Probability distribution Questions in English

(B) The expected value $E[X^2]$ is calculated using the formula $E[X^2] = \sum x_i^2 P(X=x_i)$.
Given the probability distribution:

$X$	$0$	$1$	$2$
$P(X)$	$0.2$	$0.5$	$0.3$

$E[X^2] = (0^2 \times 0.2) + (1^2 \times 0.5) + (2^2 \times 0.3)$
$E[X^2] = (0 \times 0.2) + (1 \times 0.5) + (4 \times 0.3)$
$E[X^2] = 0 + 0.5 + 1.2$
$E[X^2] = 1.7$

(B) We know that the sum of all probabilities in a probability distribution is $1$.
Thus,$\sum_{i=1}^{100} P(X=x_i) = 1$.
Substituting the given probability $P(X=x_i) = K i(i+1)$,we get:
$K \sum_{i=1}^{100} (i^2 + i) = 1$.
Using the summation formulas $\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$ for $n=100$:
$K \left[ \frac{100(101)(201)}{6} + \frac{100(101)}{2} \right] = 1$.
$K \left[ \frac{100 \times 101}{2} \left( \frac{201}{3} + 1 \right) \right] = 1$.
$K \left[ 5050 \times (67 + 1) \right] = 1$.
$K \times 5050 \times 68 = 1$.
$K = \frac{1}{5050 \times 68} = \frac{1}{343400}$.
Therefore,$200 K = \frac{200}{343400} = \frac{2}{3434} = \frac{1}{1717}$.
Hence,option $(B)$ is correct.

(A) Given the probability distribution:
$\sum P(X) = 1$
$K + 2K + K^2 + 2K^2 + 5K^2 + K + K + 2K = 1$
$8K^2 + 7K - 1 = 0$
$(8K - 1)(K + 1) = 0$
Since $K > 0$,we have $K = \frac{1}{8}$.
The events are $E = \{2, 3, 5, 7\}$ and $F = \{1, 2, 3\}$.
$E \cup F = \{1, 2, 3, 5, 7\}$.
$P(E \cup F) = P(1) + P(2) + P(3) + P(5) + P(7)$
$P(E \cup F) = K + 2K + K^2 + 5K^2 + K = 6K^2 + 4K$
Substituting $K = \frac{1}{8}$:
$P(E \cup F) = 6(\frac{1}{64}) + 4(\frac{1}{8}) = \frac{6}{64} + \frac{32}{64} = \frac{38}{64}$.

(A) The number of accidents follows a Poisson distribution with parameter $\lambda = 3$.
The probability mass function of a Poisson random variable $X$ is given by $P(X = x) = \frac{\lambda^x e^{-\lambda}}{x!}$,where $x = 0, 1, 2, \dots$.
We want to find the probability that no accidents occur today,which corresponds to $P(X = 0)$.
Substituting $\lambda = 3$ and $x = 0$ into the formula:
$P(X = 0) = \frac{3^0 e^{-3}}{0!}$.
Since $3^0 = 1$ and $0! = 1$,we get:
$P(X = 0) = \frac{1 \times e^{-3}}{1} = e^{-3} = \frac{1}{e^3}$.
Thus,the probability that no accidents occur today is $\frac{1}{e^3}$.

(C) For a probability distribution,the sum of all probabilities must be equal to $1$.
Therefore,$0.1 + k + 0.2 + 2k + 0.3 + k = 1$.
Combining the terms,we get $0.6 + 4k = 1$.
$4k = 0.4$,which implies $k = 0.1$.
The mean of $X$,denoted by $E(X)$,is given by $\sum x_i P(x_i)$.
$E(X) = (-2 \times 0.1) + (-1 \times 0.1) + (0 \times 0.2) + (1 \times 0.2) + (2 \times 0.3) + (3 \times 0.1)$.
$E(X) = -0.2 - 0.1 + 0 + 0.2 + 0.6 + 0.3$.
$E(X) = 0.8$.

(D) For a probability distribution,the sum of all probabilities must be equal to $1$.
Therefore,$\sum P(X=x) = 1$.
$0.1 + k + 0.2 + 2k + 0.3 + k = 1$
Combine the terms involving $k$ and the constant terms:
$(0.1 + 0.2 + 0.3) + (k + 2k + k) = 1$
$0.6 + 4k = 1$
$4k = 1 - 0.6$
$4k = 0.4$
$k = \frac{0.4}{4}$
$k = 0.1$

(C) The given probability distribution for the random variable $X$ is:
$\begin{array}{|c|c|c|c|c|c|c|}\hline X=x_i & 1 & 2 & 3 & 4 & 5 & 6 \\\hline P(X=x_i) & 0.2 & 0.3 & 0.12 & 0.1 & 0.2 & 0.08 \\\hline \end{array}$
We define the events $A$ and $B$ as follows:
$A = \{x_i \mid x_i \text{ is a prime number}\} = \{2, 3, 5\}$
$B = \{x_i \mid x_i < 4\} = \{1, 2, 3\}$
The union of these two events is $A \cup B = \{1, 2, 3, 5\}$.
Therefore,the probability $P(A \cup B)$ is the sum of the probabilities of these individual outcomes:
$P(A \cup B) = P(X=1) + P(X=2) + P(X=3) + P(X=5)$
$P(A \cup B) = 0.2 + 0.3 + 0.12 + 0.2 = 0.82$
Thus,the correct option is $C$.

(B) Given the probability function $P(X=k)=c k^2$,where $c$ is a constant and $k \in\{0,1,2,3,4\}$.
Since the sum of probabilities must be $1$,we have $\sum_{k=0}^{4} P(X=k) = 1$.
$c(0^2) + c(1^2) + c(2^2) + c(3^2) + c(4^2) = 1$
$c(0 + 1 + 4 + 9 + 16) = 1$
$30c = 1 \implies c = \frac{1}{30}$.
We know that the variance $\sigma^2 = E(X^2) - \mu^2$,where $\mu = E(X)$.
Therefore,$\sigma^2 + \mu^2 = E(X^2)$.
$E(X^2) = \sum_{k=0}^{4} k^2 P(X=k) = \sum_{k=0}^{4} k^2 (c k^2) = c \sum_{k=0}^{4} k^4$.
$E(X^2) = c(0^4 + 1^4 + 2^4 + 3^4 + 4^4) = c(0 + 1 + 16 + 81 + 256) = 354c$.
Substituting $c = \frac{1}{30}$,we get:
$\sigma^2 + \mu^2 = 354 \times \frac{1}{30} = \frac{354}{30} = 11.8$.
Thus,option $(B)$ is correct.

(A) Given that the total number of pages is $250$ and the total number of errors is $200$.
The average number of errors per page,denoted by $\lambda$,is calculated as:
$\lambda = \frac{200}{250} = \frac{4}{5} = 0.8$.
The Poisson probability distribution is given by $P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$.
The probability that a single page contains no error $(x=0)$ is:
$P(X=0) = \frac{e^{-0.8} (0.8)^0}{0!} = e^{-0.8} = e^{-4/5}$.
For a random sample of $5$ pages,the probability that none of them contain any error is:
$P = (P(X=0))^5 = (e^{-4/5})^5 = e^{-4}$.
Thus,the correct option is $A$.

(D) Given,the probability distribution of $X$ is defined by $P(X=0)=3C^3$,$P(X=2)=5C-10C^2$,and $P(X=4)=4C-1$.
We know that the sum of all probabilities in a distribution must be $1$,so $\Sigma P(X)=1$.
$3C^3 + (5C-10C^2) + (4C-1) = 1$
$3C^3 - 10C^2 + 9C - 2 = 0$
Factoring the cubic equation: $(C-1)(3C^2-7C+2) = 0$
$(C-1)(3C-1)(C-2) = 0$
This gives $C = 1, \frac{1}{3}, 2$.
Since probabilities must be between $0$ and $1$,we test the values. If $C=1$,$P(X=4)=4(1)-1=3 > 1$ (impossible). If $C=2$,$P(X=4)=4(2)-1=7 > 1$ (impossible). Thus,$C=\frac{1}{3}$.
Substituting $C=\frac{1}{3}$ into the probabilities:
$P(X=0) = 3(\frac{1}{3})^3 = \frac{3}{27} = \frac{1}{9}$
$P(X=2) = 5(\frac{1}{3}) - 10(\frac{1}{3})^2 = \frac{5}{3} - \frac{10}{9} = \frac{15-10}{9} = \frac{5}{9}$
$P(X=4) = 4(\frac{1}{3}) - 1 = \frac{4}{3} - 1 = \frac{1}{3}$
The mean $E(X) = \Sigma X P(X) = (0 \times \frac{1}{9}) + (2 \times \frac{5}{9}) + (4 \times \frac{1}{3}) = 0 + \frac{10}{9} + \frac{4}{3} = \frac{10+12}{9} = \frac{22}{9}$.
The mean of squares $E(X^2) = \Sigma X^2 P(X) = (0^2 \times \frac{1}{9}) + (2^2 \times \frac{5}{9}) + (4^2 \times \frac{1}{3}) = 0 + \frac{20}{9} + \frac{16}{3} = \frac{20+48}{9} = \frac{68}{9}$.
Variance $Var(X) = E(X^2) - [E(X)]^2 = \frac{68}{9} - (\frac{22}{9})^2 = \frac{68}{9} - \frac{484}{81} = \frac{612-484}{81} = \frac{128}{81}$.

(D) Let $P(X=-1) = a$,$P(X=0) = b$,and $P(X=1) = c$.
Since the sum of probabilities is $1$,we have $a + b + c = 1$.
Given $P(X=0) = b = 0.2$.
Substituting $b$ into the sum: $a + 0.2 + c = 1 \Rightarrow a + c = 0.8 \Rightarrow a = 0.8 - c$.
The mean of the random variable $X$ is given by $E(X) = \sum x_i P(X=x_i) = 0.2$.
So,$(-1)(a) + (0)(b) + (1)(c) = 0.2$.
$-a + c = 0.2$.
Substitute $a = 0.8 - c$ into the equation: $-(0.8 - c) + c = 0.2$.
$-0.8 + c + c = 0.2$.
$2c = 1.0$.
$c = 0.5$.
Thus,$P(X=1) = 0.5$.

(A) Let $X$ be the random variable representing the number of red balls drawn. The total number of balls is $4 + 5 = 9$. The number of ways to draw $3$ balls from $9$ is ${}^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
The probability distribution of $X$ is:
$P(X=0) = \frac{{}^4C_3}{{}^9C_3} = \frac{4}{84}$
$P(X=1) = \frac{{}^4C_2 \times {}^5C_1}{{}^9C_3} = \frac{6 \times 5}{84} = \frac{30}{84}$
$P(X=2) = \frac{{}^4C_1 \times {}^5C_2}{{}^9C_3} = \frac{4 \times 10}{84} = \frac{40}{84}$
$P(X=3) = \frac{{}^4C_0 \times {}^5C_3}{{}^9C_3} = \frac{1 \times 10}{84} = \frac{10}{84}$
The mean $E(X) = \sum x_i P(x_i) = 0 \times \frac{4}{84} + 1 \times \frac{30}{84} + 2 \times \frac{40}{84} + 3 \times \frac{10}{84}$
$E(X) = \frac{0 + 30 + 80 + 30}{84} = \frac{140}{84} = \frac{5}{3}$.
Thus,the mean is $\frac{5}{3}$. Hence,option $(A)$ is correct.

(D) The variance of a random variable $X$ is given by $\text{Var}(X) = E(X^2) - (E(X))^2$.
Since the sum of probabilities in a probability distribution is $1$,we have:
$\frac{1}{8} + \frac{3}{8} + 3K + K = 1$
$\frac{4}{8} + 4K = 1$
$\frac{1}{2} + 4K = 1 \Rightarrow 4K = \frac{1}{2} \Rightarrow K = \frac{1}{8}$.
Now,we calculate $E(X)$ and $E(X^2)$:
$E(X) = \sum x_i P(x_i) = (0 \times \frac{1}{8}) + (1 \times \frac{3}{8}) + (2 \times \frac{3}{8}) + (3 \times \frac{1}{8}) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{12}{8} = \frac{3}{2}$.
$E(X^2) = \sum x_i^2 P(x_i) = (0^2 \times \frac{1}{8}) + (1^2 \times \frac{3}{8}) + (2^2 \times \frac{3}{8}) + (3^2 \times \frac{1}{8}) = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = \frac{24}{8} = 3$.
Therefore,$\text{Var}(X) = E(X^2) - (E(X))^2 = 3 - (\frac{3}{2})^2 = 3 - \frac{9}{4} = \frac{12 - 9}{4} = \frac{3}{4}$.

(D) The variance of a random variable $X$ is given by the formula:
$\operatorname{Var}(X) = E(X^2) - [E(X)]^2$
First,we calculate the expected value $E(X) = \sum P_i x_i$:
$E(X) = (-2) \times \frac{1}{6} + (-1) \times \frac{1}{6} + 0 \times \frac{1}{3} + 1 \times \frac{1}{6} + 2 \times \frac{1}{6}$
$E(X) = -\frac{2}{6} - \frac{1}{6} + 0 + \frac{1}{6} + \frac{2}{6} = 0$
Next,we calculate $E(X^2) = \sum P_i x_i^2$:
$E(X^2) = (-2)^2 \times \frac{1}{6} + (-1)^2 \times \frac{1}{6} + 0^2 \times \frac{1}{3} + 1^2 \times \frac{1}{6} + 2^2 \times \frac{1}{6}$
$E(X^2) = 4 \times \frac{1}{6} + 1 \times \frac{1}{6} + 0 + 1 \times \frac{1}{6} + 4 \times \frac{1}{6}$
$E(X^2) = \frac{4+1+0+1+4}{6} = \frac{10}{6} = \frac{5}{3}$
Finally,the variance is:
$\operatorname{Var}(X) = \frac{5}{3} - (0)^2 = \frac{5}{3}$

(A) The probability distribution is given by $P(X=j) = (\frac{1}{2})^j$ for $j = 1, 2, 3, \ldots$.
This is a geometric distribution with $p = \frac{1}{2}$ and $q = 1 - p = \frac{1}{2}$.
The mean $E[X]$ of a geometric distribution starting from $j=1$ is given by $\frac{1}{p} = \frac{1}{1/2} = 2$.
The variance $Var(X)$ of a geometric distribution is given by $\frac{q}{p^2}$.
Substituting the values,$Var(X) = \frac{1/2}{(1/2)^2} = \frac{1/2}{1/4} = 2$.
Thus,the variance of $X$ is $2$.

(D) Since the sum of probabilities is $1$:
$\frac{1}{6} + k + \frac{1}{4} + k + \frac{1}{6} = 1$
$2k + \frac{7}{12} = 1 \implies 2k = \frac{5}{12} \implies k = \frac{5}{24}$
Calculation for Variance:
$\begin{array}{|c|c|c|c|} \hline x_i & P(X=x_i) & x_i P(x_i) & x_i^2 P(x_i) \\ \hline -2 & 1/6 & -1/3 & 2/3 \\ \hline -1 & 5/24 & -5/24 & 5/24 \\ \hline 0 & 1/4 & 0 & 0 \\ \hline 1 & 5/24 & 5/24 & 5/24 \\ \hline 2 & 1/6 & 1/3 & 2/3 \\ \hline \text{Total} & 1 & 0 & 21/12 \\ \hline \end{array}$
$\text{Variance} = E(X^2) - [E(X)]^2$
$= \frac{21}{12} - (0)^2 = \frac{7}{4}$

(A) The mean $E(X)$ is calculated as $\sum x_i P(x_i) = (-3 \times \frac{1}{6}) + (6 \times \frac{1}{2}) + (9 \times \frac{1}{3}) = -0.5 + 3 + 3 = 5.5 = \frac{11}{2}$.
The expectation of the square $E(X^2)$ is $\sum x_i^2 P(x_i) = ((-3)^2 \times \frac{1}{6}) + (6^2 \times \frac{1}{2}) + (9^2 \times \frac{1}{3}) = (9 \times \frac{1}{6}) + (36 \times \frac{1}{2}) + (81 \times \frac{1}{3}) = 1.5 + 18 + 27 = 46.5 = \frac{93}{2}$.
The variance $Var(X)$ is given by $E(X^2) - [E(X)]^2 = \frac{93}{2} - (\frac{11}{2})^2 = \frac{93}{2} - \frac{121}{4} = \frac{186 - 121}{4} = \frac{65}{4}$.

(C) For a probability distribution,the sum of all probabilities must be equal to $1$.
Given $P(X=k) = \frac{3^{ck}}{k!}$ for $k = 1, 2, 3, \ldots$,we have:
$\sum_{k=1}^{\infty} P(X=k) = \sum_{k=1}^{\infty} \frac{(3^c)^k}{k!} = 1$.
Recall the Taylor series expansion for the exponential function: $e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} = 1 + \sum_{k=1}^{\infty} \frac{x^k}{k!}$.
Therefore,$\sum_{k=1}^{\infty} \frac{x^k}{k!} = e^x - 1$.
Substituting $x = 3^c$,we get:
$e^{3^c} - 1 = 1$,which implies $e^{3^c} = 2$.
Taking the natural logarithm on both sides:
$3^c = \log_e 2$.
Taking the logarithm base $3$ on both sides:
$c = \log_3(\log_e 2)$.
Thus,the correct option is $C$.

(C) For a Poisson distribution with mean $\lambda = 2$,the probability mass function is given by $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} = \frac{e^{-2} 2^k}{k!}$.
We need to find $P(X > \frac{3}{2})$. Since $X$ takes only non-negative integer values,$X > \frac{3}{2}$ is equivalent to $X \geq 2$.
Using the complement rule,$P(X \geq 2) = 1 - P(X < 2) = 1 - [P(X = 0) + P(X = 1)]$.
Calculating the individual probabilities:
$P(X = 0) = \frac{e^{-2} 2^0}{0!} = e^{-2} = \frac{1}{e^2}$.
$P(X = 1) = \frac{e^{-2} 2^1}{1!} = 2e^{-2} = \frac{2}{e^2}$.
Therefore,$P(X \geq 2) = 1 - [\frac{1}{e^2} + \frac{2}{e^2}] = 1 - \frac{3}{e^2} = \frac{e^2 - 3}{e^2}$.
Thus,the correct option is $C$.

(B) The sum of probabilities for all possible values of a random variable must be equal to $1$.
Thus,$\sum_{k=0}^{\infty} P(X=k) = 1$.
Substituting the given expression: $\sum_{k=0}^{\infty} \frac{(k+1)a}{3^k} = 1$.
Taking $a$ as a constant: $a \sum_{k=0}^{\infty} (k+1) \left(\frac{1}{3}\right)^k = 1$.
Let $S = \sum_{k=0}^{\infty} (k+1) x^k$ where $x = 1/3$.
This is an arithmetico-geometric series: $S = 1 + 2x + 3x^2 + 4x^3 + \ldots$.
We know that $\sum_{k=0}^{\infty} x^k = \frac{1}{1-x}$.
Differentiating both sides with respect to $x$: $\sum_{k=1}^{\infty} k x^{k-1} = \frac{1}{(1-x)^2}$.
Multiplying by $x$: $\sum_{k=1}^{\infty} k x^k = \frac{x}{(1-x)^2}$.
Then $S = \sum_{k=0}^{\infty} (k+1) x^k = \sum_{k=0}^{\infty} k x^k + \sum_{k=0}^{\infty} x^k = \frac{x}{(1-x)^2} + \frac{1}{1-x} = \frac{x + 1 - x}{(1-x)^2} = \frac{1}{(1-x)^2}$.
For $x = 1/3$,$S = \frac{1}{(1 - 1/3)^2} = \frac{1}{(2/3)^2} = \frac{1}{4/9} = 9/4$.
Therefore,$a \times (9/4) = 1$,which gives $a = 4/9$.

(A) Let $X$ be the number of defective bulbs in a consignment of $n = 600$ bulbs.
The probability of a bulb being defective is $p = \frac{1}{100} = 0.01$.
Since $n$ is large and $p$ is small,we use the Poisson distribution with parameter $\lambda = np = 600 \times 0.01 = 6$.
The probability of having $k$ defective bulbs is given by $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} = \frac{e^{-6} 6^k}{k!}$.
We need to find the probability of at least two defective bulbs,which is $P(X \ge 2) = 1 - [P(X = 0) + P(X = 1)]$.
Calculating the probabilities:
$P(X = 0) = \frac{e^{-6} 6^0}{0!} = e^{-6}$.
$P(X = 1) = \frac{e^{-6} 6^1}{1!} = 6e^{-6}$.
Therefore,$P(X \ge 2) = 1 - (e^{-6} + 6e^{-6}) = 1 - 7e^{-6}$.
The correct option is $A$.

(D) For a probability distribution,the sum of all probabilities must be $1$.
$\sum P(X=x) = \lambda + 2\lambda + 3\lambda + 4\lambda = 10\lambda = 1$.
Thus,$\lambda = \frac{1}{10}$.
Now,calculate $\alpha$ and $\beta$:
$\alpha = P(X < 3) = P(X=1) + P(X=2) = \lambda + 2\lambda = 3\lambda$.
$\beta = P(X > 2) = P(X=3) + P(X=4) = 3\lambda + 4\lambda = 7\lambda$.
Therefore,the ratio $\alpha : \beta = 3\lambda : 7\lambda = 3 : 7$.

(B) For a probability distribution,the sum of all probabilities must be $1$.
$\sum P(X = x_i) = k + 2k + 3k + 4k = 10k = 1 \implies k = 0.1$.
The mean $\mu = E(X) = \sum x_i P(X = x_i) = (3 \times 0.1) + (5 \times 0.2) + (7 \times 0.3) + (9 \times 0.4) = 0.3 + 1.0 + 2.1 + 3.6 = 7.0$.
Now,$E(X^2) = \sum x_i^2 P(X = x_i) = (3^2 \times 0.1) + (5^2 \times 0.2) + (7^2 \times 0.3) + (9^2 \times 0.4) = (9 \times 0.1) + (25 \times 0.2) + (49 \times 0.3) + (81 \times 0.4) = 0.9 + 5.0 + 14.7 + 32.4 = 53.0$.
The variance $\sigma^2 = E(X^2) - [E(X)]^2 = 53.0 - (7.0)^2 = 53.0 - 49.0 = 4.0$.
The standard deviation $\sigma = \sqrt{\sigma^2} = \sqrt{4.0} = 2$.

(C) For a probability distribution,the sum of all probabilities must be equal to $1$,i.e.,$\sum P(X=x) = 1$.
Summing the given probabilities:
$0 + K + 2K + 2K + 3K + K^2 + 2K^2 + (7K^2 + K) = 1$
Combining like terms:
$(K + 2K + 2K + 3K + K) + (K^2 + 2K^2 + 7K^2) = 1$
$9K + 10K^2 = 1$
$10K^2 + 9K - 1 = 0$
Factoring the quadratic equation:
$10K^2 + 10K - K - 1 = 0$
$10K(K + 1) - 1(K + 1) = 0$
$(10K - 1)(K + 1) = 0$
This gives $K = \frac{1}{10}$ or $K = -1$. Since probability cannot be negative,$K$ must be positive,so $K = \frac{1}{10}$.
We need to find $P(0 < X < 5)$,which is:
$P(X=1) + P(X=2) + P(X=3) + P(X=4)$
$= K + 2K + 2K + 3K = 8K$
Substituting $K = \frac{1}{10}$:
$P(0 < X < 5) = 8 \times \frac{1}{10} = \frac{8}{10}$.

(A) For a Poisson distribution with parameter $\lambda$,the probability mass function is given by $P(X=x) = \frac{\lambda^x e^{-\lambda}}{x!}$.
Given $P(X=1) = 2P(X=2)$.
Substituting the formula:
$\frac{\lambda^1 e^{-\lambda}}{1!} = 2 \times \frac{\lambda^2 e^{-\lambda}}{2!}$
$\lambda e^{-\lambda} = 2 \times \frac{\lambda^2 e^{-\lambda}}{2}$
$\lambda e^{-\lambda} = \lambda^2 e^{-\lambda}$
Since $e^{-\lambda} \neq 0$,we divide both sides by $e^{-\lambda}$:
$\lambda = \lambda^2$
$\lambda^2 - \lambda = 0 \Rightarrow \lambda(\lambda - 1) = 0$.
Since $\lambda > 0$ for a Poisson distribution,we have $\lambda = 1$.
Now,we calculate $P(X=3)$:
$P(X=3) = \frac{\lambda^3 e^{-\lambda}}{3!} = \frac{1^3 e^{-1}}{3 \times 2 \times 1} = \frac{e^{-1}}{6}$.

(A) The mean $E(X) = \sum x_i P(x_i) = 0 \times \frac{1}{10} + 1 \times \frac{2}{10} + 2 \times \frac{3}{10} + 3 \times \frac{4}{10} = 0 + \frac{2}{10} + \frac{6}{10} + \frac{12}{10} = \frac{20}{10} = 2$.
The variance $Var(X) = E(X^2) - [E(X)]^2$.
$E(X^2) = \sum x_i^2 P(x_i) = 0^2 \times \frac{1}{10} + 1^2 \times \frac{2}{10} + 2^2 \times \frac{3}{10} + 3^2 \times \frac{4}{10} = 0 + \frac{2}{10} + \frac{12}{10} + \frac{36}{10} = \frac{50}{10} = 5$.
Therefore,$Var(X) = 5 - (2)^2 = 5 - 4 = 1$.

(D) Given: $p = 0.001$,$n = 2000$.
Since $n$ is large and $p$ is very small,we use the Poisson distribution with parameter $\lambda = np$.
$\lambda = 2000 \times 0.001 = 2$.
The probability mass function for Poisson distribution is $P(X = x) = \frac{\lambda^x e^{-\lambda}}{x!}$.
We need to find the probability for exactly $x = 3$ individuals:
$P(X = 3) = \frac{2^3 e^{-2}}{3!} = \frac{8 \times e^{-2}}{6} = \frac{4}{3 e^2}$.
Thus,the correct option is $D$.

(A) For a probability distribution,the sum of all probabilities must be equal to $1$.
Therefore,$\sum P(X = x) = 1$.
$\frac{1}{10} + k + \frac{1}{5} + 2k + \frac{3}{10} + k = 1$
Combine the constant terms and the terms with $k$:
$(\frac{1}{10} + \frac{2}{10} + \frac{3}{10}) + (k + 2k + k) = 1$
$\frac{6}{10} + 4k = 1$
$4k = 1 - \frac{6}{10}$
$4k = \frac{10 - 6}{10}$
$4k = \frac{4}{10}$
$k = \frac{1}{10}$

(C) For a Poisson distribution,the probability mass function is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.
Given $P(X=1) = P(X=2)$,we have:
$\frac{e^{-\lambda} \lambda^1}{1!} = \frac{e^{-\lambda} \lambda^2}{2!}$
Dividing both sides by $e^{-\lambda} \lambda$ (assuming $\lambda \neq 0$):
$1 = \frac{\lambda}{2}$
$\lambda = 2$
Now,we need to find $P(X=4)$:
$P(X=4) = \frac{e^{-\lambda} \lambda^4}{4!} = \frac{e^{-2} (2)^4}{24}$
$P(X=4) = \frac{16}{24 e^2} = \frac{2}{3 e^2}$

(A) To find the variance of the random variable $X$,we use the formula: $\text{Var}(X) = E(X^2) - [E(X)]^2$.
First,we calculate the mean $E(X) = \sum x_i P(x_i)$:
$E(X) = (0 \times 0.4) + (1 \times 0.3) + (2 \times 0.1) + (3 \times 0.1) + (4 \times 0.1)$
$E(X) = 0 + 0.3 + 0.2 + 0.3 + 0.4 = 1.2$
Next,we calculate $E(X^2) = \sum x_i^2 P(x_i)$:
$E(X^2) = (0^2 \times 0.4) + (1^2 \times 0.3) + (2^2 \times 0.1) + (3^2 \times 0.1) + (4^2 \times 0.1)$
$E(X^2) = 0 + 0.3 + 0.4 + 0.9 + 1.6 = 3.2$
Now,calculate the variance:
$\text{Var}(X) = E(X^2) - [E(X)]^2$
$\text{Var}(X) = 3.2 - (1.2)^2$
$\text{Var}(X) = 3.2 - 1.44 = 1.76$
Thus,the variance of $X$ is $1.76$.

(B) Given that $P(X=k) = \frac{(k+1)a}{3^k}$ for $k \in \{0, 1, 2, \ldots, \infty\}$.
As we know that the sum of all probabilities in a probability distribution is $1$,so $\sum_{k=0}^{\infty} P(X=k) = 1$.
Substituting the given expression:
$a \left( 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \ldots \infty \right) = 1 \quad \dots (i)$
Let $S = 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \ldots \infty$.
Then $\frac{1}{3}S = \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \ldots \infty$.
Subtracting the two equations:
$S - \frac{1}{3}S = 1 + \left( \frac{2}{3} - \frac{1}{3} \right) + \left( \frac{3}{3^2} - \frac{2}{3^2} \right) + \ldots \infty$
$\frac{2}{3}S = 1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \ldots \infty$
This is an infinite geometric series with first term $1$ and common ratio $r = \frac{1}{3}$.
$\frac{2}{3}S = \frac{1}{1 - \frac{1}{3}} = \frac{1}{2/3} = \frac{3}{2}$.
$S = \frac{3}{2} \times \frac{3}{2} = \frac{9}{4}$.
From equation $(i)$,$a \times S = 1 \implies a \times \frac{9}{4} = 1$.
Therefore,$a = \frac{4}{9}$.

(C) For a Poisson distribution with parameter $m$,the probability mass function is given by $P(X=x) = \frac{e^{-m} m^x}{x!}$.
Given $P(X=0) = 0.8$.
Substituting $x=0$ in the formula,we get $P(X=0) = \frac{e^{-m} m^0}{0!} = e^{-m}$.
So,$e^{-m} = 0.8$.
Taking the natural logarithm on both sides,$-m = \log_e(0.8) = \log_e(8/10) = \log_e(4/5)$.
Therefore,$m = -\log_e(4/5) = \log_e(5/4)$.
In a Poisson distribution,the variance is equal to the parameter $m$.
Thus,the variance is $\log_e(5/4)$.

(A) The probability mass function of a Poisson distribution is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$,where $\lambda$ is the mean of the distribution.
Given the condition $P(X=2) = 3 P(X=3)$.
Substituting the formula:
$\frac{e^{-\lambda} \lambda^2}{2!} = 3 \cdot \frac{e^{-\lambda} \lambda^3}{3!}$
Dividing both sides by $e^{-\lambda} \lambda^2$ (assuming $\lambda \neq 0$):
$\frac{1}{2} = 3 \cdot \frac{\lambda}{3 \cdot 2 \cdot 1}$
$\frac{1}{2} = \frac{3 \lambda}{6}$
$\frac{1}{2} = \frac{\lambda}{2}$
$\lambda = 1$.
Thus,the mean of the Poisson distribution is $1$.

(D) Given that the mean of the random variable $X$ is $1.3$.
The formula for the mean is $\Sigma x_i P(X=x_i) = 1.3$.
Substituting the values: $0 \cdot P(X=0) + 1 \cdot P(X=1) + 2 \cdot P(X=2) + 3 \cdot P(X=3) = 1.3$.
Given $P(X=2) = 0.3$ and $P(X=3) = 2 P(X=1)$,we substitute these into the equation:
$0 + P(X=1) + 2(0.3) + 3(2 P(X=1)) = 1.3$.
$P(X=1) + 0.6 + 6 P(X=1) = 1.3$.
$7 P(X=1) = 0.7$,which gives $P(X=1) = 0.1$.
Now,$P(X=3) = 2 P(X=1) = 2(0.1) = 0.2$.
Since the sum of all probabilities is $1$:
$P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1$.
$P(X=0) + 0.1 + 0.3 + 0.2 = 1$.
$P(X=0) + 0.6 = 1$.
Therefore,$P(X=0) = 0.4$.

(B) Let $X$ be the random variable representing the number of accidents in a day. $X \sim \text{Poisson}(\lambda)$.
Given that $10$ accidents occurred in $50$ days,the mean rate $\lambda$ per day is $\lambda = \frac{10}{50} = 0.2$.
The probability of $X$ accidents in a day is given by $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$.
We need to find $P(X \geq 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]$.
$P(X \geq 3) = 1 - \left[ \frac{e^{-0.2} (0.2)^0}{0!} + \frac{e^{-0.2} (0.2)^1}{1!} + \frac{e^{-0.2} (0.2)^2}{2!} \right]$.
$P(X \geq 3) = 1 - e^{-0.2} \left[ 1 + 0.2 + \frac{0.04}{2} \right]$.
$P(X \geq 3) = 1 - e^{-0.2} [1 + 0.2 + 0.02] = 1 - 1.22 e^{-0.2}$.

(C) The Poisson distribution is given by $P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$,where $\lambda = 6$.
We need to find $P(1 \leq X \leq 5) = P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)$.
$P(1 \leq X \leq 5) = e^{-6} \left[ \frac{6^1}{1!} + \frac{6^2}{2!} + \frac{6^3}{3!} + \frac{6^4}{4!} + \frac{6^5}{5!} \right]$
$= e^{-6} \left[ 6 + \frac{36}{2} + \frac{216}{6} + \frac{1296}{24} + \frac{7776}{120} \right]$
$= e^{-6} [6 + 18 + 36 + 54 + 64.8]$
$= e^{-6} [178.8]$
$= 6 \times e^{-6} \left( \frac{178.8}{6} \right) = 6 \times e^{-6} (29.8)$.

(C) For a Poisson distribution,the mean and variance are both equal to $\lambda$. Given variance $\lambda = 2$.
The probability mass function is $P(X=n) = \frac{\lambda^n e^{-\lambda}}{n!}$.
We need to find $P(X \geq 3) = 1 - \{P(X=0) + P(X=1) + P(X=2)\}$.
$P(X=0) = \frac{2^0 e^{-2}}{0!} = e^{-2}$.
$P(X=1) = \frac{2^1 e^{-2}}{1!} = 2e^{-2}$.
$P(X=2) = \frac{2^2 e^{-2}}{2!} = \frac{4 e^{-2}}{2} = 2e^{-2}$.
Summing these,$P(X < 3) = e^{-2} + 2e^{-2} + 2e^{-2} = 5e^{-2} = \frac{5}{e^2}$.
Therefore,$P(X \geq 3) = 1 - \frac{5}{e^2} = \frac{e^2-5}{e^2}$.

(C) The random variable $X$ takes values $x_i \in \{1, 2, 3, 4, 5, 6\}$ with probabilities $P_i = \frac{1}{6}$ for each.
Mean $\mu = E(X) = \sum x_i P_i = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = \frac{7}{2}$.
$E(X^2) = \sum x_i^2 P_i = \frac{1^2+2^2+3^2+4^2+5^2+6^2}{6} = \frac{1+4+9+16+25+36}{6} = \frac{91}{6}$.
Variance $\sigma^2 = E(X^2) - (E(X))^2 = \frac{91}{6} - \left(\frac{7}{2}\right)^2 = \frac{91}{6} - \frac{49}{4} = \frac{182 - 147}{12} = \frac{35}{12}$.
Therefore,$\frac{\text{Variance of } X}{\text{Mean of } X} = \frac{35/12}{7/2} = \frac{35}{12} \times \frac{2}{7} = \frac{5}{6}$.

(D) For a probability distribution,the sum of probabilities must be $1$. Here,the given values are $P(X=x) = \frac{2}{20}, \frac{4}{20}, \frac{6}{20}, \frac{8}{20}$.
The mean $\mu = E(X) = \sum x \cdot P(X=x) = 1(\frac{2}{20}) + 2(\frac{4}{20}) + 3(\frac{6}{20}) + 4(\frac{8}{20}) = \frac{2+8+18+32}{20} = \frac{60}{20} = 3$.
The variance $\sigma^2 = E(X^2) - [E(X)]^2$.
$E(X^2) = \sum x^2 \cdot P(X=x) = 1^2(\frac{2}{20}) + 2^2(\frac{4}{20}) + 3^2(\frac{6}{20}) + 4^2(\frac{8}{20}) = \frac{2 + 16 + 54 + 128}{20} = \frac{200}{20} = 10$.
Variance $\sigma^2 = 10 - (3)^2 = 10 - 9 = 1$.
Standard deviation $\sigma = \sqrt{1} = 1$.

(B) The mean of $X$ is given by $\bar{X} = E(X) = \sum_{n=1}^{m} n \cdot P(X=n) = \sum_{n=1}^{m} n \cdot \frac{1}{m} = \frac{1}{m} \cdot \frac{m(m+1)}{2} = \frac{m+1}{2}$.
The variance of $X$ is given by $\text{Var}(X) = E(X^2) - [E(X)]^2$.
First,calculate $E(X^2) = \sum_{n=1}^{m} n^2 \cdot P(X=n) = \frac{1}{m} \sum_{n=1}^{m} n^2 = \frac{1}{m} \cdot \frac{m(m+1)(2m+1)}{6} = \frac{(m+1)(2m+1)}{6}$.
Now,$\text{Var}(X) = \frac{(m+1)(2m+1)}{6} - \left( \frac{m+1}{2} \right)^2 = \frac{(m+1)(2m+1)}{6} - \frac{(m+1)^2}{4}$.
Taking $\frac{m+1}{2}$ as a common factor: $\text{Var}(X) = \frac{m+1}{2} \left[ \frac{2m+1}{3} - \frac{m+1}{2} \right] = \frac{m+1}{2} \left[ \frac{4m+2 - 3m - 3}{6} \right] = \frac{m+1}{2} \cdot \frac{m-1}{6} = \frac{m^2-1}{12}$.