Probability distribution Questions in English

(B) The mathematical expectation $E(X)$ is calculated as the sum of the products of each outcome and its corresponding probability: $E(X) = \sum x_i p_i$.
Given data:
- Clear: $x = 300, p = 0.50$
- Fair: $x = 250, p = 0.30$
- Dubious: $x = 150, p = 0.15$
- Rainy: $x = 100, p = 0.05$
Calculation:
$E(X) = (300 \times 0.50) + (250 \times 0.30) + (150 \times 0.15) + (100 \times 0.05)$
$E(X) = 150 + 75 + 22.5 + 5$
$E(X) = 252.5$
Thus,the mathematical expectation is ₹ $252.5$.

(C) The probability distribution is given as:
$P(X=1)=0.15, P(X=2)=0.23, P(X=3)=0.12, P(X=4)=0.10, P(X=5)=0.20, P(X=6)=0.08, P(X=7)=0.07, P(X=8)=0.05$.
Event $E = \{X \text{ is a prime number}\} = \{2, 3, 5, 7\}$.
$P(E) = P(X=2) + P(X=3) + P(X=5) + P(X=7) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62$.
Event $F = \{X < 4\} = \{1, 2, 3\}$.
$P(F) = P(X=1) + P(X=2) + P(X=3) = 0.15 + 0.23 + 0.12 = 0.50$.
Event $E \cap F = \{X \text{ is a prime number and } X < 4\} = \{2, 3\}$.
$P(E \cap F) = P(X=2) + P(X=3) = 0.23 + 0.12 = 0.35$.
Using the addition theorem of probability:
$P(E \cup F) = P(E) + P(F) - P(E \cap F)$
$P(E \cup F) = 0.62 + 0.50 - 0.35 = 0.77$.

(B) In a game,$3$ coins are tossed. The total number of outcomes is $2^3 = 8$.
$P(\text{getting all heads or all tails}) = P(HHH, TTT) = \frac{2}{8} = \frac{1}{4}$.
$P(\text{getting one head or two heads}) = P(HTT, THT, TTH, HHT, HTH, THH) = \frac{6}{8} = \frac{3}{4}$.
Let $X$ be the random variable representing the amount won or lost.
$P(X = 100) = \frac{1}{4}$.
$P(X = -40) = \frac{3}{4}$.
The expected value $E(X) = \sum x_i p_i$.
$E(X) = 100 \times \frac{1}{4} + (-40) \times \frac{3}{4}$.
$E(X) = 25 - 30 = -5$.
Since the result is negative,the person expects a loss of ₹ $5$ per game.

(C) The probability distribution is given by the table:
| $X=x$ | $1$ | $2$ | $3$ | $\ldots$ | $n$ |
|---|---|---|---|---|---|
| $P(X=x)$ | $\frac{1}{n}$ | $\frac{1}{n}$ | $\frac{1}{n}$ | $\ldots$ | $\frac{1}{n}$ |
$E(X) = \sum x P(X=x) = \frac{1}{n} (1 + 2 + \ldots + n) = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2}$
$E(X^2) = \sum x^2 P(X=x) = \frac{1}{n} (1^2 + 2^2 + \ldots + n^2) = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6}$
$\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2$
$= \frac{(n+1)}{2} \left[ \frac{2n+1}{3} - \frac{n+1}{2} \right] = \frac{(n+1)}{2} \left[ \frac{4n+2 - 3n - 3}{6} \right] = \frac{(n+1)(n-1)}{12} = \frac{n^2-1}{12}$
Given $\frac{\operatorname{Var}(X)}{E(X)} = \frac{4}{1}$,we have:
$\frac{(n^2-1)/12}{(n+1)/2} = 4$
$\frac{(n+1)(n-1)}{12} \cdot \frac{2}{n+1} = 4$
$\frac{n-1}{6} = 4$
$n-1 = 24$
$n = 25$

(C) Let the random variable $X$ denote the number of heads obtained when three fair coins are tossed. The probability distribution of $X$ is given by the following table:
| $X = x$ | $0$ | $1$ | $2$ | $3$ |
| :--- | :--- | :--- | :--- | :--- |
| $P(X = x)$ | $\frac{1}{8}$ | $\frac{3}{8}$ | $\frac{3}{8}$ | $\frac{1}{8}$ |
The expected value $E(X)$ is calculated as:
$E(X) = \sum x P(X=x) = 0 \times \frac{1}{8} + 1 \times \frac{3}{8} + 2 \times \frac{3}{8} + 3 \times \frac{1}{8} = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{12}{8} = \frac{3}{2} = 1.5$
The expected value $E(X^2)$ is calculated as:
$E(X^2) = \sum x^2 P(X=x) = 0^2 \times \frac{1}{8} + 1^2 \times \frac{3}{8} + 2^2 \times \frac{3}{8} + 3^2 \times \frac{1}{8} = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = \frac{24}{8} = 3$
The variance $V(X)$ is given by the formula:
$V(X) = E(X^2) - [E(X)]^2$
$V(X) = 3 - (\frac{3}{2})^2 = 3 - \frac{9}{4} = \frac{12 - 9}{4} = \frac{3}{4} = 0.75$

(D) The given probability mass function is $P(X=x) = \frac{\binom{5}{x}}{2^5}$ for $x \in \{0, 1, 2, 3, 4, 5\}$.
$P(X=0) = \frac{\binom{5}{0}}{32} = \frac{1}{32}$ and $P(X=5) = \frac{\binom{5}{5}}{32} = \frac{1}{32}$. Thus,$P(X=0) = P(X=5)$ is correct.
$P(X \leq 1) = P(X=0) + P(X=1) = \frac{1}{32} + \frac{5}{32} = \frac{6}{32}$.
$P(X \geq 4) = P(X=4) + P(X=5) = \frac{5}{32} + \frac{1}{32} = \frac{6}{32}$. Thus,$P(X \leq 1) = P(X \geq 4)$ is correct.
$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) = \frac{1}{32} + \frac{5}{32} + \frac{10}{32} = \frac{16}{32} = \frac{1}{2}$.
$P(X \geq 3) = P(X=3) + P(X=4) + P(X=5) = \frac{10}{32} + \frac{5}{32} + \frac{1}{32} = \frac{16}{32} = \frac{1}{2}$.
Since $P(X \leq 2) = \frac{1}{2}$ and $P(X \geq 3) = \frac{1}{2}$,the statement $P(X \leq 2) > P(X \geq 3)$ is incorrect.

(D) The probability distribution is given as:
$P(X=1)=0.15, P(X=2)=0.23, P(X=3)=0.10, P(X=4)=0.12, P(X=5)=0.20, P(X=6)=0.08, P(X=7)=0.07, P(X=8)=0.05$.
Event $E = \{ X \text{ is a prime number} \} = \{ 2, 3, 5, 7 \}$.
$P(E) = P(X=2) + P(X=3) + P(X=5) + P(X=7) = 0.23 + 0.10 + 0.20 + 0.07 = 0.60$.
Event $F = \{ X < 4 \} = \{ 1, 2, 3 \}$.
$P(F) = P(X=1) + P(X=2) + P(X=3) = 0.15 + 0.23 + 0.10 = 0.48$.
Event $E \cap F = \{ X \text{ is a prime number and } X < 4 \} = \{ 2, 3 \}$.
$P(E \cap F) = P(X=2) + P(X=3) = 0.23 + 0.10 = 0.33$.
Using the addition rule for probability:
$P(E \cup F) = P(E) + P(F) - P(E \cap F)$
$P(E \cup F) = 0.60 + 0.48 - 0.33 = 0.75$.

(A) We know that the sum of all probabilities in a probability distribution is $1$,i.e.,$\sum P(X=x) = 1$.
Given the distribution:
$k^2 + 2k + k + 2k + 5k^2 = 1$
$6k^2 + 5k - 1 = 0$
$(6k - 1)(k + 1) = 0$
Since $P(X) \geq 0$,we must have $k > 0$,so $k = \frac{1}{6}$.
Now,we need to find $P(X \geq 2)$:
$P(X \geq 2) = P(X=2) + P(X=3) + P(X=4) + P(X=5)$
$P(X \geq 2) = 2k + k + 2k + 5k^2 = 5k + 5k^2$
Substituting $k = \frac{1}{6}$:
$P(X \geq 2) = 5(\frac{1}{6}) + 5(\frac{1}{6})^2$
$P(X \geq 2) = \frac{5}{6} + \frac{5}{36} = \frac{30 + 5}{36} = \frac{35}{36}$.

(C) The sum of all probabilities in a probability distribution is always $1$.
$\therefore k + 0.3 + 0.15 + 0.15 + 0.1 + 2k = 1$
$3k + 0.7 = 1$
$3k = 0.3$
$k = 0.1$
The expected value $E(X)$ is given by $\sum x_i \cdot P(x_i)$:
$E(X) = 0(k) + 1(0.3) + 2(0.15) + 3(0.15) + 4(0.1) + 5(2k)$
$E(X) = 0(0.1) + 0.3 + 0.3 + 0.45 + 0.4 + 5(0.2)$
$E(X) = 0 + 0.3 + 0.3 + 0.45 + 0.4 + 1.0$
$E(X) = 2.45$

(B) Let $X$ denote the number of heads. When three coins are tossed,the sample space is $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$,so the total number of outcomes is $8$.
The possible values of $X$ are $0, 1, 2,$ and $3$.
$P(X=0) = P(\{TTT\}) = \frac{1}{8}$
$P(X=1) = P(\{HTT, THT, TTH\}) = \frac{3}{8}$
$P(X=2) = P(\{HHT, HTH, THH\}) = \frac{3}{8}$
$P(X=3) = P(\{HHH\}) = \frac{1}{8}$
Thus,the probability distribution is:

$X=x$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$

Therefore,Option $(B)$ is the correct answer.

(D) Let $P(X=0)=p_0, P(X=1)=p_1, P(X=2)=p_2, P(X=3)=p_3$.
Given $p_2 = 0.3$ and $p_3 = 2p_1$.
The sum of probabilities is $p_0 + p_1 + p_2 + p_3 = 1$.
Substituting the values,$p_0 + p_1 + 0.3 + 2p_1 = 1 \Rightarrow p_0 + 3p_1 = 0.7 \dots (i)$.
The mean is given by $E(X) = \sum x_i p_i = 0(p_0) + 1(p_1) + 2(p_2) + 3(p_3) = 1.3$.
Substituting the values,$0 + p_1 + 2(0.3) + 3(2p_1) = 1.3$.
$p_1 + 0.6 + 6p_1 = 1.3 \Rightarrow 7p_1 = 0.7 \Rightarrow p_1 = 0.1$.
From equation $(i)$,$p_0 + 3(0.1) = 0.7 \Rightarrow p_0 + 0.3 = 0.7 \Rightarrow p_0 = 0.4$.
Thus,$P(X=0) = 0.4$.

(A) Since the sum of all probabilities in a probability distribution is $1$,we have:
$\sum_{x=0}^7 P(X=x) = 1$
$0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1$
$10k^2 + 9k - 1 = 0$
$(10k - 1)(k + 1) = 0$
Since $k \geq 0$,we have $k = \frac{1}{10}$.
Now,we need to find $P(X \geq 6)$:
$P(X \geq 6) = P(X=6) + P(X=7)$
$P(X \geq 6) = 2k^2 + (7k^2 + k) = 9k^2 + k$
Substituting $k = \frac{1}{10}$:
$P(X \geq 6) = 9\left(\frac{1}{10}\right)^2 + \frac{1}{10} = \frac{9}{100} + \frac{10}{100} = \frac{19}{100}$.

(B) To find $E(X^2)$,we first determine the probability mass function $P(X=x)$ from the cumulative distribution function $F(x)$ using the relation $P(X=x_i) = F(x_i) - F(x_{i-1})$.
$P(X=-1) = F(-1) = 0.3$
$P(X=0) = F(0) - F(-1) = 0.7 - 0.3 = 0.4$
$P(X=1) = F(1) - F(0) = 0.8 - 0.7 = 0.1$
$P(X=2) = F(2) - F(1) = 1 - 0.8 = 0.2$
Now,we calculate the expected value $E(X^2)$ using the formula $E(X^2) = \sum x_i^2 \cdot P(X=x_i)$:
$E(X^2) = (-1)^2(0.3) + (0)^2(0.4) + (1)^2(0.1) + (2)^2(0.2)$
$E(X^2) = (1)(0.3) + (0)(0.4) + (1)(0.1) + (4)(0.2)$
$E(X^2) = 0.3 + 0 + 0.1 + 0.8$
$E(X^2) = 1.2$

(C) We know that for a probability distribution,the sum of all probabilities must be $1$,i.e.,$\sum_{x=0}^{\infty} P(X=x) = 1$.
Substituting the given expression: $k \sum_{x=0}^{\infty} (x+1) (\frac{1}{5})^x = 1$.
Expanding the series: $k [ 1 + 2(\frac{1}{5}) + 3(\frac{1}{5})^2 + 4(\frac{1}{5})^3 + \ldots ] = 1$.
This is an arithmetico-geometric series of the form $\sum_{n=0}^{\infty} (a+nd)r^n = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$ where $a=1, d=1, r=\frac{1}{5}$.
Calculating the sum: $k [ \frac{1}{1 - 1/5} + \frac{1 \times 1/5}{(1 - 1/5)^2} ] = 1$.
$k [ \frac{5}{4} + \frac{1/5}{16/25} ] = k [ \frac{5}{4} + \frac{5}{16} ] = 1$.
$k [ \frac{20+5}{16} ] = 1 \Rightarrow \frac{25k}{16} = 1$.
Therefore,$k = \frac{16}{25}$.

(B) Three fair coins numbered $1$ and $0$ are tossed.
The sample space $S = \{111, 110, 101, 011, 100, 010, 001, 000\}$.
The total number of outcomes $n(S) = 8$.
$X$ represents the sum of the numbers on the uppermost faces.
The probability distribution of $X$ is:

$X$	$0$	$1$	$2$	$3$
$P(X)$	$1/8$	$3/8$	$3/8$	$1/8$

$E(X) = \sum x_i P(x_i) = 0 \times \frac{1}{8} + 1 \times \frac{3}{8} + 2 \times \frac{3}{8} + 3 \times \frac{1}{8} = \frac{0+3+6+3}{8} = \frac{12}{8} = \frac{3}{2}$.
$E(X^2) = \sum x_i^2 P(x_i) = 0^2 \times \frac{1}{8} + 1^2 \times \frac{3}{8} + 2^2 \times \frac{3}{8} + 3^2 \times \frac{1}{8} = \frac{0+3+12+9}{8} = \frac{24}{8} = 3$.
$\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = 3 - (\frac{3}{2})^2 = 3 - \frac{9}{4} = \frac{12-9}{4} = \frac{3}{4} = 0.75$.

(B) We know that the sum of all probabilities in a probability mass function must be equal to $1$.
Thus,$\sum_{r = 0}^n P(X = r) = 1$.
$\frac{^n C_0 + ^n C_1 + ^n C_2 + \dots + ^n C_n}{32} = 1$.
Using the identity $\sum_{r = 0}^n n C_r = 2^n$,we get $\frac{2^n}{32} = 1$.
$2^n = 32$,which implies $2^n = 2^5$,so $n = 5$.
Now,we need to calculate $P[X \leq 2] = P(X = 0) + P(X = 1) + P(X = 2)$.
$P[X \leq 2] = \frac{^5 C_0}{32} + \frac{^5 C_1}{32} + \frac{^5 C_2}{32}$.
Substituting the values: $^5 C_0 = 1$,$^5 C_1 = 5$,and $^5 C_2 = \frac{5 \times 4}{2} = 10$.
$P[X \leq 2] = \frac{1 + 5 + 10}{32} = \frac{16}{32} = \frac{1}{2}$.

(C) Given that $X$ takes values $1, 2, \ldots, n$ with equal probability $P(X) = \frac{1}{n}$.
$E(X) = \sum_{i=1}^{n} x_i p_i = \frac{1}{n} \sum_{i=1}^{n} i = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2}$.
$E(X^2) = \sum_{i=1}^{n} x_i^2 p_i = \frac{1}{n} \sum_{i=1}^{n} i^2 = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6}$.
$\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2$.
Given $\operatorname{Var}(X) = E(X)$,we have:
$\frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4} = \frac{n+1}{2}$.
Dividing by $(n+1)$ (since $n \neq -1$):
$\frac{2n+1}{6} - \frac{n+1}{4} = \frac{1}{2}$.
Multiply by $12$:
$2(2n+1) - 3(n+1) = 6$.
$4n + 2 - 3n - 3 = 6$.
$n - 1 = 6 \implies n = 7$.

(C) When a player tosses $2$ fair coins,the sample space is $S = \{HH, HT, TH, TT\}$.
Let $X$ be a random variable that denotes the amount received by the player.
The possible values for $X$ are $5, 2,$ and $1$.
The probabilities are:
$P(X=5) = P(\{HH\}) = \frac{1}{4}$
$P(X=2) = P(\{HT, TH\}) = \frac{2}{4} = \frac{1}{2}$
$P(X=1) = P(\{TT\}) = \frac{1}{4}$
The probability distribution is:
$E(X) = \sum X P(X) = 5 \times \frac{1}{4} + 2 \times \frac{1}{2} + 1 \times \frac{1}{4} = \frac{5}{4} + 1 + \frac{1}{4} = \frac{10}{4} = 2.5$
$E(X^2) = \sum X^2 P(X) = 5^2 \times \frac{1}{4} + 2^2 \times \frac{1}{2} + 1^2 \times \frac{1}{4} = \frac{25}{4} + 2 + \frac{1}{4} = \frac{34}{4} = 8.5$
$\text{Variance}(X) = E(X^2) - [E(X)]^2 = \frac{34}{4} - \left(\frac{10}{4}\right)^2 = \frac{34}{4} - \frac{100}{16} = \frac{136 - 100}{16} = \frac{36}{16} = \frac{9}{4}$.

(A) For a discrete random variable,the sum of probabilities must be $1$.
$\sum_{x=0}^{6} f(x) = 1$
$k(0^2 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2) = 1$
$k(0 + 1 + 4 + 9 + 16 + 25 + 36) = 1$
$91k = 1 \implies k = \frac{1}{91}$
The cumulative distribution function $F(4)$ is defined as $P(X \leq 4)$.
$F(4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)$
$F(4) = k(0^2 + 1^2 + 2^2 + 3^2 + 4^2)$
$F(4) = k(0 + 1 + 4 + 9 + 16) = 30k$
Substituting $k = \frac{1}{91}$,we get:
$F(4) = 30 \times \frac{1}{91} = \frac{30}{91}$

(C) The random variable $X$ represents the number of sixes in $2$ tosses of a fair die. The possible values for $X$ are $0, 1, 2$.
The probability of getting a six in a single toss is $p = \frac{1}{6}$,and the probability of not getting a six is $q = 1 - \frac{1}{6} = \frac{5}{6}$.
For $X = 0$ (no sixes): $P(X = 0) = q \times q = \frac{5}{6} \times \frac{5}{6} = \frac{25}{36}$.
For $X = 1$ (one six): $P(X = 1) = (p \times q) + (q \times p) = (\frac{1}{6} \times \frac{5}{6}) + (\frac{5}{6} \times \frac{1}{6}) = \frac{5}{36} + \frac{5}{36} = \frac{10}{36} = \frac{5}{18}$.
For $X = 2$ (two sixes): $P(X = 2) = p \times p = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$.
The probability distribution is:

$X = x$	$0$	$1$	$2$
$P(X = x)$	$\frac{25}{36}$	$\frac{5}{18}$	$\frac{1}{36}$

(A) Let $X$ be the random variable representing the number of defective baskets drawn. Since we draw $2$ baskets without replacement,$X$ can take values $0, 1, 2$.
Total baskets = $20$,Defective = $6$,Non-defective = $14$.
$P(X=0) = \frac{14}{20} \times \frac{13}{19} = \frac{182}{380}$
$P(X=1) = \frac{6}{20} \times \frac{14}{19} + \frac{14}{20} \times \frac{6}{19} = \frac{84+84}{380} = \frac{168}{380}$
$P(X=2) = \frac{6}{20} \times \frac{5}{19} = \frac{30}{380}$
The expected value $E(X) = \sum x_i P(x_i) = 0 \times P(X=0) + 1 \times P(X=1) + 2 \times P(X=2)$
$E(X) = 0 + \frac{168}{380} + 2 \times \frac{30}{380} = \frac{168 + 60}{380} = \frac{228}{380} = 0.6$.

(C) fair die is tossed twice in succession. The total number of outcomes is $6 \times 6 = 36$.
Let $X$ be the number of fours in $2$ tosses.
The possible values for $X$ are $0, 1, 2$.
$1$. For $X = 0$: The outcomes are all pairs where neither die shows a $4$. There are $5 \times 5 = 25$ such outcomes. Thus,$P(X = 0) = \frac{25}{36}$.
$2$. For $X = 1$: The outcomes are $(4, \text{not } 4)$ or $(\text{not } 4, 4)$. There are $1 \times 5 + 5 \times 1 = 10$ such outcomes. Thus,$P(X = 1) = \frac{10}{36} = \frac{5}{18}$.
$3$. For $X = 2$: The only outcome is $(4, 4)$. There is $1$ such outcome. Thus,$P(X = 2) = \frac{1}{36}$.
The probability distribution is:

$X = x_i$	$0$	$1$	$2$
$P_i$	$\frac{25}{36}$	$\frac{5}{18}$	$\frac{1}{36}$

(B) The expected value $E(X)$ is defined as $\sum_{x=1}^{n} x \cdot P(X=x)$.
Given $P(X=x) = \frac{2x}{n(n+1)}$ for $x = 1, 2, \ldots, n$.
Thus,$E(X) = \sum_{x=1}^{n} x \cdot \frac{2x}{n(n+1)}$.
$E(X) = \frac{2}{n(n+1)} \sum_{x=1}^{n} x^2$.
Using the formula for the sum of squares of the first $n$ natural numbers,$\sum_{x=1}^{n} x^2 = \frac{n(n+1)(2n+1)}{6}$.
Substituting this into the expression:
$E(X) = \frac{2}{n(n+1)} \cdot \frac{n(n+1)(2n+1)}{6}$.
$E(X) = \frac{2(2n+1)}{6} = \frac{2n+1}{3}$.

(B) Let $X$ be the random variable representing the sum of the numbers on the uppermost faces. The possible values of $X$ are $0, 1, 2, 3$.
Since there are $3$ coins,the total number of outcomes is $2^3 = 8$.
The probability distribution is:
$P(X=0) = \frac{1}{8}$
$P(X=1) = \frac{3}{8}$
$P(X=2) = \frac{3}{8}$
$P(X=3) = \frac{1}{8}$
$E(X) = \sum x_i p_i = 0 \times \frac{1}{8} + 1 \times \frac{3}{8} + 2 \times \frac{3}{8} + 3 \times \frac{1}{8} = \frac{0+3+6+3}{8} = \frac{12}{8} = 1.5$.
$E(X^2) = \sum x_i^2 p_i = 0^2 \times \frac{1}{8} + 1^2 \times \frac{3}{8} + 2^2 \times \frac{3}{8} + 3^2 \times \frac{1}{8} = \frac{0+3+12+9}{8} = \frac{24}{8} = 3$.
$Variance(X) = E(X^2) - [E(X)]^2 = 3 - (1.5)^2 = 3 - 2.25 = 0.75$.

(A) The probability distribution is given by:
$E(X) = \sum_{i=1}^{k} x_i p_i = \frac{1}{k} + \frac{2}{k} + \dots + \frac{k}{k} = \frac{1}{k} \sum_{i=1}^{k} i = \frac{1}{k} \cdot \frac{k(k+1)}{2} = \frac{k+1}{2}$
$E(X^2) = \sum_{i=1}^{k} x_i^2 p_i = \frac{1^2 + 2^2 + \dots + k^2}{k} = \frac{1}{k} \cdot \frac{k(k+1)(2k+1)}{6} = \frac{(k+1)(2k+1)}{6}$
Variance $= E(X^2) - [E(X)]^2$
$= \frac{(k+1)(2k+1)}{6} - \left( \frac{k+1}{2} \right)^2$
$= \frac{2k^2 + 3k + 1}{6} - \frac{k^2 + 2k + 1}{4}$
$= \frac{2(2k^2 + 3k + 1) - 3(k^2 + 2k + 1)}{12}$
$= \frac{4k^2 + 6k + 2 - 3k^2 - 6k - 3}{12}$
$= \frac{k^2 - 1}{12}$

(C) The sum of probabilities in a probability distribution must be equal to $1$.
$\sum P(X=x) = P(X=0) + P(X=1) + P(X=2) = 1$
$(4k - 10k^2) + (5k - 1) + 3k^3 = 1$
$3k^3 - 10k^2 + 9k - 2 = 0$
By testing values,we find that $k = \frac{1}{3}$ is a root:
$3(\frac{1}{27}) - 10(\frac{1}{9}) + 9(\frac{1}{3}) - 2 = \frac{1}{9} - \frac{10}{9} + 3 - 2 = -1 + 1 = 0$
Factoring the polynomial,we get $(k - \frac{1}{3})(3k^2 - 9k + 6) = 0$,which simplifies to $(k - \frac{1}{3})(k - 1)(k - 2) = 0$.
If $k = 1$,$P(X=0) = 4(1) - 10(1)^2 = -6$,which is impossible as probability cannot be negative.
If $k = 2$,$P(X=0) = 4(2) - 10(4) = 8 - 40 = -32$,which is impossible.
Thus,$k = \frac{1}{3}$ is the only valid value.
We need to find $P(X < 2) = P(X=0) + P(X=1)$.
$P(X < 2) = (4k - 10k^2) + (5k - 1) = 9k - 10k^2 - 1$
Substituting $k = \frac{1}{3}$:
$P(X < 2) = 9(\frac{1}{3}) - 10(\frac{1}{9}) - 1 = 3 - \frac{10}{9} - 1 = 2 - \frac{10}{9} = \frac{18 - 10}{9} = \frac{8}{9}$.

(C) We are given the probability density function $f(x) = 3(1 - 2x^2)$ for $0 < x < 1$.
To find $P\left(\frac{1}{4} < x < \frac{1}{3}\right)$,we integrate the function over the given interval:
$P\left(\frac{1}{4} < x < \frac{1}{3}\right) = \int_{\frac{1}{4}}^{\frac{1}{3}} 3(1 - 2x^2) dx$
$= \int_{\frac{1}{4}}^{\frac{1}{3}} (3 - 6x^2) dx$
$= [3x - 2x^3]_{\frac{1}{4}}^{\frac{1}{3}}$
$= \left(3\left(\frac{1}{3}\right) - 2\left(\frac{1}{3}\right)^3\right) - \left(3\left(\frac{1}{4}\right) - 2\left(\frac{1}{4}\right)^3\right)$
$= \left(1 - \frac{2}{27}\right) - \left(\frac{3}{4} - \frac{2}{64}\right)$
$= \left(\frac{25}{27}\right) - \left(\frac{3}{4} - \frac{1}{32}\right)$
$= \frac{25}{27} - \frac{3}{4} + \frac{1}{32}$
Finding a common denominator $(864)$:
$= \frac{25 \times 32 - 3 \times 216 + 1 \times 27}{864}$
$= \frac{800 - 648 + 27}{864} = \frac{179}{864}$

(D) The expected value $E(X)$ is defined as $\sum x_i \cdot P(x_i)$.
Given $P(x) = \frac{2x}{n(n+1)}$ for $x = 1, 2, \ldots, n$.
$E(X) = \sum_{x=1}^{n} x \cdot \frac{2x}{n(n+1)}$
$E(X) = \frac{2}{n(n+1)} \sum_{x=1}^{n} x^2$
Using the formula for the sum of squares of the first $n$ natural numbers,$\sum_{x=1}^{n} x^2 = \frac{n(n+1)(2n+1)}{6}$.
$E(X) = \frac{2}{n(n+1)} \cdot \frac{n(n+1)(2n+1)}{6}$
$E(X) = \frac{2n+1}{3}$

(A) The sample space of the experiment is $S = \{1, 2, 3, 4, 5, 6\}$.
We determine the number of factors $(X)$ for each outcome:
$X(1) = 1$ (factor: $1$)
$X(2) = 2$ (factors: $1, 2$)
$X(3) = 2$ (factors: $1, 3$)
$X(4) = 3$ (factors: $1, 2, 4$)
$X(5) = 2$ (factors: $1, 5$)
$X(6) = 4$ (factors: $1, 2, 3, 6$)
Now,we calculate the probabilities for each value of $X$:
$P(X = 1) = P(\{1\}) = 1/6$
$P(X = 2) = P(\{2, 3, 5\}) = 3/6 = 1/2$
$P(X = 3) = P(\{4\}) = 1/6$
$P(X = 4) = P(\{6\}) = 1/6$
Thus,the probability distribution is as shown in option $A$.

(B) For a probability distribution,the sum of all probabilities must be equal to $1$.
Therefore,$\sum P(X) = 1$.
$K + \frac{1}{6} + \frac{3}{8} + 2K + \frac{1}{12} = 1$
Combine the terms with $K$ and the constant fractions:
$3K + (\frac{1}{6} + \frac{3}{8} + \frac{1}{12}) = 1$
Find a common denominator for the fractions,which is $24$:
$3K + (\frac{4}{24} + \frac{9}{24} + \frac{2}{24}) = 1$
$3K + \frac{15}{24} = 1$
$3K + \frac{5}{8} = 1$
$3K = 1 - \frac{5}{8}$
$3K = \frac{3}{8}$
$K = \frac{1}{8}$

(D) When a coin is tossed twice,the sample space $S$ is given by $S = \{HH, HT, TH, TT\}$.
Here,$X$ represents the number of tails.
For $X=0$ (no tails),the outcome is $\{HH\}$,so $P(X=0) = 1/4$.
For $X=1$ (one tail),the outcomes are $\{HT, TH\}$,so $P(X=1) = 2/4 = 1/2$.
For $X=2$ (two tails),the outcome is $\{TT\}$,so $P(X=2) = 1/4$.
The probability distribution is:

$X=x_i$	$0$	$1$	$2$
$P_i$	$1/4$	$1/2$	$1/4$

This matches option $D$.

(A) For a probability distribution,the sum of all probabilities must be $1$.
$\sum P(X) = k + 3k + 5k + 7k + 9k + 11k + 13k = 49k = 1$
Therefore,$k = \frac{1}{49}$.
We need to find $P(X \ge 2)$.
$P(X \ge 2) = 1 - P(X < 2)$
$P(X < 2) = P(X = 0) + P(X = 1) = k + 3k = 4k$
$P(X \ge 2) = 1 - 4k = 1 - 4(\frac{1}{49}) = 1 - \frac{4}{49} = \frac{45}{49}$.

(C) The total number of ways to select $2$ numbers from the first $6$ positive integers is $^6C_2 = \frac{6 \times 5}{2} = 15$.
Let $X$ be the random variable representing the larger of the two numbers. The possible values for $X$ are $2, 3, 4, 5, 6$.
The probability distribution is as follows:
For $X=2$: Pairs are $(1,2)$,so $P(X=2) = \frac{1}{15}$.
For $X=3$: Pairs are $(1,3), (2,3)$,so $P(X=3) = \frac{2}{15}$.
For $X=4$: Pairs are $(1,4), (2,4), (3,4)$,so $P(X=4) = \frac{3}{15}$.
For $X=5$: Pairs are $(1,5), (2,5), (3,5), (4,5)$,so $P(X=5) = \frac{4}{15}$.
For $X=6$: Pairs are $(1,6), (2,6), (3,6), (4,6), (5,6)$,so $P(X=6) = \frac{5}{15}$.
Calculating $E(X) = \sum x_i P_i = 2(\frac{1}{15}) + 3(\frac{2}{15}) + 4(\frac{3}{15}) + 5(\frac{4}{15}) + 6(\frac{5}{15}) = \frac{2+6+12+20+30}{15} = \frac{70}{15} = \frac{14}{3}$.
Calculating $E(X^2) = \sum x_i^2 P_i = 4(\frac{1}{15}) + 9(\frac{2}{15}) + 16(\frac{3}{15}) + 25(\frac{4}{15}) + 36(\frac{5}{15}) = \frac{4+18+48+100+180}{15} = \frac{350}{15} = \frac{70}{3}$.
$\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = \frac{70}{3} - (\frac{14}{3})^2 = \frac{70}{3} - \frac{196}{9} = \frac{210-196}{9} = \frac{14}{9}$.

(A) For a probability distribution,the sum of all probabilities must be equal to $1$.
$\sum P(x) = k + 0.3 + 0.15 + 0.15 + 0.1 + 2k = 1$
$3k + 0.7 = 1$
$3k = 0.3$
$k = 0.1$
Now,the expected value $E(x)$ is calculated as $\sum x_i P(x_i)$:
$E(x) = (0 \times k) + (1 \times 0.3) + (2 \times 0.15) + (3 \times 0.15) + (4 \times 0.1) + (5 \times 2k)$
Substitute $k = 0.1$:
$E(x) = (0 \times 0.1) + (1 \times 0.3) + (2 \times 0.15) + (3 \times 0.15) + (4 \times 0.1) + (5 \times 0.2)$
$E(x) = 0 + 0.3 + 0.3 + 0.45 + 0.4 + 1.0$
$E(x) = 2.45$

(D) The random variable $X$ takes values $1, 2, \ldots, n$ with probability $p_i = \frac{1}{n}$ for each $i$.
The expected value $E(X) = \sum p_i x_i = \frac{1}{n} \sum_{i=1}^{n} i = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2}$.
The variance $V(X) = E(X^2) - [E(X)]^2 = \sum p_i x_i^2 - [E(X)]^2$.
$E(X^2) = \frac{1}{n} \sum_{i=1}^{n} i^2 = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6}$.
$V(X) = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2 = \frac{(n+1)}{2} \left[ \frac{2n+1}{3} - \frac{n+1}{2} \right] = \frac{(n+1)}{2} \left[ \frac{4n+2-3n-3}{6} \right] = \frac{(n+1)(n-1)}{12} = \frac{n^2-1}{12}$.
Given the ratio $\frac{V(X)}{E(X)} = 4$,we have $\frac{(n^2-1)/12}{(n+1)/2} = 4$.
$\frac{(n-1)(n+1)}{12} \cdot \frac{2}{n+1} = 4$.
$\frac{n-1}{6} = 4 \Rightarrow n-1 = 24 \Rightarrow n = 25$.

(A) The sample space for tossing $2$ coins is $\{HH, HT, TH, TT\}$.
The probabilities are:
$P(X=5) = P(HH) = \frac{1}{4}$
$P(X=2) = P(HT, TH) = \frac{2}{4} = \frac{1}{2}$
$P(X=1) = P(TT) = \frac{1}{4}$
We calculate the mean $E(X) = \sum p_i x_i = (5 \times \frac{1}{4}) + (2 \times \frac{1}{2}) + (1 \times \frac{1}{4}) = \frac{5}{4} + 1 + \frac{1}{4} = \frac{6}{4} + 1 = \frac{3}{2} + 1 = \frac{5}{2}$.
Next,we calculate $E(X^2) = \sum p_i x_i^2 = (5^2 \times \frac{1}{4}) + (2^2 \times \frac{1}{2}) + (1^2 \times \frac{1}{4}) = \frac{25}{4} + 2 + \frac{1}{4} = \frac{26}{4} + 2 = \frac{13}{2} + 2 = \frac{17}{2}$.
Variance $Var(X) = E(X^2) - [E(X)]^2 = \frac{17}{2} - (\frac{5}{2})^2 = \frac{17}{2} - \frac{25}{4} = \frac{34-25}{4} = \frac{9}{4}$.

(C) For a function to be a probability mass function (p.m.f.),the sum of all probabilities must be equal to $1$.
$\sum_{x=0}^{3} P(X=x) = 1$
$P(0) + P(1) + P(2) + P(3) = 1$
$K \cdot \frac{2^0}{0!} + K \cdot \frac{2^1}{1!} + K \cdot \frac{2^2}{2!} + K \cdot \frac{2^3}{3!} = 1$
$K \cdot (\frac{1}{1} + \frac{2}{1} + \frac{4}{2} + \frac{8}{6}) = 1$
$K \cdot (1 + 2 + 2 + \frac{4}{3}) = 1$
$K \cdot (5 + \frac{4}{3}) = 1$
$K \cdot (\frac{15+4}{3}) = 1$
$K \cdot \frac{19}{3} = 1$
$K = \frac{3}{19}$

(B) The variance of a random variable $X$ is given by $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$.
First,we calculate $E(X) = \sum_{x=1}^{10} x P(X=x) = \frac{1}{10} (1 + 2 + 3 + \ldots + 10) = \frac{1}{10} \times \frac{10 \times 11}{2} = 5.5$.
Next,we calculate $E(X^2) = \sum_{x=1}^{10} x^2 P(X=x) = \frac{1}{10} (1^2 + 2^2 + 3^2 + \ldots + 10^2) = \frac{1}{10} \times \frac{10 \times 11 \times 21}{6} = \frac{231}{6} = 38.5$.
Finally,$\operatorname{Var}(X) = 38.5 - (5.5)^2 = 38.5 - 30.25 = 8.25$.
Since $8.25 = \frac{33}{4}$,the correct option is $B$.

(B) The expected value $E(X)$ of a random variable $X$ is given by the formula $E(X) = \sum p_i x_i$.
Using the provided table:
$E(X) = (0 \times 0.35) + (500 \times 0.25) + (1000 \times 0.15) + (1500 \times 0.10) + (2000 \times 0.08) + (2500 \times 0.05) + (3000 \times 0.02)$
Calculating each term:
$E(X) = 0 + 125 + 150 + 150 + 160 + 125 + 60$
Summing the values:
$E(X) = 770$
Thus,the expected value of the maintenance cost is Rs. $770$.

(D) The sum of all probabilities in a probability 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 $k \geq 0$,we have $k = \frac{1}{10}$.
The cumulative distribution function $F(4)$ is defined as $P(X \leq 4)$.
$F(4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$
$F(4) = 0 + k + 2k + 2k + 3k = 8k$
Substituting $k = \frac{1}{10}$:
$F(4) = 8 \times \frac{1}{10} = \frac{8}{10} = \frac{4}{5}$.

(D) Given the p.m.f $P(X=x) = \frac{1}{32} \binom{5}{x}$ for $x \in \{0, 1, 2, 3, 4, 5\}$.
$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$
$= \frac{1}{32} \left[ \binom{5}{0} + \binom{5}{1} + \binom{5}{2} \right] = \frac{1}{32} (1 + 5 + 10) = \frac{16}{32} = \frac{1}{2}$.
$P(X \geq 3) = P(X=3) + P(X=4) + P(X=5)$
$= \frac{1}{32} \left[ \binom{5}{3} + \binom{5}{4} + \binom{5}{5} \right] = \frac{1}{32} (10 + 5 + 1) = \frac{16}{32} = \frac{1}{2}$.
Since $P(X \leq 2) = \frac{1}{2}$ and $P(X \geq 3) = \frac{1}{2}$,it follows that $P(X \leq 2) = P(X \geq 3)$.

(D) To find the variance of the probability distribution,we use the formula: $\text{Variance} = E(X^2) - [E(X)]^2$,where $E(X) = \sum p_i x_i$ and $E(X^2) = \sum p_i x_i^2$.
First,calculate $E(X)$:
$E(X) = (0 \times \frac{9}{16}) + (1 \times \frac{3}{8}) + (2 \times \frac{1}{16}) = 0 + \frac{3}{8} + \frac{2}{16} = \frac{6}{16} + \frac{2}{16} = \frac{8}{16} = \frac{1}{2}$.
Next,calculate $E(X^2)$:
$E(X^2) = (0^2 \times \frac{9}{16}) + (1^2 \times \frac{3}{8}) + (2^2 \times \frac{1}{16}) = 0 + \frac{3}{8} + \frac{4}{16} = \frac{6}{16} + \frac{4}{16} = \frac{10}{16} = \frac{5}{8}$.
Now,calculate the variance:
$\text{Variance} = E(X^2) - [E(X)]^2 = \frac{5}{8} - (\frac{1}{2})^2 = \frac{5}{8} - \frac{1}{4} = \frac{5}{8} - \frac{2}{8} = \frac{3}{8}$.

(B) The sum of probabilities in a probability distribution must be $1$,i.e.,$\sum P(X = x_i) = 1$.
Given $P(X=0) = \frac{5}{16}$,$P(X=1) = \frac{5}{16}$,$P(X=2) = \frac{2k}{48}$,and $P(X=3) = \frac{1}{4} = \frac{12}{48}$.
Summing these: $\frac{5}{16} + \frac{5}{16} + \frac{2k}{48} + \frac{12}{48} = 1$.
Converting all to denominator $48$: $\frac{15}{48} + \frac{15}{48} + \frac{2k}{48} + \frac{12}{48} = 1$.
$\frac{42 + 2k}{48} = 1 \Rightarrow 42 + 2k = 48 \Rightarrow 2k = 6 \Rightarrow k = 3$.
Now,the probability distribution is:
$P(X=0) = \frac{15}{48}$,$P(X=1) = \frac{15}{48}$,$P(X=2) = \frac{6}{48} = \frac{1}{8}$,$P(X=3) = \frac{12}{48}$.
The expected value $E(X) = \sum x_i P(x_i) = (0 \times \frac{15}{48}) + (1 \times \frac{15}{48}) + (2 \times \frac{6}{48}) + (3 \times \frac{12}{48})$.
$E(X) = 0 + \frac{15}{48} + \frac{12}{48} + \frac{36}{48} = \frac{63}{48} = \frac{21}{16} = 1.3125$.

(A) The expected profit per cake is calculated using the formula for the expected value: $E(X) = \sum p_i x_i$.
Here,$x_i$ represents the profit for each type of cake and $p_i$ represents the probability (demand) of each type.
Given values:
$x_1 = 2, p_1 = 0.20$
$x_2 = 2.5, p_2 = 0.05$
$x_3 = 3, p_3 = 0.10$
$x_4 = 1.5, p_4 = 0.50$
$x_5 = 1, p_5 = 0.15$
Expected Profit $= (2 \times 0.20) + (2.5 \times 0.05) + (3 \times 0.10) + (1.5 \times 0.50) + (1 \times 0.15)$
$= 0.4 + 0.125 + 0.3 + 0.75 + 0.15$
$= 1.725$
Therefore,the expected profit per cake is $Rs \ 1.725$.

(B) For a probability distribution,the sum of all probabilities must be $1$.
$\sum P(X = x) = k + 3k + 5k + 7k + 8k + k = 1$
$25k = 1 \Rightarrow k = \frac{1}{25}$
We need to find $P(2 \leq X < 5)$,which is $P(X = 2) + P(X = 3) + P(X = 4)$.
$P(2 \leq X < 5) = 3k + 5k + 7k = 15k$
Substituting $k = \frac{1}{25}$:
$P(2 \leq X < 5) = 15 \times \frac{1}{25} = \frac{15}{25} = \frac{3}{5}$

(C) The cumulative distribution function $F(x)$ is defined as $F(x) = P(X \le x)$.
To find the probability mass function $P(X=x)$,we use the relation $P(X=x) = F(x) - F(x-1)$.
For $x=4$,$P(X=4) = F(4) - F(3) = 0.62 - 0.48 = 0.14$.
For $x=5$,$P(X=5) = F(5) - F(4) = 0.85 - 0.62 = 0.23$.
Therefore,$P(X=4) + P(X=5) = 0.14 + 0.23 = 0.37$.

(D) For a probability distribution,the sum of all probabilities must be equal to $1$.
$\sum P(X) = K + 2K + 3K + 4K + 5K + 6K = 1$
$21K = 1 \Rightarrow K = \frac{1}{21}$
We need to find $P(2 < X < 6)$,which corresponds to $P(X=3) + P(X=4) + P(X=5)$.
$P(2 < X < 6) = 3K + 4K + 5K = 12K$
Substituting the value of $K = \frac{1}{21}$:
$P(2 < X < 6) = 12 \times \frac{1}{21} = \frac{12}{21} = \frac{4}{7}$

(B) The cumulative distribution function $F(x)$ is defined as $F(x) = P(X \leq x)$.
Therefore,$F(0) = P(X \leq 0) = P(X = -2) + P(X = -1) + P(X = 0)$.
From the given table:
$P(X = -2) = 0.2$
$P(X = -1) = 0.3$
$P(X = 0) = 0.15$
So,$F(0) = 0.2 + 0.3 + 0.15 = 0.65$.
Alternatively,we know that the sum of all probabilities is $1$.
$P(X \leq 0) + P(X > 0) = 1$
$F(0) = 1 - P(X > 0)$.
Thus,$F(0) = 1 - (P(X = 1) + P(X = 2)) = 1 - (0.25 + 0.1) = 1 - 0.35 = 0.65$.

(A) The sum of all probabilities in a probability distribution must be equal to $1$.
$k + 2k + 3k + 4k + 4k + 3k + 2k + k + k = 1$
$21k = 1$
$k = \frac{1}{21}$
We need to find $P(3 < X \leq 6)$,which is $P(X=4) + P(X=5) + P(X=6)$.
From the table:
$P(X=4) = 4k$
$P(X=5) = 3k$
$P(X=6) = 2k$
Therefore,$P(3 < X \leq 6) = 4k + 3k + 2k = 9k$.
Substituting the value of $k = \frac{1}{21}$:
$P(3 < X \leq 6) = 9 \times \frac{1}{21} = \frac{9}{21} = \frac{3}{7}$.