series with the first term $a = r$ and common ratio $r$. The sum is given by $\frac{a}{1-r}$ for $|r| < 1$.
$5 \left( \frac{r}{1 - r} \right) = 1$
$5r = 1 - r$
$6r = 1$
$r = \frac{1}{6}$

(D) For any probability distribution,the sum of all probabilities must be equal to $1$.
$\sum_{x=0}^{7} P(x) = 1$
$\Rightarrow 0.01 + 0.10 + 0.26 + 0.33 + 0.18 + 0.06 + K + 0.04 = 1$
$\Rightarrow 0.98 + K = 1$
$\Rightarrow K = 1 - 0.98 = 0.02$
Now,we need to calculate $P(X \geq 3) - P(X < 6)$.
$P(X \geq 3) = P(3) + P(4) + P(5) + P(6) + P(7) = 0.33 + 0.18 + 0.06 + 0.02 + 0.04 = 0.63$
$P(X < 6) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5) = 0.01 + 0.10 + 0.26 + 0.33 + 0.18 + 0.06 = 0.94$
Therefore,$P(X \geq 3) - P(X < 6) = 0.63 - 0.94 = -0.31$.

(A) The probability of a mask being defective is $p = \frac{1}{500} = 0.002$.
In a packet of $n = 100$ masks,the expected number of defective masks is $\lambda = np = 100 \times 0.002 = 0.2$.
Using Poisson distribution,the probability that a packet contains $r$ defective masks is $P(X = r) = \frac{e^{-\lambda} \lambda^r}{r!}$.
For a packet to contain no defective masks,we set $r = 0$:
$P(X = 0) = \frac{e^{-0.2} (0.2)^0}{0!} = e^{-0.2}$.
In a consignment of $10,000$ packets,the number of packets with no defective masks is $10,000 \times P(X = 0) = 10,000 \times e^{-0.2} = \frac{10,000}{e^{0.2}}$.

(D) Given the probability function $P(X=r) = K r^2$ for $r \in \{-2, -1, 0, 1, 2, 3\}$.
Since the sum of all probabilities must be $1$,we have:
$\sum P(X=r) = 1$
$K((-2)^2 + (-1)^2 + 0^2 + 1^2 + 2^2 + 3^2) = 1$
$K(4 + 1 + 0 + 1 + 4 + 9) = 1$
$19K = 1 \Rightarrow K = \frac{1}{19}$
We need to find the sum of the variance $\sigma^2$ and the square of the mean $\mu^2$. We know that $\sigma^2 = E(X^2) - \mu^2$,so $\sigma^2 + \mu^2 = E(X^2)$.
$E(X^2) = \sum r^2 P(X=r) = \sum r^2 (K r^2) = K \sum r^4$
$E(X^2) = K((-2)^4 + (-1)^4 + 0^4 + 1^4 + 2^4 + 3^4)$
$E(X^2) = K(16 + 1 + 0 + 1 + 16 + 81) = K(115)$
Substituting $K = \frac{1}{19}$,we get:
$E(X^2) = \frac{115}{19}$
Thus,$\sigma^2 + \mu^2 = \frac{115}{19}$.

(B) For a Poisson distribution,the probability mass function is given by $P(X = x) = \frac{e^{-\lambda} \cdot \lambda^x}{x!}$.
Given that $P(X = 2) = 2 \cdot P(X = 1)$.
Substituting the formula:
$\frac{e^{-\lambda} \cdot \lambda^2}{2!} = 2 \cdot \frac{e^{-\lambda} \cdot \lambda^1}{1!}$.
Dividing both sides by $e^{-\lambda} \cdot \lambda$ (assuming $\lambda \neq 0$):
$\frac{\lambda}{2} = 2 \cdot 1$.
$\lambda = 4$.
In a Poisson distribution,the variance is equal to the parameter $\lambda$,so $\sigma^2 = \lambda = 4$.
The standard deviation $\sigma$ is $\sqrt{\lambda} = \sqrt{4} = 2$.

(C) We know that the sum of all probabilities in a distribution must be equal to $1$.
$\sum P(X = x) = 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 + k = 1$
$10k^2 + 9k - 1 = 0$
$(10k - 1)(k + 1) = 0$
Since $k$ must be positive for probabilities to be valid,we have $k = \frac{1}{10}$.
We need to find $P(0 < X < 4) = P(X = 1) + P(X = 2) + P(X = 3)$.
$P(0 < X < 4) = k + 2k + 2k = 5k$.
Substituting $k = \frac{1}{10}$,we get $P(0 < X < 4) = 5 \times \frac{1}{10} = \frac{1}{2}$.

(B) The probability mass function of a Poisson distribution is given by $p(x) = \frac{e^{-\lambda} \lambda^x}{x!}$ for $x = 0, 1, 2, \dots$.
To find the value of $x$ for which $p(x)$ is maximum,we examine the ratio $\frac{p(x)}{p(x-1)}$.
$\frac{p(x)}{p(x-1)} = \frac{e^{-\lambda} \lambda^x / x!}{e^{-\lambda} \lambda^{x-1} / (x-1)!} = \frac{\lambda}{x}$.
For $p(x)$ to be maximum,we need $\frac{p(x)}{p(x-1)} \geq 1$ and $\frac{p(x+1)}{p(x)} \leq 1$.
This implies $\frac{\lambda}{x} \geq 1 \implies x \leq \lambda$ and $\frac{\lambda}{x+1} \leq 1 \implies x+1 \geq \lambda$.
Thus,the mode $x$ satisfies $\lambda - 1 \leq x \leq \lambda$.
Given $\lambda = 3.725$,we have $3.725 - 1 \leq x \leq 3.725$,which means $2.725 \leq x \leq 3.725$.
Since $x$ must be an integer,the value of $x$ that maximizes $p(x)$ is $3$.

(B) We know that the sum of all probabilities in a probability distribution is $1$,so $\sum P(X=x_i) = 1$.
$0.1 + 0.15 + 0.3 + 0.25 + k + k = 1$
$0.8 + 2k = 1 \implies 2k = 0.2 \implies k = 0.1$.
Now,we calculate the mean $E(X) = \sum x_i P(x_i)$ and $E(X^2) = \sum x_i^2 P(x_i)$:

$x_i$	$P(x_i)$	$x_i P(x_i)$	$x_i^2 P(x_i)$
$1$	$0.1$	$0.1$	$0.1$
$2$	$0.15$	$0.3$	$0.6$
$3$	$0.3$	$0.9$	$2.7$
$4$	$0.25$	$1.0$	$4.0$
$5$	$0.1$	$0.5$	$2.5$
$6$	$0.1$	$0.6$	$3.6$
Total	$1.0$	$3.4$	$13.5$

The variance is given by $Var(X) = E(X^2) - [E(X)]^2$.
$Var(X) = 13.5 - (3.4)^2 = 13.5 - 11.56 = 1.94$.

(C) For a Poisson distribution,the variance is equal to the parameter $\lambda$. Given that the variance is $3$,we have $\lambda = 3$.
The probability mass function is given by $P(X=r) = \frac{e^{-\lambda} \lambda^r}{r!}$.
For a Poisson distribution,the probability $P(X=r)$ is maximum when $r = \lfloor \lambda \rfloor$ if $\lambda$ is not an integer,and it takes two maximum values at $r = \lambda$ and $r = \lambda - 1$ if $\lambda$ is an integer.
Here,$\lambda = 3$,which is an integer.
Therefore,$P(X=r)$ is maximum at $r = 3$ and $r = 3 - 1 = 2$.
Comparing this with the given options,$r=2$ and $r=3$ are both valid,but since $r=2$ is provided in the options,it is a correct value.