Probability distribution Questions in English

(A) Given that three coins are tossed.
The sample space $S$ is given by:
$S = \{TTT, TTH, THT, HTT, HHT, HTH, THH, HHH\}$
Therefore,$n(S) = 8$.
Let $X$ represent the number of heads.
Then $X$ can take values $0, 1, 2, 3$.
The probability distribution of $X$ is:
| $X$ | $0$ | $1$ | $2$ | $3$ |
|---|---|---|---|---|
| $P(X)$ | $1/8$ | $3/8$ | $3/8$ | $1/8$ |
The mean $\mu$ is given by $\Sigma X_i P(X_i)$:
$\mu = 0 \times \left(\frac{1}{8}\right) + 1 \times \left(\frac{3}{8}\right) + 2 \times \left(\frac{3}{8}\right) + 3 \times \left(\frac{1}{8}\right)$
$\mu = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8}$
$\mu = \frac{12}{8} = 1.5$

(C) For a probability distribution,the sum of all probabilities must be equal to $1$,i.e.,$\Sigma P(X) = 1$.
$(k-1) + 3k + k + 3k + 3k^2 + k^2 + (k^2+k) = 1$
Combining like terms:
$5k^2 + 9k - 1 = 1$
$5k^2 + 9k - 2 = 0$
Factoring the quadratic equation:
$5k^2 + 10k - k - 2 = 0$
$5k(k+2) - 1(k+2) = 0$
$(5k-1)(k+2) = 0$
This gives $k = \frac{1}{5}$ or $k = -2$.
Since the probability $P(X)$ cannot be negative,$k$ must be positive. For $k = -2$,$P(X=1) = -2-1 = -3$,which is impossible.
Therefore,$k = \frac{1}{5}$.

(B) The mean $(\mu)$ of the probability distribution is given by $\mu = \sum x_{i} P_{i}$.
$\mu = (0 \times \frac{25}{36}) + (1 \times \frac{5}{18}) + (2 \times \frac{1}{36}) = 0 + \frac{10}{36} + \frac{2}{36} = \frac{12}{36} = \frac{1}{3}$.
The variance $(\sigma^2)$ is given by $\sigma^2 = \sum x_{i}^2 P_{i} - \mu^2$.
First,calculate $\sum x_{i}^2 P_{i} = (0^2 \times \frac{25}{36}) + (1^2 \times \frac{5}{18}) + (2^2 \times \frac{1}{36}) = 0 + \frac{10}{36} + \frac{4}{36} = \frac{14}{36} = \frac{7}{18}$.
Now,$\sigma^2 = \frac{7}{18} - (\frac{1}{3})^2 = \frac{7}{18} - \frac{1}{9} = \frac{7-2}{18} = \frac{5}{18}$.
Therefore,the standard deviation $\sigma = \sqrt{\frac{5}{18}} = \sqrt{\frac{5 \times 2}{18 \times 2}} = \sqrt{\frac{10}{36}} = \frac{\sqrt{10}}{6}$.
Alternatively,$\sigma = \sqrt{\frac{5}{18}} = \frac{1}{3} \sqrt{\frac{5}{2}}$.
Thus,option $B$ is correct.

(C) Given that the mean of a Poisson variate $X$ is $\lambda = 1$.
The probability mass function is $P(X=r) = \frac{e^{-\lambda} \lambda^r}{r!} = \frac{e^{-1}}{r!}$.
We need to calculate $\sum_{r=0}^{\infty} |r-1| P(X=r)$.
$\sum_{r=0}^{\infty} |r-1| \frac{e^{-1}}{r!} = e^{-1} \left[ |0-1| \frac{1}{0!} + |1-1| \frac{1}{1!} + |2-1| \frac{1}{2!} + |3-1| \frac{1}{3!} + \dots \right]$
$= e^{-1} \left[ 1 \cdot 1 + 0 \cdot 1 + 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{6} + \dots \right]$
$= e^{-1} \left[ 1 + 0 + \frac{1}{2} + \frac{1}{3} + \dots \right]$
Note that $\sum_{r=0}^{\infty} |r-1| P(X=r) = E[|X-1|]$.
For a Poisson distribution with $\lambda=1$,$E[|X-1|] = \sum_{r=0}^{\infty} |r-1| \frac{e^{-1}}{r!} = e^{-1} \left( \sum_{r=0}^{\infty} r \frac{1}{r!} - \sum_{r=0}^{\infty} \frac{1}{r!} + P(X=0) \right)$ is not the direct path.
Calculating the sum: $e^{-1} [ |0-1| + |1-1|\frac{1}{1!} + |2-1|\frac{1}{2!} + |3-1|\frac{1}{3!} + \dots ] = e^{-1} [ 1 + 0 + \frac{1}{2} + \frac{2}{6} + \dots ] = e^{-1} [ 1 + 0 + 0.5 + 0.333 + \dots ] = \frac{2}{e}$.
Thus,the correct option is $C$.

(C) For a Poisson distribution with unit mean,$\bar{x} = 1$. The probability mass function is $P(X=x) = \frac{e^{-1}}{x!}$.
We need to calculate $\sum_{x=0}^{\infty} |x-1| \frac{e^{-1}}{x!}$.
$= \frac{1}{e} \left[ |0-1| \frac{1}{0!} + |1-1| \frac{1}{1!} + |2-1| \frac{1}{2!} + |3-1| \frac{1}{3!} + \dots \right]$
$= \frac{1}{e} \left[ 1 + 0 + \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \dots \right]$
$= \frac{1}{e} \left[ 1 + \sum_{x=2}^{\infty} \frac{x-1}{x!} \right] = \frac{1}{e} \left[ 1 + \sum_{x=2}^{\infty} \frac{1}{(x-1)!} - \sum_{x=2}^{\infty} \frac{1}{x!} \right]$
$= \frac{1}{e} \left[ 1 + (e-1) - (e-1-1) \right] = \frac{1}{e} [1 + e - 1 - e + 2] = \frac{2}{e}$.
Thus,the correct option is $C$.

(D) The probability mass function of a Poisson distribution is given by $P(X=r) = \frac{e^{-\lambda} \lambda^r}{r!}$.
Given the equation $2 P(X=1) = 5 P(X=5) + 2 P(X=3)$.
Substituting the formula: $2 \frac{e^{-\lambda} \lambda^1}{1!} = 5 \frac{e^{-\lambda} \lambda^5}{5!} + 2 \frac{e^{-\lambda} \lambda^3}{3!}$.
Dividing both sides by $e^{-\lambda}$ (since $e^{-\lambda} \neq 0$):
$2\lambda = \frac{5 \lambda^5}{120} + \frac{2 \lambda^3}{6}$.
$2\lambda = \frac{\lambda^5}{24} + \frac{\lambda^3}{3}$.
Dividing by $\lambda$ (assuming $\lambda > 0$):
$2 = \frac{\lambda^4}{24} + \frac{\lambda^2}{3}$.
Multiply by $24$: $48 = \lambda^4 + 8\lambda^2$.
$\lambda^4 + 8\lambda^2 - 48 = 0$.
Let $u = \lambda^2$,then $u^2 + 8u - 48 = 0$.
$(u + 12)(u - 4) = 0$.
Since $\lambda^2$ must be positive,$u = 4$,so $\lambda^2 = 4$,which implies $\lambda = 2$.
The standard deviation of a Poisson distribution is $\sigma = \sqrt{\lambda}$.
Therefore,$\sigma = \sqrt{2}$.

(B) We know that the sum of probabilities for all possible values of a random variable is $1$.
$\sum_{k=0}^{5} P(X=k) = 1$
$a \left( \frac{0+1}{2^0} + \frac{1+1}{2^1} + \frac{2+1}{2^2} + \frac{3+1}{2^3} + \frac{4+1}{2^4} + \frac{5+1}{2^5} \right) = 1$
$a \left( 1 + 1 + \frac{3}{4} + \frac{4}{8} + \frac{5}{16} + \frac{6}{32} \right) = 1$
$a \left( 2 + 0.75 + 0.5 + 0.3125 + 0.1875 \right) = 1$
$a \left( \frac{64+48+32+20+12}{32} \right) = 1$ $\Rightarrow a \left( \frac{176}{32} \right) = 1$ $\Rightarrow a \left( \frac{11}{2} \right) = 1$ $\Rightarrow a = \frac{2}{11}$.
Wait,re-calculating the sum:
$a(1 + 1 + 0.75 + 0.5 + 0.3125 + 0.1875) = a(2 + 0.75 + 0.5 + 0.5) = a(3.75) = a(\frac{15}{4}) = 1$ $\Rightarrow a = \frac{4}{15}$.
The prime values for $X$ in the set $\{0, 1, 2, 3, 4, 5\}$ are $2, 3, 5$.
$P(X \in \{2, 3, 5\}) = P(X=2) + P(X=3) + P(X=5)$
$= a \left( \frac{3}{4} + \frac{4}{8} + \frac{6}{32} \right) = a \left( \frac{24+16+6}{32} \right) = a \left( \frac{46}{32} \right) = a \left( \frac{23}{16} \right)$
$= \frac{4}{15} \times \frac{23}{16} = \frac{23}{60}$.

(B) For a probability distribution,the sum of all probabilities must be $1$:
$\sum P(X=x_i) = 1$
$0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1$
$10k^2 + 9k - 1 = 0$
$(10k - 1)(k + 1) = 0$
Since $P(X=x) \ge 0$,$k$ must be positive,so $k = \frac{1}{10}$.
Now,we need to find $P(0 < X < 6) = P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)$.
$P(0 < X < 6) = k + 2k + 2k + 3k + k^2 = 8k + k^2$.
Substituting $k = \frac{1}{10}$:
$P(0 < X < 6) = 8\left(\frac{1}{10}\right) + \left(\frac{1}{10}\right)^2 = \frac{8}{10} + \frac{1}{100} = \frac{80+1}{100} = \frac{81}{100} = \left(\frac{9}{10}\right)^2$.

(B) Given that $X$ and $Y$ are independent discrete random variables.
For the binomial distribution $X \sim B(n, p)$,the variance is given by $Var(X) = npq$,where $q = 1 - p$.
Here,$n = 16$ and $p = 0.25$,so $q = 1 - 0.25 = 0.75$.
Thus,$Var(X) = 16 \times 0.25 \times 0.75 = 4 \times 0.75 = 3$.
For the Poisson distribution $Y \sim P(\lambda)$,the variance is given by $Var(Y) = \lambda$.
Here,$\lambda = 2$,so $Var(Y) = 2$.
The sum of the variances is $Var(X) + Var(Y) = 3 + 2 = 5$.

(C) For a Poisson distribution,the mean $\lambda$ is equal to the variance. Given variance $= 2$,so $\lambda = 2$.
The probability mass function is $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$.
We need to find $P(X \geq 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]$.
$P(X=0) = \frac{e^{-2} 2^0}{0!} = e^{-2}$.
$P(X=1) = \frac{e^{-2} 2^1}{1!} = 2e^{-2}$.
$P(X=2) = \frac{e^{-2} 2^2}{2!} = \frac{4e^{-2}}{2} = 2e^{-2}$.
Sum $= e^{-2} + 2e^{-2} + 2e^{-2} = 5e^{-2} = \frac{5}{e^2}$.
Therefore,$P(X \geq 3) = 1 - \frac{5}{e^2} = \frac{e^2 - 5}{e^2}$.

(C) The number of accidents follows a Poisson distribution with parameter $\lambda = 5$.
The probability mass function for a Poisson distribution is given by $P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}$.
We need to find the probability that at most one accident occurs,which is $P(X \le 1) = P(X = 0) + P(X = 1)$.
For $x = 0$,$P(X = 0) = \frac{e^{-5} 5^0}{0!} = \frac{e^{-5} \times 1}{1} = e^{-5}$.
For $x = 1$,$P(X = 1) = \frac{e^{-5} 5^1}{1!} = \frac{e^{-5} \times 5}{1} = 5e^{-5}$.
Therefore,$P(X \le 1) = e^{-5} + 5e^{-5} = 6e^{-5} = \frac{6}{e^5}$.

(D) Let $X_1, X_2, X_3, X_4$ be the values of the coins $1, 2, 5, 10$ respectively.
Let $I_k$ be an indicator random variable such that $I_k = 1$ if the $k$-th coin shows heads and $I_k = 0$ if it shows tails.
The probability of getting heads for each coin is $P(I_k = 1) = \frac{1}{2}$.
The expected value of each indicator variable is $E[I_k] = 1 \times \frac{1}{2} + 0 \times \frac{1}{2} = \frac{1}{2}$.
The total sum $S$ of the values of the coins showing heads is $S = 1 \cdot I_1 + 2 \cdot I_2 + 5 \cdot I_3 + 10 \cdot I_4$.
By the linearity of expectation,$E[S] = 1 \cdot E[I_1] + 2 \cdot E[I_2] + 5 \cdot E[I_3] + 10 \cdot E[I_4]$.
$E[S] = 1 \times \frac{1}{2} + 2 \times \frac{1}{2} + 5 \times \frac{1}{2} + 10 \times \frac{1}{2}$.
$E[S] = \frac{1+2+5+10}{2} = \frac{18}{2} = 9$.

(C) When two coins are tossed,the sample space is $S = \{HH, HT, TH, TT\}$. The total number of outcomes is $4$.
Let $X$ be the random variable representing the prize money.
The possible outcomes are:
$1$. Two heads $(HH)$: $P(X = 2) = 1/4$. Prize = $Rs. 2$.
$2$. One head ($HT$ or $TH$): $P(X = 1) = 2/4 = 1/2$. Prize = $Rs. 1$.
$3$. No head $(TT)$: $P(X = -3) = 1/4$. Prize = $-Rs. 3$ (loss).
The mean (expected value) $E(X)$ is calculated as:
$E(X) = \sum x_i p_i$
$E(X) = (2 \times 1/4) + (1 \times 1/2) + (-3 \times 1/4)$
$E(X) = 2/4 + 1/2 - 3/4$
$E(X) = 1/2 + 1/2 - 3/4 = 1 - 3/4 = 1/4$.
Thus,the mean of the prize money is $1/4$.

(A) For a Poisson distribution,the probability mass function is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k !}$.
Here,$\lambda$ is the mean number of accidents per day.
Given that $10$ accidents occur in $50$ days,the average number of accidents per day is $\lambda = \frac{10}{50} = 0.2$.
We need to find the probability that three or more accidents occur in a day,which is $P(X \geq 3)$.
$P(X \geq 3) = P(X=3) + P(X=4) + P(X=5) + \dots$
This can be written as $\sum_{k=3}^{\infty} \frac{e^{-\lambda} \lambda^k}{k !}$ with $\lambda = 0.2$.
Thus,the correct expression is $\sum_{k=3}^{\infty} \frac{e^{-0.2} (0.2)^k}{k !}$.

(D) Given: $n = 300$,$p = 0.01$.
Let $m$ be the mean number of defective items.
$m = n \times p = 300 \times 0.01 = 3$.
Since $n$ is large and $p$ is small,we use the Poisson distribution with parameter $m = 3$.
The probability of $X$ defective items is given by $P(X = k) = \frac{e^{-m} m^k}{k!}$.
We need to find the probability that the number of defective items is at most one,i.e.,$P(X \leq 1)$.
$P(X \leq 1) = P(X = 0) + P(X = 1)$.
$P(X = 0) = \frac{e^{-3} 3^0}{0!} = \frac{e^{-3} \times 1}{1} = \frac{1}{e^3}$.
$P(X = 1) = \frac{e^{-3} 3^1}{1!} = \frac{3}{e^3}$.
Therefore,$P(X \leq 1) = \frac{1}{e^3} + \frac{3}{e^3} = \frac{4}{e^3}$.

(C) Given that $P(X=x)=k\left(\frac{3}{8}\right)^X$ for $x=1, 2, 3, \ldots$ is a probability distribution function.
Since the sum of all probabilities must be equal to $1$,we have $\sum_{x=1}^{\infty} P(X=x) = 1$.
Substituting the given expression: $k \sum_{x=1}^{\infty} \left(\frac{3}{8}\right)^x = 1$.
This is an infinite geometric series with the first term $a = \frac{3}{8}$ and common ratio $r = \frac{3}{8}$.
The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$.
Thus,$k \left( \frac{3/8}{1-3/8} \right) = 1$.
$k \left( \frac{3/8}{5/8} \right) = 1$.
$k \left( \frac{3}{5} \right) = 1$.
Therefore,$k = \frac{5}{3}$.

(B) For a discrete random variable $X$,the sum of all probabilities must be equal to $1$,i.e.,$\sum P(x) = 1$.
Given $P(x) = c\left(\frac{2}{3}\right)^x$ for $x = 1, 2, 3, \ldots$.
Therefore,$\sum_{x=1}^{\infty} c\left(\frac{2}{3}\right)^x = 1$.
$c \left[ \frac{2}{3} + \left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^3 + \ldots \right] = 1$.
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 $S = \frac{a}{1-r}$.
$c \left[ \frac{2/3}{1 - 2/3} \right] = 1$.
$c \left[ \frac{2/3}{1/3} \right] = 1$.
$c(2) = 1$.
$c = \frac{1}{2}$.

(D) The random variable $X$ takes values $\{1, 2, 3, 4, 5, 6\}$ with probability $P(X=x_i) = \frac{1}{6}$ for each $i$.
Mean $E(X) = \sum x_i P(x_i) = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = \frac{7}{2}$.
Variance $Var(X) = E(X^2) - [E(X)]^2$.
$E(X^2) = \sum x_i^2 P(x_i) = \frac{1^2+2^2+3^2+4^2+5^2+6^2}{6} = \frac{1+4+9+16+25+36}{6} = \frac{91}{6}$.
$Var(X) = \frac{91}{6} - (\frac{7}{2})^2 = \frac{91}{6} - \frac{49}{4} = \frac{182 - 147}{12} = \frac{35}{12}$.
Thus,the mean is $\frac{7}{2}$ and the variance is $\frac{35}{12}$.

(A) Given,the mean of the Poisson distribution is $\lambda = 6$.
The probability mass function of a Poisson distribution is given by $P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$.
We need to find $P(X \geq 3)$.
Using the complement rule,$P(X \geq 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]$.
Substituting the values:
$P(X=0) = \frac{e^{-6} 6^0}{0!} = e^{-6}$.
$P(X=1) = \frac{e^{-6} 6^1}{1!} = 6e^{-6}$.
$P(X=2) = \frac{e^{-6} 6^2}{2!} = \frac{36e^{-6}}{2} = 18e^{-6}$.
Therefore,$P(X \geq 3) = 1 - [e^{-6} + 6e^{-6} + 18e^{-6}] = 1 - 25e^{-6} = 1 - \frac{25}{e^6}$.

(C) For a Poisson distribution,the variance is given by $\lambda$. Here,$\lambda = 3$.
The probability mass function is $P(x=n) = \frac{\lambda^n \cdot e^{-\lambda}}{n!}$.
We need to find $P(1 < x < 4) = P(x=2) + P(x=3)$.
$P(x=2) = \frac{3^2 \cdot e^{-3}}{2!} = \frac{9 \cdot e^{-3}}{2}$.
$P(x=3) = \frac{3^3 \cdot e^{-3}}{3!} = \frac{27 \cdot e^{-3}}{6} = \frac{9 \cdot e^{-3}}{2}$.
Therefore,$P(1 < x < 4) = \frac{9 \cdot e^{-3}}{2} + \frac{9 \cdot e^{-3}}{2} = 9 \cdot e^{-3}$.

(C) The sum of all probabilities in a probability distribution must be equal to $1$.
Therefore,$\sum_{x=0}^{\infty} P(X=x) = 1$.
Substituting the given expression: $\sum_{x=0}^{\infty} k \frac{2^{2x+1}}{(2x+1)!} = 1$.
$k \sum_{x=0}^{\infty} \frac{2^{2x+1}}{(2x+1)!} = 1$.
Let $n = 2x+1$. As $x$ goes from $0$ to $\infty$,$n$ takes odd values $1, 3, 5, \ldots$.
So,$k \sum_{n=1, 3, 5, \ldots}^{\infty} \frac{2^n}{n!} = 1$.
The Taylor series expansion for $\sinh(z)$ is $\sum_{n=1, 3, 5, \ldots}^{\infty} \frac{z^n}{n!} = \sinh(z)$.
Here,$z = 2$,so $\sum_{n=1, 3, 5, \ldots}^{\infty} \frac{2^n}{n!} = \sinh(2)$.
Thus,$k \sinh(2) = 1$.
$k = \frac{1}{\sinh(2)} = \text{cosech } 2$.

(A) The sum of probabilities in a distribution is $1$:
$\alpha + \alpha + \alpha + \beta + \beta + 0.3 = 1 \implies 3\alpha + 2\beta = 0.7$ (Equation $1$).
The mean $\mu$ is given by $\sum x_i P(x_i) = 4.2$:
$1(\alpha) + 2(\alpha) + 3(\alpha) + 4(\beta) + 5(\beta) + 6(0.3) = 4.2$
$6\alpha + 9\beta + 1.8 = 4.2 \implies 6\alpha + 9\beta = 2.4 \implies 2\alpha + 3\beta = 0.8$ (Equation $2$).
Solving Equations $1$ and $2$:
Multiply Eq $1$ by $2$: $6\alpha + 4\beta = 1.4$.
Multiply Eq $2$ by $3$: $6\alpha + 9\beta = 2.4$.
Subtracting: $5\beta = 1.0 \implies \beta = 0.2$.
Substituting $\beta = 0.2$ into Eq $1$: $3\alpha + 2(0.2) = 0.7 \implies 3\alpha = 0.3 \implies \alpha = 0.1$.
We need to find $\sigma^2 + \mu^2$. Since $\sigma^2 = E(X^2) - \mu^2$,then $\sigma^2 + \mu^2 = E(X^2)$.
$E(X^2) = \sum x_i^2 P(x_i) = 1^2(0.1) + 2^2(0.1) + 3^2(0.1) + 4^2(0.2) + 5^2(0.2) + 6^2(0.3)$
$E(X^2) = 0.1 + 0.4 + 0.9 + 3.2 + 5.0 + 10.8 = 20.4$.

(D) For a probability distribution,the sum of all probabilities must be $1$.
$\sum P(X = x_i) = 3K + 5K + 3k^2 + (4k^2 + k) + 3k^2 = 1$
$10k^2 + 9K - 1 = 0$
$(10k - 1)(k + 1) = 0$
Since $P(X = x_i) \geq 0$,we must have $k > 0$,so $k = \frac{1}{10}$.
We need to find $P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$.
$P(X \leq 2) = 3K + 5K + 3k^2 = 8K + 3k^2$.
Substituting $k = \frac{1}{10}$:
$P(X \leq 2) = 8(\frac{1}{10}) + 3(\frac{1}{10})^2 = \frac{8}{10} + \frac{3}{100} = \frac{80 + 3}{100} = \frac{83}{100}$.

(B) The sum of probabilities is $\sum_{k=0}^{\infty} P(X=k) = 1$.
Given $P(X=k) = \frac{3k+1}{2^{k+c}}$,we have $\frac{1}{2^c} \sum_{k=0}^{\infty} (3k+1) \left(\frac{1}{2}\right)^k = 1$.
Let $S = \sum_{k=0}^{\infty} (3k+1) x^k$ where $x = \frac{1}{2}$.
$S = 3 \sum_{k=0}^{\infty} k x^k + \sum_{k=0}^{\infty} x^k = 3 \frac{x}{(1-x)^2} + \frac{1}{1-x}$.
Substituting $x = \frac{1}{2}$: $S = 3 \frac{1/2}{(1/2)^2} + \frac{1}{1/2} = 3(2) + 2 = 8$.
Thus,$\frac{1}{2^c} (8) = 1 \implies 2^3 = 2^c \implies c = 3$.
We need to find $P(X \leq 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)$.
$P(X=k) = \frac{3k+1}{2^{k+3}}$.
$P(X=0) = \frac{1}{8}, P(X=1) = \frac{4}{16} = \frac{2}{8}, P(X=2) = \frac{7}{32}, P(X=3) = \frac{10}{64} = \frac{5}{32}$.
Sum $= \frac{4}{32} + \frac{8}{32} + \frac{7}{32} + \frac{5}{32} = \frac{24}{32} = \frac{3}{4}$.
Since $c=3$,the option $\frac{c}{4} = \frac{3}{4}$ matches option $B$.

(A) Step $1$: Calculate $E(X) = \Sigma x P(X=x)$.
$E(X) = (-1)(\frac{1}{3}) + (0)(\frac{1}{6}) + (1)(\frac{1}{6}) + (2)(\frac{1}{3}) = -\frac{1}{3} + 0 + \frac{1}{6} + \frac{2}{3} = \frac{-2+1+4}{6} = \frac{3}{6} = \frac{1}{2}$.
Step $2$: Calculate $E(X^2) = \Sigma x^2 P(X=x)$.
$E(X^2) = (-1)^2(\frac{1}{3}) + (0)^2(\frac{1}{6}) + (1)^2(\frac{1}{6}) + (2)^2(\frac{1}{3}) = \frac{1}{3} + 0 + \frac{1}{6} + \frac{4}{3} = \frac{2+1+8}{6} = \frac{11}{6}$.
Step $3$: Calculate $\operatorname{var}(X) = E(X^2) - [E(X)]^2$.
$\operatorname{var}(X) = \frac{11}{6} - (\frac{1}{2})^2 = \frac{11}{6} - \frac{1}{4} = \frac{22-3}{12} = \frac{19}{12}$.
Step $4$: Calculate $6 \Sigma(x^2) P(X=x) - \operatorname{var}(X)$.
$6 E(X^2) - \operatorname{var}(X) = 6(\frac{11}{6}) - \frac{19}{12} = 11 - \frac{19}{12} = \frac{132-19}{12} = \frac{113}{12}$.

(C) The sum of all probabilities in a probability distribution must be equal to $1$.
Therefore,$k^3 + (2k^3 + k) + (4k - 10k^2) + (4k - 1) = 1$.
Simplifying the equation: $3k^3 - 10k^2 + 9k - 2 = 0$.
By testing values,we find that $k = \frac{1}{3}$ is a root: $3(\frac{1}{27}) - 10(\frac{1}{9}) + 9(\frac{1}{3}) - 2 = \frac{1}{9} - \frac{10}{9} + 3 - 2 = -1 + 1 = 0$.
The probabilities are: $P(-1) = (\frac{1}{3})^3 = \frac{1}{27}$,$P(0) = 2(\frac{1}{27}) + \frac{1}{3} = \frac{11}{27}$,$P(1) = 4(\frac{1}{3}) - 10(\frac{1}{9}) = \frac{12-10}{9} = \frac{2}{9} = \frac{6}{27}$,$P(2) = 4(\frac{1}{3}) - 1 = \frac{1}{3} = \frac{9}{27}$.
Sum check: $\frac{1+11+6+9}{27} = \frac{27}{27} = 1$.
The mean $E(X) = \sum x_i P(x_i) = (-1)(\frac{1}{27}) + (0)(\frac{11}{27}) + (1)(\frac{6}{27}) + (2)(\frac{9}{27}) = \frac{-1 + 0 + 6 + 18}{27} = \frac{23}{27}$.

(A) For a probability distribution,the sum of all probabilities must be equal to $1$.
$\sum P(X = x_{i}) = 1$
$\frac{k^2}{3} + k^2 + \frac{2k^2}{3} + \frac{k}{2} + \frac{k}{2} = 1$
$\frac{k^2 + 3k^2 + 2k^2}{3} + k = 1$
$\frac{6k^2}{3} + k = 1$
$2k^2 + k - 1 = 0$
$(2k - 1)(k + 1) = 0$
Since $k$ must be positive for probabilities to be valid,$k = 1/2$.
The mean $E(X) = \sum x_{i} P(X = x_{i})$
$E(X) = (-2) \cdot \frac{k^2}{3} + (-1) \cdot k^2 + (0) \cdot \frac{2k^2}{3} + (1) \cdot \frac{k}{2} + (2) \cdot \frac{k}{2}$
$E(X) = -\frac{2k^2}{3} - k^2 + 0 + \frac{k}{2} + k$
$E(X) = -\frac{5k^2}{3} + \frac{3k}{2}$
Substituting $k = 1/2$:
$E(X) = -\frac{5(1/4)}{3} + \frac{3(1/2)}{2}$
$E(X) = -\frac{5}{12} + \frac{3}{4} = -\frac{5}{12} + \frac{9}{12} = \frac{4}{12} = 1/3$.

(A) The random variable $X$ follows a discrete uniform distribution on the set $\{1, 2, \ldots, n\}$.
The mean $E[X]$ is given by:
$E[X] = \sum_{k=1}^{n} k \cdot P(X=k) = \sum_{k=1}^{n} k \cdot \frac{1}{n} = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2}$.
The expected value of $X^2$ is given by:
$E[X^2] = \sum_{k=1}^{n} k^2 \cdot P(X=k) = \sum_{k=1}^{n} k^2 \cdot \frac{1}{n} = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6}$.
The variance $Var(X)$ is given by:
$Var(X) = E[X^2] - (E[X])^2 = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2$.
Factoring out $\frac{n+1}{2}$:
$Var(X) = \frac{n+1}{2} \left[ \frac{2n+1}{3} - \frac{n+1}{2} \right] = \frac{n+1}{2} \left[ \frac{4n+2 - 3n - 3}{6} \right] = \frac{n+1}{2} \cdot \frac{n-1}{6} = \frac{n^2-1}{12}$.

(D) For a Poisson distribution with mean $\lambda = 5$,the probability mass function is given by $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$.
We need to find $P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$.
Substituting the values:
$P(X = 0) = \frac{e^{-5} 5^0}{0!} = e^{-5}$
$P(X = 1) = \frac{e^{-5} 5^1}{1!} = 5e^{-5}$
$P(X = 2) = \frac{e^{-5} 5^2}{2!} = \frac{25}{2}e^{-5}$
Summing these probabilities:
$P(X < 3) = e^{-5} \left( 1 + 5 + \frac{25}{2} \right) = e^{-5} \left( 6 + 12.5 \right) = 18.5 e^{-5} = \frac{37}{2} e^{-5}$.

(B) Let $X$ be the random variable representing the number of kings drawn.
Total number of cards = $52$. Number of kings = $4$. Number of non-kings = $48$.
We draw $2$ cards. The possible values for $X$ are $0, 1, 2$.
$P(X=0) = \frac{{}^{48}C_2}{{}^{52}C_2} = \frac{48 \times 47}{52 \times 51} = \frac{1128}{1326} = \frac{188}{221}$.
$P(X=1) = \frac{{}^{4}C_1 \times {}^{48}C_1}{{}^{52}C_2} = \frac{4 \times 48}{1326} = \frac{192}{1326} = \frac{32}{221}$.
$P(X=2) = \frac{{}^{4}C_2}{{}^{52}C_2} = \frac{6}{1326} = \frac{1}{221}$.
The mean $E(X) = \sum x_i P(x_i) = 0 \times P(X=0) + 1 \times P(X=1) + 2 \times P(X=2)$.
$E(X) = 0 + \frac{32}{221} + 2 \times \frac{1}{221} = \frac{32+2}{221} = \frac{34}{221} = \frac{2}{13}$.

(B) The sum of all probabilities in a probability distribution must be $1$.
$\sum P(X=x) = \frac{1}{10} + (K + \frac{2}{10}) + (K + \frac{3}{10}) + (K + \frac{3}{10}) + (K + \frac{4}{10}) + (K + \frac{2}{10}) = 1$
$\Rightarrow 5K + \frac{15}{10} = 1$
$\Rightarrow 5K + 1.5 = 1$
$\Rightarrow 5K = -0.5$
$\Rightarrow K = -0.1 = -\frac{1}{10}$
Now,substitute $K = -\frac{1}{10}$ into the table:
For $x = -2, P = \frac{1}{10}$
For $x = -1, P = -\frac{1}{10} + \frac{2}{10} = \frac{1}{10}$
For $x = 0, P = -\frac{1}{10} + \frac{3}{10} = \frac{2}{10}$
For $x = 1, P = -\frac{1}{10} + \frac{3}{10} = \frac{2}{10}$
For $x = 2, P = -\frac{1}{10} + \frac{4}{10} = \frac{3}{10}$
For $x = 3, P = -\frac{1}{10} + \frac{2}{10} = \frac{1}{10}$
The mean $\mu = \sum x P(x) = (-2)(\frac{1}{10}) + (-1)(\frac{1}{10}) + (0)(\frac{2}{10}) + (1)(\frac{2}{10}) + (2)(\frac{3}{10}) + (3)(\frac{1}{10})$
$\mu = \frac{-2 - 1 + 0 + 2 + 6 + 3}{10} = \frac{8}{10} = \frac{4}{5}$

(B) Let $X$ be the number of red balls drawn. The total number of balls is $3 + 5 = 8$. We draw $3$ balls from $8$,so the total number of ways is ${}^8 C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
The random variable $X$ can take values $0, 1, 2, 3$.
$P(X=0) = \frac{{}^5 C_0 \times {}^3 C_3}{56} = \frac{1 \times 1}{56} = \frac{1}{56}$
$P(X=1) = \frac{{}^5 C_1 \times {}^3 C_2}{56} = \frac{5 \times 3}{56} = \frac{15}{56}$
$P(X=2) = \frac{{}^5 C_2 \times {}^3 C_1}{56} = \frac{10 \times 3}{56} = \frac{30}{56}$
$P(X=3) = \frac{{}^5 C_3 \times {}^3 C_0}{56} = \frac{10 \times 1}{56} = \frac{10}{56}$
The mean $E(X) = \sum x_i P(x_i) = 0 \times \frac{1}{56} + 1 \times \frac{15}{56} + 2 \times \frac{30}{56} + 3 \times \frac{10}{56} = \frac{0 + 15 + 60 + 30}{56} = \frac{105}{56} = \frac{15}{8}$.

(D) Given $P(X=k) = k^2 P$ for $k = 1, 2, 3, 4, 5, 6$.
Since the sum of all probabilities must be $1$,we have:
$\sum_{k=1}^6 P(X=k) = 1$
$P(1^2) + P(2^2) + P(3^2) + P(4^2) + P(5^2) + P(6^2) = 1$
$P(1 + 4 + 9 + 16 + 25 + 36) = 1$
$91P = 1 \Rightarrow P = \frac{1}{91}$.
The mean of $X$ is given by $E(X) = \sum_{k=1}^6 k \cdot P(X=k)$.
$E(X) = \sum_{k=1}^6 k \cdot (k^2 P) = P \sum_{k=1}^6 k^3$.
$E(X) = \frac{1}{91} (1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3)$.
$E(X) = \frac{1}{91} (1 + 8 + 27 + 64 + 125 + 216) = \frac{441}{91}$.

(A) For a probability distribution,the sum of all probabilities must be $1$.
$\sum P(X=x) = 1$
$0.05 + 0.1 + 2K + 0 + 0.3 + K + 0.1 = 1$
$0.55 + 3K = 1 \Rightarrow 3K = 0.45 \Rightarrow K = 0.15$.
Now,the distribution is:

$X$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X)$	$0.05$	$0.1$	$0.3$	$0$	$0.3$	$0.15$	$0.1$

Mean $\mu = E(X) = \sum x_i P(x_i) = (-3)(0.05) + (-2)(0.1) + (-1)(0.3) + (0)(0) + (1)(0.3) + (2)(0.15) + (3)(0.1)$
$\mu = -0.15 - 0.2 - 0.3 + 0 + 0.3 + 0.3 + 0.3 = 0.25$.
Variance $\sigma^2 = E(X^2) - \mu^2 = \sum x_i^2 P(x_i) - (0.25)^2$.
$E(X^2) = (-3)^2(0.05) + (-2)^2(0.1) + (-1)^2(0.3) + (0)^2(0) + (1)^2(0.3) + (2)^2(0.15) + (3)^2(0.1)$
$E(X^2) = 9(0.05) + 4(0.1) + 1(0.3) + 0 + 1(0.3) + 4(0.15) + 9(0.1)$
$E(X^2) = 0.45 + 0.4 + 0.3 + 0 + 0.3 + 0.6 + 0.9 = 2.95$.
Variance $= 2.95 - (0.25)^2 = 2.95 - 0.0625 = 2.8875$.

(A) For a probability distribution,the sum of all probabilities must be equal to $1$.
$\Sigma P(X=x) = 1$
$2k + 4k + 3k + k = 1$
$10k = 1$
$k = \frac{1}{10}$

(B) For a probability distribution,the sum of all probabilities must be equal to $1$.
$P(X=0) + P(X=1) + P(X=2) = 1$
$3C^3 + (4C - 10C^2) + (5C - 1) = 1$
$3C^3 - 10C^2 + 9C - 2 = 0$
By testing values,we find that $(C - 1)$ is a factor. Dividing the polynomial,we get $(C - 1)(3C^2 - 7C + 2) = 0$.
Factoring further: $(C - 1)(3C - 1)(C - 2) = 0$.
Thus,the possible values for $C$ are $1, \frac{1}{3}, 2$.
We must check if these values result in probabilities between $0$ and $1$:
If $C = 2$,$P(X=2) = 5(2) - 1 = 9$,which is $> 1$. So $C \neq 2$.
If $C = 1$,$P(X=1) = 4(1) - 10(1)^2 = -6$,which is $< 0$. So $C \neq 1$.
If $C = \frac{1}{3}$,$P(X=0) = 3(\frac{1}{27}) = \frac{1}{9}$,$P(X=1) = 4(\frac{1}{3}) - 10(\frac{1}{9}) = \frac{12-10}{9} = \frac{2}{9}$,and $P(X=2) = 5(\frac{1}{3}) - 1 = \frac{2}{3}$.
Since all probabilities are between $0$ and $1$ and sum to $1$,the correct value is $C = \frac{1}{3}$.

(C) The mean (expected value) of a random variable $X$ is given by the formula $E(X) = \sum x_i P(X=x_i)$.
Given the probability distribution:
$P(X=-2) = \frac{1}{6}$
$P(X=-1) = \frac{1}{6}$
$P(X=0) = \frac{1}{3}$
$P(X=1) = \frac{1}{6}$
$P(X=2) = \frac{1}{6}$
Calculating the mean:
$E(X) = (-2) \times \frac{1}{6} + (-1) \times \frac{1}{6} + 0 \times \frac{1}{3} + 1 \times \frac{1}{6} + 2 \times \frac{1}{6}$
$E(X) = -\frac{2}{6} - \frac{1}{6} + 0 + \frac{1}{6} + \frac{2}{6}$
$E(X) = \frac{-2 - 1 + 0 + 1 + 2}{6} = \frac{0}{6} = 0$
Thus,the mean of $X$ is $0$.

(B) We know that the sum of all probabilities in a distribution is $1$,so $\sum_{k=0}^{\infty} P(X=k) = 1$.
Given $P(X=k) = \frac{(k+1)c}{2^k}$,we have $c \sum_{k=0}^{\infty} \frac{k+1}{2^k} = 1$.
Let $S = \sum_{k=0}^{\infty} \frac{k+1}{2^k} = 1 + \frac{2}{2} + \frac{3}{4} + \frac{4}{8} + \ldots$.
Then $\frac{1}{2}S = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \ldots$.
Subtracting the two equations: $S - \frac{1}{2}S = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots = \frac{1}{1 - 1/2} = 2$.
Thus,$\frac{1}{2}S = 2$,which means $S = 4$.
Since $c \cdot S = 1$,we get $4c = 1$,so $c = \frac{1}{4}$.
Now,$P(X \geq 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]$.
$P(X=0) = \frac{(0+1)c}{2^0} = c = \frac{1}{4}$.
$P(X=1) = \frac{(1+1)c}{2^1} = c = \frac{1}{4}$.
$P(X=2) = \frac{(2+1)c}{2^2} = \frac{3c}{4} = \frac{3}{16}$.
$P(X \geq 3) = 1 - [\frac{1}{4} + \frac{1}{4} + \frac{3}{16}] = 1 - [\frac{4+4+3}{16}] = 1 - \frac{11}{16} = \frac{5}{16}$.

(B) For a Poisson distribution,the mean is equal to the variance.
Given that mean $= l$ and variance $= m$,we have $l = m$.
Given $l + m = 8$,substituting $l = m$ gives $2l = 8$,so $l = 4$ and $m = 4$.
We need to evaluate $e^4[1 - P(X > 2)]$.
Since $1 - P(X > 2) = P(X \leq 2)$,we have:
$e^4[P(X = 0) + P(X = 1) + P(X = 2)]$.
Using the Poisson probability formula $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$ with $\lambda = 4$:
$P(X = 0) = \frac{e^{-4} \times 4^0}{0!} = e^{-4}$.
$P(X = 1) = \frac{e^{-4} \times 4^1}{1!} = 4e^{-4}$.
$P(X = 2) = \frac{e^{-4} \times 4^2}{2!} = \frac{16e^{-4}}{2} = 8e^{-4}$.
Summing these probabilities:
$P(X \leq 2) = e^{-4}(1 + 4 + 8) = 13e^{-4}$.
Finally,$e^4 \times 13e^{-4} = 13$.

(C) The total number of pages is $600$ and the total number of mistakes is $40$.
The average number of mistakes per page,denoted by $\lambda$,is $\lambda = \frac{40}{600} = \frac{1}{15}$.
The number of errors per page follows a Poisson distribution with parameter $\lambda = \frac{1}{15}$.
The probability that a single page has no mistakes is given by $P(X=0) = \frac{e^{-\lambda} \lambda^0}{0!} = e^{-\lambda} = e^{-1/15}$.
Since we are selecting $10$ pages at random,the probability that all $10$ pages have no mistakes is $(P(X=0))^{10} = (e^{-1/15})^{10} = e^{-10/15} = e^{-2/3}$.

(B) For a Poisson distribution,the probability mass function is given by $P(X=k) = \frac{\lambda^k \cdot e^{-\lambda}}{k!}$.
Given $\lambda = 3$,we need to find $P(|X-3| < 2)$.
The inequality $|X-3| < 2$ implies $-2 < X-3 < 2$,which simplifies to $1 < X < 5$.
Since $X$ is a discrete random variable taking non-negative integer values,the possible values for $X$ are $2, 3, 4$.
Thus,$P(|X-3| < 2) = P(X=2) + P(X=3) + P(X=4)$.
Substituting the values into the formula:
$P(X=2) = \frac{3^2 \cdot e^{-3}}{2!} = \frac{9}{2} e^{-3}$
$P(X=3) = \frac{3^3 \cdot e^{-3}}{3!} = \frac{27}{6} e^{-3} = \frac{9}{2} e^{-3}$
$P(X=4) = \frac{3^4 \cdot e^{-3}}{4!} = \frac{81}{24} e^{-3} = \frac{27}{8} e^{-3}$
Summing these probabilities:
$P(|X-3| < 2) = e^{-3} \left( \frac{9}{2} + \frac{9}{2} + \frac{27}{8} \right) = e^{-3} \left( 9 + \frac{27}{8} \right) = e^{-3} \left( \frac{72+27}{8} \right) = \frac{99}{8 e^3}$.

(C) The expected value $E(X)$ of a discrete random variable $X$ is calculated using the formula $E(X) = \sum x_i P(x_i)$.
Given values of $X$ are $10, 20, 30, 40$ and their respective probabilities are $0.3, 0.3, 0.2, 0.2$.
Substituting these values into the formula:
$E(X) = (10 \times 0.3) + (20 \times 0.3) + (30 \times 0.2) + (40 \times 0.2)$
$E(X) = 3 + 6 + 6 + 8$
$E(X) = 23$
Therefore,the expected value of $X$ is $23$.

(D) Given the probability distribution table:

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$K$	$K + \frac{1}{7}$	$2K$	$\frac{2}{5}$

Since the sum of all probabilities must be $1$,we have:
$K + (K + \frac{1}{7}) + 2K + \frac{2}{5} = 1$
$4K + \frac{5 + 14}{35} = 1$
$4K + \frac{19}{35} = 1$
$4K = 1 - \frac{19}{35} = \frac{16}{35}$
$K = \frac{4}{35}$
Now,substituting $K$ back into the table:
$P(X=0) = \frac{4}{35}$
$P(X=1) = \frac{4}{35} + \frac{5}{35} = \frac{9}{35}$
$P(X=2) = 2 \times \frac{4}{35} = \frac{8}{35}$
$P(X=3) = \frac{2}{5} = \frac{14}{35}$
The mean $\mu = E(X) = \sum x_i P(x_i)$:
$\mu = 0 \times \frac{4}{35} + 1 \times \frac{9}{35} + 2 \times \frac{8}{35} + 3 \times \frac{14}{35}$
$\mu = 0 + \frac{9}{35} + \frac{16}{35} + \frac{42}{35}$
$\mu = \frac{67}{35}$

(C) We know that for any probability distribution,the sum of all probabilities must be equal to $1$.
Given $P(X=x) = \frac{c^x}{x!}$ for $x = 1, 2, 3, \ldots$.
Therefore,$\sum_{x=1}^{\infty} P(X=x) = 1$.
Substituting the given expression: $\sum_{x=1}^{\infty} \frac{c^x}{x!} = 1$.
We know the Taylor series expansion for the exponential function is $e^c = \sum_{x=0}^{\infty} \frac{c^x}{x!} = 1 + \frac{c}{1!} + \frac{c^2}{2!} + \frac{c^3}{3!} + \ldots$.
Thus,$\sum_{x=1}^{\infty} \frac{c^x}{x!} = e^c - 1$.
Setting this equal to $1$: $e^c - 1 = 1$.
$e^c = 2$.
Taking the natural logarithm on both sides: $c = \ln(2)$.

(B) Given that $X$ follows a Poisson distribution with parameter $\lambda$. The probability mass function is given by $P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$ for $x = 0, 1, 2, \dots$.
Given the equation: $3 P(X=4) = \frac{1}{2} P(X=2) + P(X=0)$.
Substituting the formula: $3 \frac{e^{-\lambda} \lambda^4}{4!} = \frac{1}{2} \frac{e^{-\lambda} \lambda^2}{2!} + \frac{e^{-\lambda} \lambda^0}{0!}$.
Dividing both sides by $e^{-\lambda}$: $\frac{3 \lambda^4}{24} = \frac{\lambda^2}{4} + 1$.
Simplifying: $\frac{\lambda^4}{8} = \frac{\lambda^2}{4} + 1$.
Multiplying by $8$: $\lambda^4 = 2 \lambda^2 + 8$,which gives $\lambda^4 - 2 \lambda^2 - 8 = 0$.
Let $u = \lambda^2$. Then $u^2 - 2u - 8 = 0$.
Factoring the quadratic: $(u - 4)(u + 2) = 0$.
This gives $u = 4$ or $u = -2$.
Since $\lambda^2 = u$ and $\lambda^2$ must be non-negative,we have $\lambda^2 = 4$,which implies $\lambda = 2$ (as $\lambda > 0$).
The mean of a Poisson distribution is $\lambda$,so the mean is $2$.

(D) Given,the mean of the Poisson distribution is $\lambda = \frac{1}{3}$.
For a Poisson distribution,the probability mass function is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.
For $k=1$,$P(X=1) = \frac{e^{-1/3} (1/3)^1}{1!} = \frac{1}{3} e^{-1/3}$.
For $k=2$,$P(X=2) = \frac{e^{-1/3} (1/3)^2}{2!} = \frac{1}{2} \times \frac{1}{9} e^{-1/3} = \frac{1}{18} e^{-1/3}$.
Now,the ratio $P(X=1) : P(X=2) = \frac{\frac{1}{3} e^{-1/3}}{\frac{1}{18} e^{-1/3}} = \frac{1/3}{1/18} = \frac{18}{3} = 6$.
Thus,the ratio is $6 : 1$.

(C) Given,$P(X=x) = K(x+1)\left(\frac{1}{5}\right)^x$.
Since the sum of all probabilities in a probability distribution is $1$,we have $\sum_{x=0}^{\infty} P(X=x) = 1$.
Substituting the given expression:
$\sum_{x=0}^{\infty} K(x+1)\left(\frac{1}{5}\right)^x = 1$
$K \left[ 1 + 2\left(\frac{1}{5}\right) + 3\left(\frac{1}{5}\right)^2 + 4\left(\frac{1}{5}\right)^3 + \ldots \right] = 1$.
This is an arithmetico-geometric series of the form $\sum_{n=1}^{\infty} n r^{n-1} = (1-r)^{-2}$ where $r = \frac{1}{5}$.
Thus,$K(1 - \frac{1}{5})^{-2} = 1$.
$K(\frac{4}{5})^{-2} = 1 \Rightarrow K(\frac{5}{4})^2 = 1$.
$K(\frac{25}{16}) = 1 \Rightarrow K = \frac{16}{25}$.
Now,$P(X=0) = K(0+1)\left(\frac{1}{5}\right)^0 = K(1)(1) = K$.
Therefore,$P(X=0) = \frac{16}{25}$.

(A) $X$ is a Poisson variate.
Given $P(X=2) = P(X=3)$.
Using the formula $P(X=r) = \frac{e^{-\lambda} \cdot \lambda^r}{r!}$,we have:
$\frac{e^{-\lambda} \cdot \lambda^2}{2!} = \frac{e^{-\lambda} \cdot \lambda^3}{3!}$
$\Rightarrow \frac{\lambda^2}{2} = \frac{\lambda^3}{6}$
$\Rightarrow \lambda = \frac{6}{2} = 3$.
Now,we calculate $P(X=4)$:
$P(X=4) = \frac{e^{-\lambda} \cdot \lambda^4}{4!} = \frac{e^{-3} \cdot 3^4}{4!}$.
Finally,we find $e^3 P(X=4)$:
$e^3 P(X=4) = e^3 \cdot \frac{e^{-3} \cdot 3^4}{4!} = \frac{3^4}{4!} = \frac{81}{24} = \frac{27}{8} = \left(\frac{3}{2}\right)^3$.

(C) Given the probability distribution function $P(X=x)=K\left(\frac{2}{5}\right)^x$ for $x=1, 2, 3, \ldots$.
We know that the sum of all probabilities in a distribution must be equal to $1$,i.e.,$\sum_{x=1}^{\infty} P(X=x) = 1$.
Substituting the given function,we get $K \sum_{x=1}^{\infty} \left(\frac{2}{5}\right)^x = 1$.
This is an infinite geometric series with the first term $a = \frac{2}{5}$ and common ratio $r = \frac{2}{5}$.
The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$.
Therefore,$K \left( \frac{2/5}{1 - 2/5} \right) = 1$.
$K \left( \frac{2/5}{3/5} \right) = 1$.
$K \left( \frac{2}{3} \right) = 1$.
Thus,$K = \frac{3}{2}$.

(A) The sample space $S$ for tossing three fair coins is given by:
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
The total number of outcomes is $n(S) = 8$.
$X$ is a random variable representing the number of heads.
We need to find $P(X = 2)$,which is the probability of getting exactly two heads.
The outcomes with exactly two heads are $\{HHT, HTH, THH\}$.
Thus,the number of favorable outcomes is $n(X = 2) = 3$.
Therefore,$P(X = 2) = \frac{n(X = 2)}{n(S)} = \frac{3}{8}$.