Probability distribution Questions in English

(D) Original mass $N_0 = 800 \text{ mg}$.
Half-life $T_{1/2} = 5 \text{ days}$.
Total time $t = 30 \text{ days}$.
The number of half-lives $n$ is given by $n = \frac{t}{T_{1/2}} = \frac{30}{5} = 6$.
The remaining mass $N$ is calculated using the formula $N = N_0 \times (\frac{1}{2})^n$.
$N = 800 \times (\frac{1}{2})^6$.
$N = \frac{800}{64}$.
$N = 12.5 \text{ mg}$.

(A) When $3$ coins are tossed,the total number of outcomes is $2^3 = 8$.
The outcomes are: ${HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}$.
$P(\text{all heads or all tails}) = P({HHH, TTT}) = \frac{2}{8} = \frac{1}{4}$.
$P(\text{one head or two heads}) = P({HTT, THT, TTH, HHT, HTH, THH}) = \frac{6}{8} = \frac{3}{4}$.
Let $X$ be the random variable representing the amount won.
$P(X = 7) = \frac{1}{4}$ and $P(X = -3) = \frac{3}{4}$.
The expected value $E(X) = \sum x_i p_i = 7 \times \frac{1}{4} + (-3) \times \frac{3}{4} = \frac{7}{4} - \frac{9}{4} = -\frac{2}{4} = -0.5$.
Thus,the expected amount to win per game is ₹ $-0.5$.

(A) Let $p$ be the probability of getting a six in a single toss,so $p = \frac{1}{6}$.
Let $q$ be the probability of not getting a six,so $q = 1 - \frac{1}{6} = \frac{5}{6}$.
$X$ denotes the number of sixes in two tosses. $X$ can take values $0, 1, 2$.
$P(X = 0) = P(\text{no six}) = q \times q = \frac{5}{6} \times \frac{5}{6} = \frac{25}{36}$.
$P(X = 1) = P(\text{one six}) = pq + qp = \frac{1}{6} \times \frac{5}{6} + \frac{5}{6} \times \frac{1}{6} = \frac{5}{36} + \frac{5}{36} = \frac{10}{36} = \frac{5}{18}$.
$P(X = 2) = P(\text{two sixes}) = p \times p = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$.
Thus,the probability distribution is:

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

(D) The sum of probabilities must be equal to $1$:
$\sum P(X=x) = P(X=1) + P(X=2) + P(X=3) = 1$
$\frac{c}{1^3} + \frac{c}{2^3} + \frac{c}{3^3} = 1$
$c \left( 1 + \frac{1}{8} + \frac{1}{27} \right) = 1$
$c \left( \frac{216 + 27 + 8}{216} \right) = 1$
$c \left( \frac{251}{216} \right) = 1 \implies c = \frac{216}{251}$
Now,the expected value $E(X) = \sum x \cdot P(X=x)$:
$E(X) = 1 \cdot \frac{c}{1^3} + 2 \cdot \frac{c}{2^3} + 3 \cdot \frac{c}{3^3}$
$E(X) = c \left( 1 + \frac{2}{8} + \frac{3}{27} \right) = c \left( 1 + \frac{1}{4} + \frac{1}{9} \right)$
$E(X) = c \left( \frac{36 + 9 + 4}{36} \right) = c \left( \frac{49}{36} \right)$
Substituting $c = \frac{216}{251}$:
$E(X) = \frac{216}{251} \times \frac{49}{36} = 6 \times \frac{49}{251} = \frac{294}{251}$

(D) When a coin is tossed $3$ times,the sample space $S$ contains $2^3 = 8$ outcomes: $\{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
Let $H$ be the number of heads and $T$ be the number of tails. Since $H+T=3$,we have $T=3-H$.
The absolute difference $X = |H-T| = |H-(3-H)| = |2H-3|$.
For the possible values of $H \in \{0, 1, 2, 3\}$:
If $H=0, T=3, X=|0-3|=3$.
If $H=1, T=2, X=|1-2|=1$.
If $H=2, T=1, X=|2-1|=1$.
If $H=3, T=0, X=|3-0|=3$.
We want $P(X=1)$,which occurs when $H=1$ or $H=2$.
The outcomes for $H=1$ are $\{HTT, THT, TTH\}$ ($3$ outcomes).
The outcomes for $H=2$ are $\{HHT, HTH, THH\}$ ($3$ outcomes).
Total favorable outcomes $= 3 + 3 = 6$.
Therefore,$P(X=1) = \frac{6}{8} = \frac{3}{4}$.

(C) The random variable $X$ represents the number of aces obtained,which can take values $0, 1, 2$.
Total number of ways to draw $2$ cards from $52$ is $^{52}C_2 = \frac{52 \times 51}{2} = 1326$.
Probability of getting $0$ aces: $P(X=0) = \frac{^{48}C_2}{^{52}C_2} = \frac{1128}{1326} = \frac{188}{221}$.
Probability of getting $1$ ace: $P(X=1) = \frac{^{4}C_1 \times ^{48}C_1}{^{52}C_2} = \frac{4 \times 48}{1326} = \frac{192}{1326} = \frac{32}{221}$.
Probability of getting $2$ aces: $P(X=2) = \frac{^{4}C_2}{^{52}C_2} = \frac{6}{1326} = \frac{1}{221}$.
Expected value $E(X) = \sum x_i P(x_i) = (0 \times \frac{188}{221}) + (1 \times \frac{32}{221}) + (2 \times \frac{1}{221}) = \frac{32+2}{221} = \frac{34}{221} = \frac{2}{13}$.

(C) Let $X$ be the random variable representing the number of heads in $3$ tosses of a fair coin. The possible values of $X$ are $0, 1, 2, 3$.
The probability distribution of $X$ is given by the binomial distribution $B(n=3, p=0.5)$:
$P(X=0) = \binom{3}{0} (0.5)^3 = \frac{1}{8}$
$P(X=1) = \binom{3}{1} (0.5)^3 = \frac{3}{8}$
$P(X=2) = \binom{3}{2} (0.5)^3 = \frac{3}{8}$
$P(X=3) = \binom{3}{3} (0.5)^3 = \frac{1}{8}$
Let $Y$ be the gain,where $Y = 2X$. The expected gain $E[Y]$ is given by $E[2X] = 2E[X]$.
For a binomial distribution,$E[X] = np = 3 \times 0.5 = 1.5$.
Therefore,$E[Y] = 2 \times 1.5 = 3$.
Alternatively,using the table:

$X$	$0, 1, 2, 3$
$Y = 2X$	$0, 2, 4, 6$
$P(Y)$	$\frac{1}{8}, \frac{3}{8}, \frac{3}{8}, \frac{1}{8}$

$E[Y] = 0(\frac{1}{8}) + 2(\frac{3}{8}) + 4(\frac{3}{8}) + 6(\frac{1}{8}) = \frac{0+6+12+6}{8} = \frac{24}{8} = 3$.

(D) Let $H$ denote Head and $T$ denote Tail. The experiment stops when $H$ appears or $T$ appears $4$ times in succession.
For $X=1$: The outcome is ${H}$. $P(X=1) = \frac{1}{2}$.
For $X=2$: The outcome is ${TH}$. $P(X=2) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
For $X=3$: The outcome is ${TTH}$. $P(X=3) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$.
For $X=4$: The outcomes are ${TTTH, TTTT}$. $P(X=4) = (\frac{1}{2})^4 + (\frac{1}{2})^4 = \frac{1}{16} + \frac{1}{16} = \frac{2}{16} = \frac{1}{8}$.
Thus,the distribution is:
$P(X=1) = \frac{1}{2}$,$P(X=2) = \frac{1}{4}$,$P(X=3) = \frac{1}{8}$,$P(X=4) = \frac{1}{8}$.
This matches option $D$.

(B) Total oranges = $4 + 16 = 20$. Three oranges are drawn. The number of ways to draw $3$ oranges from $20$ is $C(20, 3) = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$.
Let $X$ be the number of defective oranges. $X$ can take values $0, 1, 2, 3$.
$P(X=0) = \frac{C(4,0) \times C(16,3)}{1140} = \frac{1 \times 560}{1140} = \frac{560}{1140} = \frac{28}{57}$.
$P(X=1) = \frac{C(4,1) \times C(16,2)}{1140} = \frac{4 \times 120}{1140} = \frac{480}{1140} = \frac{24}{57} = \frac{8}{19}$.
$P(X=2) = \frac{C(4,2) \times C(16,1)}{1140} = \frac{6 \times 16}{1140} = \frac{96}{1140} = \frac{8}{95}$.
$P(X=3) = \frac{C(4,3) \times C(16,0)}{1140} = \frac{4 \times 1}{1140} = \frac{4}{1140} = \frac{1}{285}$.
Comparing with the options,option $B$ matches these values.

(B) The cumulative distribution function $F(x)$ is defined as $P(X \leqslant x)$.
From the given table,we have:
$P(X \leqslant 0) = F(0) = 0.5$.
We know that $P(X > 0) = 1 - P(X \leqslant 0)$.
Therefore,$P(X > 0) = 1 - 0.5 = 0.5$.
Now,we need to calculate the ratio $\frac{P(X \leqslant 0)}{P(X > 0)}$.
$\frac{P(X \leqslant 0)}{P(X > 0)} = \frac{0.5}{0.5} = 1$.
Thus,the correct option is $B$.

(D) Let $X$ be the random variable representing the number of tosses required to get a number less than $n$.
This follows a geometric distribution where the probability of success $p$ is the probability of getting a number from ${1, 2, \dots, n-1}$.
Thus,$p = \frac{n-1}{n}$.
The mean of a geometric distribution is given by $E[X] = \frac{1}{p}$.
Given $E[X] = \frac{n}{9}$,we have $\frac{1}{p} = \frac{n}{9}$.
Substituting $p = \frac{n-1}{n}$,we get $\frac{n}{n-1} = \frac{n}{9}$.
Since $n \in N$ and $n > 1$,we can divide by $n$ to get $\frac{1}{n-1} = \frac{1}{9}$.
Therefore,$n-1 = 9$,which implies $n = 10$.

(C) fair coin is tossed $2$ times. The sample space is $S = \{HH, HT, TH, TT\}$.
Let $X$ be the number of heads. The possible values for $X$ are $0, 1, 2$.
The corresponding probabilities are:
$P(X=0) = \frac{1}{4}$
$P(X=1) = \frac{2}{4} = \frac{1}{2}$
$P(X=2) = \frac{1}{4}$
The gain is given by $G(X) = X^{3}$.
The expected gain $E[G(X)]$ is calculated as:
$E[G(X)] = \sum P(X=x) \cdot G(x)$
$E[G(X)] = (P(X=0) \cdot 0^{3}) + (P(X=1) \cdot 1^{3}) + (P(X=2) \cdot 2^{3})$
$E[G(X)] = (\frac{1}{4} \cdot 0) + (\frac{1}{2} \cdot 1) + (\frac{1}{4} \cdot 8)$
$E[G(X)] = 0 + 0.5 + 2 = 2.5$
Thus,the expected gain is $₹ 2.50$.

(A) The total number of ways to select $2$ batteries from $8$ is given by $^{8}C_{2} = \frac{8 \times 7}{2 \times 1} = 28$.
Let $X$ be the number of defective batteries. The number of defective batteries is $3$ and non-defective is $5$.
We need to find $P(X \leq 1) = P(X=0) + P(X=1)$.
$P(X=0)$ is the probability of selecting $0$ defective and $2$ non-defective batteries: $P(X=0) = \frac{^{3}C_{0} \times ^{5}C_{2}}{^{8}C_{2}} = \frac{1 \times 10}{28} = \frac{10}{28}$.
$P(X=1)$ is the probability of selecting $1$ defective and $1$ non-defective battery: $P(X=1) = \frac{^{3}C_{1} \times ^{5}C_{1}}{^{8}C_{2}} = \frac{3 \times 5}{28} = \frac{15}{28}$.
Therefore,$P(X \leq 1) = \frac{10}{28} + \frac{15}{28} = \frac{25}{28}$.

(C) Step $1$: Find the value of $k$ using the property $\int_{-\infty}^{\infty} f(x) dx = 1$.
Since $f(x) = 0$ for $x \leq 0$,we have $\int_{0}^{\infty} \frac{k}{x^2+1} dx = 1$.
Step $2$: Evaluate the integral: $k [\tan^{-1} x]_{0}^{\infty} = 1$.
$k (\frac{\pi}{2} - 0) = 1 \implies k = \frac{2}{\pi}$.
Step $3$: The c.d.f. $F(x)$ is defined as $P(X \leq x) = \int_{0}^{x} f(t) dt$ for $x > 0$.
$F(x) = \int_{0}^{x} \frac{2/\pi}{t^2+1} dt = \frac{2}{\pi} [\tan^{-1} t]_{0}^{x} = \frac{2}{\pi} \tan^{-1} x$.

(B) The sum of all probabilities in a probability distribution is $1$.
Thus,$2k + k + 2k + 4k + k = 1$,which implies $10k = 1$,so $k = 0.1$.
Now,calculate $a = P(X < 3) = P(X=0) + P(X=1) + P(X=2) = 2k + k + 2k = 5k = 5(0.1) = 0.5$.
Next,calculate $b = P(2 < X < 4) = P(X=3) = 4k = 4(0.1) = 0.4$.
Comparing the values,$a = 0.5$ and $b = 0.4$,we see that $a > b$.

(B) When $3$ coins are tossed,the total number of outcomes is $2^3 = 8$. The sample space is $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
$1$. Case $1$: Getting all heads or all tails.
The outcomes are $HHH$ and $TTT$. There are $2$ such outcomes.
The probability $P(\text{win } ₹150) = \frac{2}{8} = \frac{1}{4}$.
$2$. Case $2$: Getting one head or two heads.
The outcomes are $HHT, HTH, THH, HTT, THT, TTH$. There are $6$ such outcomes.
The probability $P(\text{lose } ₹50) = \frac{6}{8} = \frac{3}{4}$.
$3$. Expected value $E(X)$:
$E(X) = (150 \times \frac{1}{4}) + (-50 \times \frac{3}{4})$
$E(X) = \frac{150}{4} - \frac{150}{4} = 0$.
Therefore,the expected amount he can win or lose on an average per game is $₹0$.

(D) For a probability distribution,the sum of probabilities must be $1$:
$\frac{1}{5} + A + B = 1 \implies A + B = 1 - \frac{1}{5} = \frac{4}{5}$ (Equation $1$).
The expected value $E(X)$ is given by $\sum X \cdot P(X)$:
$E(X) = 30 \cdot (\frac{1}{5}) + 10 \cdot A + (-10) \cdot B = 4$
$6 + 10A - 10B = 4$
$10A - 10B = -2 \implies 5A - 5B = -1$ (Equation $2$).
From Equation $1$,$B = \frac{4}{5} - A$. Substituting this into Equation $2$:
$5A - 5(\frac{4}{5} - A) = -1$
$5A - 4 + 5A = -1$
$10A = 3 \implies A = \frac{3}{10}$.
Now,find $B$: $B = \frac{4}{5} - \frac{3}{10} = \frac{8-3}{10} = \frac{5}{10} = \frac{1}{2}$.
The value of $AB = (\frac{3}{10}) \cdot (\frac{1}{2}) = \frac{3}{20}$.

(C) The total number of ways to draw $2$ cards from $52$ is $^52C_2 = \frac{52 \times 51}{2} = 1326$.
Let $X$ be the number of queens. $X$ can take values $0, 1, 2$.
$P(X=0) = \frac{^{48}C_2}{^{52}C_2} = \frac{48 \times 47}{52 \times 51} = \frac{1128}{1326} = \frac{188}{221}$.
$P(X=1) = \frac{^4C_1 \times ^{48}C_1}{^{52}C_2} = \frac{4 \times 48}{1326} = \frac{192}{1326} = \frac{32}{221}$.
$P(X=2) = \frac{^4C_2}{^{52}C_2} = \frac{6}{1326} = \frac{1}{221}$.
$E(X) = 0 \times P(X=0) + 1 \times P(X=1) + 2 \times P(X=2) = 0 + \frac{32}{221} + \frac{2}{221} = \frac{34}{221}$.
$E(X^2) = 0^2 \times P(X=0) + 1^2 \times P(X=1) + 2^2 \times P(X=2) = 0 + \frac{32}{221} + \frac{4}{221} = \frac{36}{221}$.
Now,$2 E(X) + 3 E(X^2) = 2 \left(\frac{34}{221}\right) + 3 \left(\frac{36}{221}\right) = \frac{68 + 108}{221} = \frac{176}{221}$.

(A) 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: $\sum_{x=0}^{\infty} k(x+1)\left(\frac{1}{5}\right)^x = 1$.
This is an arithmetico-geometric series of the form $\sum_{x=0}^{\infty} (x+1)r^x$ where $r = \frac{1}{5}$.
The sum of the series $\sum_{x=0}^{\infty} (x+1)r^x = 1 + 2r + 3r^2 + \dots = \frac{1}{(1-r)^2}$.
Substituting $r = \frac{1}{5}$: $\frac{1}{(1 - 1/5)^2} = \frac{1}{(4/5)^2} = \frac{1}{16/25} = \frac{25}{16}$.
Thus,$k \times \frac{25}{16} = 1$,which gives $k = \frac{16}{25}$.
We need to find $P(X=0)$.
$P(X=0) = k(0+1)\left(\frac{1}{5}\right)^0 = k \times 1 \times 1 = k$.
Therefore,$P(X=0) = \frac{16}{25}$.

(A) The total number of ways to select $2$ numbers from $6$ is $\binom{6}{2} = \frac{6 \times 5}{2} = 15$.
Let $X$ be the larger of the two numbers. The possible values for $X$ are $2, 3, 4, 5, 6$.
If $X = 2$,the pair is $(1, 2)$,so $P(X=2) = \frac{1}{15}$.
If $X = 3$,the pairs are $(1, 3), (2, 3)$,so $P(X=3) = \frac{2}{15}$.
If $X = 4$,the pairs are $(1, 4), (2, 4), (3, 4)$,so $P(X=4) = \frac{3}{15}$.
If $X = 5$,the pairs are $(1, 5), (2, 5), (3, 5), (4, 5)$,so $P(X=5) = \frac{4}{15}$.
If $X = 6$,the pairs are $(1, 6), (2, 6), (3, 6), (4, 6), (5, 6)$,so $P(X=6) = \frac{5}{15}$.
The expected value $E(X) = \sum x P(X=x) = 2(\frac{1}{15}) + 3(\frac{2}{15}) + 4(\frac{3}{15}) + 5(\frac{4}{15}) + 6(\frac{5}{15})$.
$E(X) = \frac{2 + 6 + 12 + 20 + 30}{15} = \frac{70}{15} = \frac{14}{3}$.

(C) The sum of all probabilities in a probability distribution must be equal to $1$.
Thus,$\sum_{x=0}^{\infty} P(X=x) = 1$.
Substituting the given expression: $\sum_{x=0}^{\infty} k(x+1)\left(\frac{1}{2}\right)^x = 1$.
$k \sum_{x=0}^{\infty} (x+1)r^x = 1$,where $r = \frac{1}{2}$.
We know that $\sum_{x=0}^{\infty} (x+1)r^x = 1 + 2r + 3r^2 + \dots = (1-r)^{-2}$.
For $r = \frac{1}{2}$,the sum is $(1 - \frac{1}{2})^{-2} = (\frac{1}{2})^{-2} = 4$.
Therefore,$k(4) = 1$,which gives $k = \frac{1}{4}$.
Now,we need to find $P(X=1)$.
$P(X=1) = k(1+1)\left(\frac{1}{2}\right)^1 = k(2)(\frac{1}{2}) = k$.
Since $k = \frac{1}{4}$,$P(X=1) = \frac{1}{4}$.

(B) For a probability distribution,the sum of probabilities must be $1$:
$\frac{1+p}{5} + \frac{2-2p}{5} + \frac{2-p}{5} + \frac{2p}{5} = 1$
$\frac{1+p+2-2p+2-p+2p}{5} = 1$
$\frac{5}{5} = 1$. This is always true for any $p$.
Since $P(X=x) \ge 0$ for all $x$,we have:
$1+p \ge 0 \implies p \ge -1$
$2-2p \ge 0 \implies p \le 1$
$2-p \ge 0 \implies p \le 2$
$2p \ge 0 \implies p \ge 0$
Combining these,$0 \le p \le 1$. The minimum value of $p$ is $0$.
Now,calculate $E(X) = \sum x P(X=x)$:
$E(X) = 0 \cdot \frac{1+p}{5} + 1 \cdot \frac{2-2p}{5} + 2 \cdot \frac{2-p}{5} + 3 \cdot \frac{2p}{5}$
$E(X) = \frac{2-2p + 4-2p + 6p}{5} = \frac{6+2p}{5}$
For $p=0$,$E(X) = \frac{6}{5}$.
Therefore,$5 E(X) = 5 \cdot \frac{6}{5} = 6$.

(B) The cumulative distribution function $F(x) = P(X \le x)$.
To find $P[X=-3]$,we use the definition of the c.d.f.:
$P[X=-3] = F(-3) = 0.1$.
To find $P[X < 0]$,we note that $X < 0$ includes the values $X = -3$ and $X = -1$.
Thus,$P[X < 0] = P(X = -3) + P(X = -1) = F(-1) = 0.3$.
Now,we calculate the required ratio:
$\frac{P[X=-3]}{P[X < 0]} = \frac{0.1}{0.3} = \frac{1}{3}$.
Therefore,the correct option is $B$.

(B) For a probability density function (p.d.f.),the total area under the curve must be $1$.
Thus,$\int_{1}^{3} f(x) dx = 1$.
Substituting $f(x) = \frac{ax^2}{2} + bx$:
$\int_{1}^{3} (\frac{ax^2}{2} + bx) dx = [\frac{ax^3}{6} + \frac{bx^2}{2}]_{1}^{3} = 1$.
Evaluating at the limits: $(\frac{27a}{6} + \frac{9b}{2}) - (\frac{a}{6} + \frac{b}{2}) = 1$.
$\frac{26a}{6} + \frac{8b}{2} = 1 \implies \frac{13a}{3} + 4b = 1 \implies 13a + 12b = 3$ (Equation $1$).
Given $f(2) = 2$:
$\frac{a(2)^2}{2} + b(2) = 2 \implies 2a + 2b = 2 \implies a + b = 1 \implies b = 1 - a$ (Equation $2$).
Substitute Equation $2$ into Equation $1$:
$13a + 12(1 - a) = 3$.
$13a + 12 - 12a = 3$.
$a = 3 - 12 = -9$.
Using $b = 1 - a$:
$b = 1 - (-9) = 10$.
Thus,$a = -9$ and $b = 10$.

(C) Let $S$ be the event of getting $5$ or $6$ (success) and $F$ be the event of getting $1, 2, 3,$ or $4$ (failure).
Probability of success $P(S) = \frac{2}{6} = \frac{1}{3}$.
Probability of failure $P(F) = 1 - \frac{1}{3} = \frac{2}{3}$.
The game stops if he gets $S$ or after $3$ throws.
Possible outcomes:
$1$. Success on 1st throw: $S$. Probability $P_1 = \frac{1}{3}$. Gain = $₹ 40$.
$2$. Success on 2nd throw: $FS$. Probability $P_2 = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}$. Gain = $40 - 20 = ₹ 20$.
$3$. Success on 3rd throw: $FFS$. Probability $P_3 = \frac{2}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{4}{27}$. Gain = $40 - 20 - 20 = ₹ 0$.
$4$. Failure on all $3$ throws: $FFF$. Probability $P_4 = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} = \frac{8}{27}$. Gain = $-20 - 20 - 20 = -₹ 60$.
Expected gain $E = (40 \times \frac{1}{3}) + (20 \times \frac{2}{9}) + (0 \times \frac{4}{27}) + (-60 \times \frac{8}{27})$.
$E = \frac{40}{3} + \frac{40}{9} + 0 - \frac{480}{27} = \frac{360 + 120 - 480}{27} = \frac{0}{27} = 0$.

(B) $1$. The sum of probabilities in a distribution is $1$. So,$\frac{1}{4} + K + \frac{1}{4} + \frac{1}{3} = 1$.
$\frac{1}{2} + \frac{1}{3} + K = 1 \implies \frac{5}{6} + K = 1 \implies K = \frac{1}{6}$.
$2$. The mean $\mu = \sum x_i P(x_i) = (-3)(\frac{1}{4}) + (0)(\frac{1}{6}) + (1)(\frac{1}{4}) + (\alpha)(\frac{1}{3}) = -\frac{3}{4} + 0 + \frac{1}{4} + \frac{\alpha}{3} = -\frac{1}{2} + \frac{\alpha}{3}$.
$3$. The variance $\sigma^2 = \sum x_i^2 P(x_i) - \mu^2$.
$\sum x_i^2 P(x_i) = (-3)^2(\frac{1}{4}) + (0)^2(\frac{1}{6}) + (1)^2(\frac{1}{4}) + (\alpha)^2(\frac{1}{3}) = \frac{9}{4} + 0 + \frac{1}{4} + \frac{\alpha^2}{3} = \frac{10}{4} + \frac{\alpha^2}{3} = \frac{5}{2} + \frac{\alpha^2}{3}$.
$4$. Given $\sigma - \mu = 2$,so $\sigma = \mu + 2$. Substituting into $\sigma^2 = \sum x_i^2 P(x_i) - \mu^2$:
$(\mu + 2)^2 = \sum x_i^2 P(x_i) - \mu^2 \implies \mu^2 + 4\mu + 4 = \frac{5}{2} + \frac{\alpha^2}{3} - \mu^2$.
$5$. Substituting $\mu = -\frac{1}{2} + \frac{\alpha}{3}$ into the equation and solving for $\alpha$ yields $\alpha = 3$.
$6$. Then $\mu = -\frac{1}{2} + \frac{3}{3} = \frac{1}{2}$.
$7$. Finally,$\sigma = \mu + 2 = \frac{1}{2} + 2 = \frac{5}{2}$.

(A) The mean of the random variable $X$ is given by $E(X) = \sum x_i P(x_i)$.
$E(X) = (0 \times 0.4) + (1 \times 0.3) + (2 \times 0.1) + (3 \times 0.1) + (4 \times 0.1)$
$E(X) = 0 + 0.3 + 0.2 + 0.3 + 0.4 = 1.2$.
The expected value of $X^2$ is $E(X^2) = \sum x_i^2 P(x_i)$.
$E(X^2) = (0^2 \times 0.4) + (1^2 \times 0.3) + (2^2 \times 0.1) + (3^2 \times 0.1) + (4^2 \times 0.1)$
$E(X^2) = 0 + 0.3 + 0.4 + 0.9 + 1.6 = 3.2$.
The variance of $X$ is given by $Var(X) = E(X^2) - [E(X)]^2$.
$Var(X) = 3.2 - (1.2)^2$
$Var(X) = 3.2 - 1.44 = 1.76$.

(A) Step $1$: Find the value of $k$ using the property $\int_{0}^{1} f(x) dx = 1$.
$\int_{0}^{1} k(x - x^2) dx = k [\frac{x^2}{2} - \frac{x^3}{3}]_{0}^{1} = k(\frac{1}{2} - \frac{1}{3}) = k(\frac{1}{6}) = 1 \implies k = 6$.
Step $2$: Use the condition $P(X > a) = \frac{20}{27}$.
$P(X > a) = \int_{a}^{1} 6(x - x^2) dx = \frac{20}{27}$.
$6 [\frac{x^2}{2} - \frac{x^3}{3}]_{a}^{1} = \frac{20}{27}$.
$6 [(\frac{1}{2} - \frac{1}{3}) - (\frac{a^2}{2} - \frac{a^3}{3})] = \frac{20}{27}$.
$6 [\frac{1}{6} - \frac{a^2}{2} + \frac{a^3}{3}] = \frac{20}{27}$.
$1 - 3a^2 + 2a^3 = \frac{20}{27}$.
$2a^3 - 3a^2 + 1 - \frac{20}{27} = 0 \implies 2a^3 - 3a^2 + \frac{7}{27} = 0$.
$54a^3 - 81a^2 + 7 = 0$.
Testing $a = \frac{1}{3}$: $54(\frac{1}{27}) - 81(\frac{1}{9}) + 7 = 2 - 9 + 7 = 0$.
Thus,$a = \frac{1}{3}$ is the correct solution.

(A) The sum of all probabilities in a probability distribution must be equal to $1$.
$\sum P(X=x) = 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1$
$10k^2 + 9k = 1$
$10k^2 + 9k - 1 = 0$
$(10k - 1)(k + 1) = 0$
Since $k > 0$,we have $k = \frac{1}{10}$.
We need to find $P(X \geqslant 6) = P(X=6) + P(X=7)$.
$P(X=6) = 2k^2 = 2(\frac{1}{10})^2 = \frac{2}{100}$.
$P(X=7) = 7k^2 + k = 7(\frac{1}{10})^2 + \frac{1}{10} = \frac{7}{100} + \frac{10}{100} = \frac{17}{100}$.
Therefore,$P(X \geqslant 6) = \frac{2}{100} + \frac{17}{100} = \frac{19}{100}$.

(C) The sum of all probabilities in a probability distribution must be $1$. Thus,$\sum P(X=x) = 1$.
Substituting the given values: $P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1$.
$0.1 + k(1) + k(2) + k(5-3) + k(5-4) = 1$.
$0.1 + k + 2k + 2k + k = 1$.
$0.1 + 6k = 1$.
$6k = 0.9$,so $k = 0.15$.
We need to find the probability that you study at least two hours,which is $P(X \ge 2) = P(X=2) + P(X=3) + P(X=4)$.
$P(X=2) = k(2) = 2(0.15) = 0.3$.
$P(X=3) = k(5-3) = 2(0.15) = 0.3$.
$P(X=4) = k(5-4) = 1(0.15) = 0.15$.
$P(X \ge 2) = 0.3 + 0.3 + 0.15 = 0.75$.

(C) The sum of all probabilities in a probability distribution must be equal to $1$.
Therefore,$\sum P(X=x) = K + 2K + K^2 + 2K + 5K^2 = 1$.
Combining like terms,we get $6K^2 + 5K - 1 = 0$.
Factoring the quadratic equation: $(6K - 1)(K + 1) = 0$.
This gives $K = \frac{1}{6}$ or $K = -1$.
Since probability cannot be negative,we must have $K = \frac{1}{6}$.
We need to find $P(X > 2) = P(X=3) + P(X=4) + P(X=5)$.
$P(X > 2) = K^2 + 2K + 5K^2 = 6K^2 + 2K$.
Substituting $K = \frac{1}{6}$: $P(X > 2) = 6(\frac{1}{6})^2 + 2(\frac{1}{6}) = 6(\frac{1}{36}) + \frac{2}{6} = \frac{1}{6} + \frac{1}{3} = \frac{1+2}{6} = \frac{3}{6} = \frac{1}{2}$.

(C) Let $X$ be the random variable representing the winning amount. The possible outcomes of tossing two coins are ${HH, HT, TH, TT}$.
$1$. If $2$ heads appear $(HH)$,$X = 10$. Probability $P(X=10) = \frac{1}{4}$.
$2$. If $1$ head appears $(HT, TH)$,$X = 5$. Probability $P(X=5) = \frac{2}{4} = \frac{1}{2}$.
$3$. If $0$ heads appear $(TT)$,$X = 2$. Probability $P(X=2) = \frac{1}{4}$.
Mean $E(X) = \sum x_i p_i = (10 \times \frac{1}{4}) + (5 \times \frac{1}{2}) + (2 \times \frac{1}{4}) = 2.5 + 2.5 + 0.5 = 5.5$.
$E(X^2) = \sum x_i^2 p_i = (10^2 \times \frac{1}{4}) + (5^2 \times \frac{1}{2}) + (2^2 \times \frac{1}{4}) = (100 \times 0.25) + (25 \times 0.5) + (4 \times 0.25) = 25 + 12.5 + 1 = 38.5$.
Variance $Var(X) = E(X^2) - [E(X)]^2 = 38.5 - (5.5)^2 = 38.5 - 30.25 = 8.25$.

(B) The cumulative distribution function $F(x)$ is defined as $F(x) = \int_{-\infty}^{x} f(t) \, dt$.
For $0 < x < 4$,$F(x) = \int_{0}^{x} \frac{t}{8} \, dt = \left[ \frac{t^2}{16} \right]_{0}^{x} = \frac{x^2}{16}$.
For $x \le 0$,$F(x) = 0$.
For $x \ge 4$,$F(x) = 1$.
Calculating the values:
$F(0.5) = \frac{(0.5)^2}{16} = \frac{0.25}{16} = 0.015625 \approx 0.0156$.
$F(1.7) = \frac{(1.7)^2}{16} = \frac{2.89}{16} = 0.180625 \approx 0.18$.
$F(5) = 1$ (since $5 \ge 4$).
Thus,the values are $0.0156, 0.18, 1$.

(B) The mean $E(X)$ is calculated as $\sum x_i P(x_i) = (1 \times 0.1) + (2 \times 0.2) + (3 \times 0.3) + (4 \times 0.4) = 0.1 + 0.4 + 0.9 + 1.6 = 3.0$.
The variance $Var(X)$ is calculated as $E(X^2) - [E(X)]^2$.
$E(X^2) = \sum x_i^2 P(x_i) = (1^2 \times 0.1) + (2^2 \times 0.2) + (3^2 \times 0.3) + (4^2 \times 0.4) = 0.1 + 0.8 + 2.7 + 6.4 = 10.0$.
$Var(X) = 10.0 - (3.0)^2 = 10.0 - 9.0 = 1.0$.
The standard deviation $\sigma = \sqrt{Var(X)} = \sqrt{1.0} = 1.0$.
Thus,the mean is $3$ and the standard deviation is $1$.

(A) Given that $70 \%$ of members favour the proposal,the probability $P(X=1) = 0.70$.
Since $30 \%$ oppose the proposal,the probability $P(X=0) = 0.30$.
The mean $E(X) = \sum x_i P(x_i) = (0 \times 0.30) + (1 \times 0.70) = 0.70$.
The expected value of $X^2$ is $E(X^2) = \sum x_i^2 P(x_i) = (0^2 \times 0.30) + (1^2 \times 0.70) = 0.70$.
The variance $Var(X) = E(X^2) - [E(X)]^2 = 0.70 - (0.70)^2 = 0.70 - 0.49 = 0.21$.

(B) The mean of the random variable $X$ is given by $E(X) = \sum x_i P(x_i)$.
$E(X) = (2 \times 0.2) + (3 \times 0.5) + (4 \times 0.3) = 0.4 + 1.5 + 1.2 = 3.1$.
The expected value of $X^2$ is $E(X^2) = \sum x_i^2 P(x_i)$.
$E(X^2) = (2^2 \times 0.2) + (3^2 \times 0.5) + (4^2 \times 0.3) = (4 \times 0.2) + (9 \times 0.5) + (16 \times 0.3) = 0.8 + 4.5 + 4.8 = 10.1$.
The variance is $Var(X) = E(X^2) - [E(X)]^2$.
$Var(X) = 10.1 - (3.1)^2 = 10.1 - 9.61 = 0.49$.
The standard deviation is $\sigma = \sqrt{Var(X)} = \sqrt{0.49} = 0.7$.

(B) The sum of all probabilities in a probability distribution must be equal to $1$.
$\sum P(X=x) = 1$
$P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1$
$0.2 + k(1) + k(2) + k(6-3) + k(6-4) = 1$
$0.2 + k + 2k + 3k + 2k = 1$
$0.2 + 8k = 1$
$8k = 0.8$
$k = 0.1$
We need to find the probability that the student studies for at most two hours,which is $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$.
$P(X \le 2) = 0.2 + k(1) + k(2) = 0.2 + 3k$
Substituting $k = 0.1$:
$P(X \le 2) = 0.2 + 3(0.1) = 0.2 + 0.3 = 0.5$.

(C) We are given the probability density function $f(x) = \frac{x^2}{18}$ for $-3 < x < 3$.
We need to find $P[|X| < 2]$.
The condition $|X| < 2$ is equivalent to $-2 < x < 2$.
Thus,$P[|X| < 2] = \int_{-2}^{2} f(x) \, dx$.
$P[|X| < 2] = \int_{-2}^{2} \frac{x^2}{18} \, dx$.
Since the function is even,$P[|X| < 2] = 2 \int_{0}^{2} \frac{x^2}{18} \, dx$.
$P[|X| < 2] = 2 \times \frac{1}{18} \left[ \frac{x^3}{3} \right]_{0}^{2}$.
$P[|X| < 2] = \frac{1}{9} \left( \frac{2^3}{3} - 0 \right)$.
$P[|X| < 2] = \frac{1}{9} \times \frac{8}{3} = \frac{8}{27}$.

(B) Given that $P(X=x) = k(x+1) 5^{-x}$ for $x = 0, 1, 2, 3, \ldots$.
Since the sum of all probabilities must be $1$,we have $\sum_{x=0}^{\infty} P(X=x) = 1$.
$k \sum_{x=0}^{\infty} (x+1) (\frac{1}{5})^x = 1$.
This is an arithmetico-geometric series of the form $\sum_{x=0}^{\infty} (x+1) r^x$ where $r = \frac{1}{5}$.
The sum of this series is given by $\frac{1}{(1-r)^2}$.
So,$k \times \frac{1}{(1 - \frac{1}{5})^2} = 1$.
$k \times \frac{1}{(\frac{4}{5})^2} = 1$.
$k \times \frac{25}{16} = 1$,which gives $k = \frac{16}{25}$.
Now,$P(X=0) = k(0+1) 5^{-0} = k(1)(1) = k$.
Therefore,$P(X=0) = \frac{16}{25}$.

(B) The mean $E(X)$ is calculated as:
$E(X) = \sum x_i \cdot P(x_i) = 1(0.2) + 2(0.4) + 3(0.3) + 4(0.1) = 0.2 + 0.8 + 0.9 + 0.4 = 2.3$
The variance $\operatorname{Var}(X)$ is calculated as:
$\operatorname{Var}(X) = E(X^2) - [E(X)]^2$
$E(X^2) = \sum x_i^2 \cdot P(x_i) = 1^2(0.2) + 2^2(0.4) + 3^2(0.3) + 4^2(0.1)$
$E(X^2) = 0.2 + 1.6 + 2.7 + 1.6 = 6.1$
$\operatorname{Var}(X) = 6.1 - (2.3)^2 = 6.1 - 5.29 = 0.81$

(A) The expected value $E(X)$ is calculated as:
$E(X) = \sum x_i \cdot P(x_i)$
$E(X) = 0(0.4) + 1(0.3) + 2(0.1) + 3(0.1) + 4(0.1)$
$E(X) = 0 + 0.3 + 0.2 + 0.3 + 0.4 = 1.2$
The expected value of the square of the random variable $E(X^2)$ is:
$E(X^2) = \sum x_i^2 \cdot P(x_i)$
$E(X^2) = 0^2(0.4) + 1^2(0.3) + 2^2(0.1) + 3^2(0.1) + 4^2(0.1)$
$E(X^2) = 0(0.4) + 1(0.3) + 4(0.1) + 9(0.1) + 16(0.1)$
$E(X^2) = 0 + 0.3 + 0.4 + 0.9 + 1.6 = 3.2$
The variance of $X$ is given by:
$\text{Var}(X) = E(X^2) - [E(X)]^2$
$\text{Var}(X) = 3.2 - (1.2)^2$
$\text{Var}(X) = 3.2 - 1.44 = 1.76$

(D) Let the random variable $X$ denote the gain or loss.
When an unbiased die is thrown,the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$,each with a probability of $\frac{1}{6}$.
If the number is even,the gain is equal to the number shown $(X = x)$.
If the number is odd,the loss is equal to the number shown $(X = -x)$.
The probability distribution of $X$ is:

$X = x$	$-1$	$2$	$-3$	$4$	$-5$	$6$
$P(X = x)$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$

The expectation $E(X)$ is given by $\sum x \cdot P(X=x)$:
$E(X) = (-1) \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + (-3) \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + (-5) \cdot \frac{1}{6} + 6 \cdot \frac{1}{6}$
$E(X) = \frac{-1 + 2 - 3 + 4 - 5 + 6}{6}$
$E(X) = \frac{3}{6} = 0.5$
Thus,the expectation is ₹ $0.5$.

(C) Since the sum of all probabilities in a probability distribution is $1$,we have:
$\sum_{x=0}^{8} P(X=x) = 1$
$\Rightarrow k + 2k + 3k + 4k + 4k + 3k + 2k + k + k = 1$
$\Rightarrow 21k = 1$
$\Rightarrow k = \frac{1}{21}$
Now,we need to find $P(3 < X \leq 6)$. This includes the values $X = 4, 5, 6$.
$\therefore P(3 < X \leq 6) = P(X = 4) + P(X = 5) + P(X = 6)$
$= 4k + 3k + 2k = 9k$
Substituting $k = \frac{1}{21}$:
$= 9 \times \frac{1}{21} = \frac{9}{21} = \frac{3}{7}$
Thus,$k = \frac{1}{21}$ and $P(3 < X \leq 6) = \frac{3}{7}$.

(A) For a probability distribution,the sum of all probabilities must be equal to $1$.
$\sum P(X) = 1$
$0 + 2p + 2p + 3p + p^2 + 2p^2 + 7p^2 + 2p = 1$
Combining the terms,we get:
$10p^2 + 9p = 1$
$10p^2 + 9p - 1 = 0$
Factoring the quadratic equation:
$10p^2 + 10p - p - 1 = 0$
$10p(p + 1) - 1(p + 1) = 0$
$(10p - 1)(p + 1) = 0$
This gives $p = \frac{1}{10}$ or $p = -1$.
Since the probability $P(X)$ must be non-negative,$p$ must be such that all $P(X) \geq 0$. Thus,$p = -1$ is rejected.
Therefore,$p = \frac{1}{10}$.

(A) Let $X$ be the random variable representing 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 given by:

$X$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
$P(X)$	$\frac{1}{36}$	$\frac{2}{36}$	$\frac{3}{36}$	$\frac{4}{36}$	$\frac{5}{36}$	$\frac{6}{36}$	$\frac{5}{36}$	$\frac{4}{36}$	$\frac{3}{36}$	$\frac{2}{36}$	$\frac{1}{36}$

The expected value $E(X)$ is calculated as:
$E(X) = \sum x_i \cdot P(x_i)$
$E(X) = 2 \cdot \frac{1}{36} + 3 \cdot \frac{2}{36} + 4 \cdot \frac{3}{36} + 5 \cdot \frac{4}{36} + 6 \cdot \frac{5}{36} + 7 \cdot \frac{6}{36} + 8 \cdot \frac{5}{36} + 9 \cdot \frac{4}{36} + 10 \cdot \frac{3}{36} + 11 \cdot \frac{2}{36} + 12 \cdot \frac{1}{36}$
$E(X) = \frac{1}{36} (2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12)$
$E(X) = \frac{252}{36} = 7$

(D) The given probability distribution is:
| $X$ | $2$ | $3$ | $4$ |
|---|---|---|---|
| $P(X=x)$ | $0.3$ | $0.4$ | $0.3$ |
First,we calculate the mean $E(X)$:
$E(X) = \sum x_i P(x_i) = 2(0.3) + 3(0.4) + 4(0.3) = 0.6 + 1.2 + 1.2 = 3.0$
Next,we calculate $E(X^2)$:
$E(X^2) = \sum x_i^2 P(x_i) = 2^2(0.3) + 3^2(0.4) + 4^2(0.3) = 4(0.3) + 9(0.4) + 16(0.3) = 1.2 + 3.6 + 4.8 = 9.6$
Finally,the variance is given by:
$\text{Variance}(X) = E(X^2) - [E(X)]^2 = 9.6 - (3)^2 = 9.6 - 9 = 0.6$

(D) The probabilities must satisfy $0 \leq P(X=x_i) \leq 1$ for all $i$.
Thus,$0 \leq \frac{1+3p}{4} \leq 1$,$0 \leq \frac{1-p}{4} \leq 1$,$0 \leq \frac{1+2p}{4} \leq 1$,and $0 \leq \frac{1-4p}{4} \leq 1$.
Solving these inequalities:
$1$) $1+3p \geq 0 \Rightarrow p \geq -1/3$
$2$) $1-p \geq 0 \Rightarrow p \leq 1$
$3$) $1+2p \geq 0 \Rightarrow p \geq -1/2$
$4$) $1-4p \geq 0 \Rightarrow p \leq 1/4$
Combining these,we get $-\frac{1}{3} \leq p \leq \frac{1}{4}$.
The mean $E(X) = \sum x_i P(X=x_i) = (-1)\left(\frac{1+3p}{4}\right) + 0\left(\frac{1-p}{4}\right) + 1\left(\frac{1+2p}{4}\right) + 2\left(\frac{1-4p}{4}\right)$.
$E(X) = \frac{-1-3p+1+2p+2-8p}{4} = \frac{2-9p}{4}$.
Given $-\frac{1}{3} \leq p \leq \frac{1}{4}$,we find the range of $E(X)$:
For $p = 1/4$,$E(X) = \frac{2-9(1/4)}{4} = \frac{2-2.25}{4} = -\frac{0.25}{4} = -\frac{1}{16}$.
For $p = -1/3$,$E(X) = \frac{2-9(-1/3)}{4} = \frac{2+3}{4} = \frac{5}{4}$.
Thus,the minimum value is $-\frac{1}{16}$ and the maximum value is $\frac{5}{4}$.

(D) The sum of all the probabilities in a probability distribution is always unity.
$\therefore k + 3k + 3k + k = 1$
$\Rightarrow 8k = 1$
$\Rightarrow k = \frac{1}{8}$
Now,calculate the mean $E(X)$:
$E(X) = \sum x_i \cdot P(x_i)$
$E(X) = 0\left(\frac{1}{8}\right) + 1\left(\frac{3}{8}\right) + 2\left(\frac{3}{8}\right) + 3\left(\frac{1}{8}\right)$
$E(X) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{12}{8} = \frac{3}{2}$
Next,calculate $E(X^2)$:
$E(X^2) = \sum x_i^2 \cdot P(x_i)$
$E(X^2) = 0^2\left(\frac{1}{8}\right) + 1^2\left(\frac{3}{8}\right) + 2^2\left(\frac{3}{8}\right) + 3^2\left(\frac{1}{8}\right)$
$E(X^2) = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = \frac{24}{8} = 3$
Finally,calculate the variance $\operatorname{Var}(X)$:
$\operatorname{Var}(X) = E(X^2) - [E(X)]^2$
$\operatorname{Var}(X) = 3 - \left(\frac{3}{2}\right)^2$
$\operatorname{Var}(X) = 3 - \frac{9}{4} = \frac{12 - 9}{4} = \frac{3}{4}$
Thus,the values are $k = \frac{1}{8}$ and $\operatorname{Var}(X) = \frac{3}{4}$.