MHT CET 2020 Mathematics Question Paper with Answer and Solution

(B) For a binomial distribution,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $q = 1-p$.
Given $n = 5$ and $np + npq = 1 \cdot 8$.
Substituting the values,we get $5p + 5pq = 1 \cdot 8$.
$5p(1 + q) = 1 \cdot 8$.
Since $q = 1 - p$,we have $5p(1 + 1 - p) = 1 \cdot 8$.
$5p(2 - p) = 1 \cdot 8$.
$10p - 5p^2 = 1 \cdot 8$.
$5p^2 - 10p + 1 \cdot 8 = 0$.
Multiplying by $5$ to simplify: $25p^2 - 50p + 9 = 0$.
Using the quadratic formula $p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$p = \frac{50 \pm \sqrt{2500 - 4(25)(9)}}{50} = \frac{50 \pm \sqrt{2500 - 900}}{50} = \frac{50 \pm \sqrt{1600}}{50} = \frac{50 \pm 40}{50}$.
$p = \frac{90}{50} = 1 \cdot 8$ (not possible as $0 \le p \le 1$) or $p = \frac{10}{50} = 0 \cdot 2$.
Thus,$p = 0 \cdot 2$.

(B) Let $X$ be the number of successes in $n = 100$ trials. This follows a binomial distribution $B(n, p)$.
Here,the probability of getting an even number in a single throw of a die is $p = \frac{3}{6} = \frac{1}{2}$.
The probability of failure is $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.
The variance of a binomial distribution is given by the formula $\text{Var}(X) = npq$.
Substituting the values,we get $\text{Var}(X) = 100 \times \frac{1}{2} \times \frac{1}{2} = 100 \times \frac{1}{4} = 25$.
Therefore,the variance is $25$.

(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,$P(X \leq 1) = \frac{10}{28} + \frac{15}{28} = \frac{25}{28}$.

(A) Since $f(x)$ is the p.d.f. of a random variable $X$,the total area under the curve must be $1$.
$\int_{0}^{1} f(x) dx = 1 \Rightarrow \int_{0}^{1} K(x-x^2) dx = 1$
$K \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} = 1 \Rightarrow K \left( \frac{1}{2} - \frac{1}{3} \right) = 1$
$K \left( \frac{1}{6} \right) = 1 \Rightarrow K = 6$
Now,we calculate $P(X < \frac{1}{2}) = \int_{0}^{\frac{1}{2}} 6(x-x^2) dx$
$= 6 \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{\frac{1}{2}} = \left[ 3x^2 - 2x^3 \right]_{0}^{\frac{1}{2}}$
$= 3 \left( \frac{1}{2} \right)^2 - 2 \left( \frac{1}{2} \right)^3 = 3 \left( \frac{1}{4} \right) - 2 \left( \frac{1}{8} \right)$
$= \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$

(B) Given the probability mass function $P(X=x) = \frac{{}^{5}C_{x}}{2^{5}}$ for $x \in \{0, 1, 2, 3, 4, 5\}$.
We need to calculate $P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$.
$P(X=0) = \frac{{}^{5}C_{0}}{2^{5}} = \frac{1}{32}$
$P(X=1) = \frac{{}^{5}C_{1}}{2^{5}} = \frac{5}{32}$
$P(X=2) = \frac{{}^{5}C_{2}}{2^{5}} = \frac{10}{32}$
Summing these values: $P(X \leq 2) = \frac{1+5+10}{32} = \frac{16}{32} = \frac{1}{2}$.
Now,check the options:
$P(X \geq 3) = P(X=3) + P(X=4) + P(X=5) = \frac{{}^{5}C_{3} + {}^{5}C_{4} + {}^{5}C_{5}}{2^{5}} = \frac{10+5+1}{32} = \frac{16}{32} = \frac{1}{2}$.
Since $P(X \leq 2) = \frac{1}{2}$ and $P(X \geq 3) = \frac{1}{2}$,we have $P(X \leq 2) = P(X \geq 3)$.

(A) For a probability density function,the total area under the curve must be $1$.
$\int_{0}^{4} f(x) dx = 1$
$\int_{0}^{4} \frac{k}{\sqrt{x}} dx = 1$
$k [2\sqrt{x}]_{0}^{4} = 1$
$k [2(2) - 0] = 1 \Rightarrow 4k = 1 \Rightarrow k = \frac{1}{4}$
Now,we calculate $P(1 < X < 4)$:
$P(1 < X < 4) = \int_{1}^{4} \frac{1/4}{\sqrt{x}} dx$
$= \frac{1}{4} [2\sqrt{x}]_{1}^{4}$
$= \frac{1}{2} [\sqrt{4} - \sqrt{1}]$
$= \frac{1}{2} [2 - 1] = \frac{1}{2}$

(D) For a probability distribution,the sum of all probabilities must be equal to $1$.
Therefore,$k + 0 + 2k + 5k + k + 3k = 1$.
$12k = 1 \Rightarrow k = \frac{1}{12}$.
We need to find $P(X \geq 4)$,which is $P(X = 4) + P(X = 5) + P(X = 6)$.
$P(X \geq 4) = 5k + k + 3k = 9k$.
Substituting the value of $k$,we get $P(X \geq 4) = 9 \times \frac{1}{12} = \frac{9}{12} = \frac{3}{4}$.

(C) The given probability mass function is of a binomial distribution $B(n, p)$ where $n = 6$ and $p = \frac{1}{3}$.
Since $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$,the distribution is $P(X=x) = \binom{6}{x} \left(\frac{1}{3}\right)^{x} \left(\frac{2}{3}\right)^{6-x}$.
The mean $\mu = np = 6 \times \frac{1}{3} = 2$.
The variance $\sigma^{2} = npq = 6 \times \frac{1}{3} \times \frac{2}{3} = \frac{12}{9} = \frac{4}{3}$.
Now,we calculate $2\mu + 12\sigma^{2} = 2(2) + 12\left(\frac{4}{3}\right) = 4 + 16 = 20$.

(D) For a probability mass function,the sum of all probabilities must be equal to $1$.
Therefore,$P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) = 1$.
Substituting the given values: $k + \frac{k}{3} + \frac{k}{4} + \frac{k}{2} + \frac{k}{2} = 1$.
Taking the least common multiple $(LCM)$ of the denominators $(3, 4, 2, 2)$,which is $12$:
$\frac{12k + 4k + 3k + 6k + 6k}{12} = 1$.
Summing the numerators: $\frac{31k}{12} = 1$.
Solving for $k$: $k = \frac{12}{31}$.

(C) The cumulative distribution function $F(x)$ is defined as $F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t) dt$.
For $-1 < x < 2$,we have:
$F(x) = \int_{-1}^{x} \frac{t^3}{3} dt$
$F(x) = \frac{1}{3} \left[ \frac{t^4}{4} \right]_{-1}^{x}$
$F(x) = \frac{1}{3} \left( \frac{x^4}{4} - \frac{(-1)^4}{4} \right)$
$F(x) = \frac{1}{3} \left( \frac{x^4}{4} - \frac{1}{4} \right)$
$F(x) = \frac{1}{12} (x^4 - 1)$

(A) Given the p.d.f. $f(x) = \frac{1}{2}$ for $0 < x < 2$.
First,we calculate $a = P(X < \frac{1}{2})$:
$a = \int_{0}^{1/2} \frac{1}{2} dx = \frac{1}{2} [x]_{0}^{1/2} = \frac{1}{2} (\frac{1}{2} - 0) = \frac{1}{4}$.
Next,we calculate $b = P(X > \frac{3}{2})$:
$b = \int_{3/2}^{2} \frac{1}{2} dx = \frac{1}{2} [x]_{3/2}^{2} = \frac{1}{2} (2 - \frac{3}{2}) = \frac{1}{2} (\frac{1}{2}) = \frac{1}{4}$.
Comparing $a$ and $b$,we have $a = \frac{1}{4}$ and $b = \frac{1}{4}$.
Therefore,$a - b = \frac{1}{4} - \frac{1}{4} = 0$.

(C) The cumulative distribution function (c.d.f.) is given by $F(x) = P(X \leq x) = \frac{\sqrt{x}}{2}$.
We need to find the probability $P[X > 1]$.
Using the property of the cumulative distribution function,we know that $P(X > x) = 1 - P(X \leq x) = 1 - F(x)$.
Therefore,$P[X > 1] = 1 - F(1)$.
Substituting $x = 1$ into the given function: $F(1) = \frac{\sqrt{1}}{2} = \frac{1}{2}$.
Thus,$P[X > 1] = 1 - \frac{1}{2} = \frac{1}{2}$.

(B) For a probability distribution,the sum of all probabilities must be equal to $1$.
Therefore,$P(0) + P(1) + P(2) + P(3) + P(4) = 1$.
Substituting the given values: $k + 2k + 4k + 2k + k = 1$.
$10k = 1 \Rightarrow k = \frac{1}{10}$.
We need to find $P(X \leq 2)$,which is $P(0) + P(1) + P(2)$.
$P(X \leq 2) = k + 2k + 4k = 7k$.
Substituting the value of $k$: $7 \times \frac{1}{10} = \frac{7}{10}$.

(D) Since $f(x)$ is a probability density function,the total area under the curve must be $1$:
$\int_{-2}^{2} k(4 - x^2) dx = 1$
Since the function is even,we have $2k \int_{0}^{2} (4 - x^2) dx = 1$.
$2k [4x - \frac{x^3}{3}]_0^2 = 1$
$2k (8 - \frac{8}{3}) = 1 \Rightarrow 2k(\frac{16}{3}) = 1 \Rightarrow k = \frac{3}{32}$.
Now,we calculate $P[-1 < X < 1]$:
$P[-1 < X < 1] = \int_{-1}^{1} \frac{3}{32}(4 - x^2) dx = 2 \times \frac{3}{32} \int_{0}^{1} (4 - x^2) dx$
$= \frac{6}{32} [4x - \frac{x^3}{3}]_0^1 = \frac{3}{16} (4 - \frac{1}{3}) = \frac{3}{16} \times \frac{11}{3} = \frac{11}{16}$.

(B) The probability $P(|X| < 2)$ is equivalent to $P(-2 < X < 2)$.
Given the p.d.f. $f(x) = \frac{x+2}{18}$ for $-2 < x < 4$,we integrate over the interval $(-2, 2)$:
$P(-2 < X < 2) = \int_{-2}^{2} \frac{x+2}{18} dx$
$= \frac{1}{18} \left[ \frac{x^2}{2} + 2x \right]_{-2}^{2}$
$= \frac{1}{18} \left[ (\frac{2^2}{2} + 2(2)) - (\frac{(-2)^2}{2} + 2(-2)) \right]$
$= \frac{1}{18} \left[ (2 + 4) - (2 - 4) \right]$
$= \frac{1}{18} [6 - (-2)]$
$= \frac{8}{18} = \frac{4}{9}$

(C) For a probability mass function (p.m.f.),the sum of all probabilities must be equal to $1$,i.e.,$\sum P(X) = 1$.
Given $P(X) = k \binom{4}{x}$ for $x = 0, 1, 2, 3, 4$.
Calculating each probability:
$P(X=0) = k \binom{4}{0} = k \times 1 = k$
$P(X=1) = k \binom{4}{1} = k \times 4 = 4k$
$P(X=2) = k \binom{4}{2} = k \times 6 = 6k$
$P(X=3) = k \binom{4}{3} = k \times 4 = 4k$
$P(X=4) = k \binom{4}{4} = k \times 1 = k$
Summing these probabilities:
$k + 4k + 6k + 4k + k = 1$
$16k = 1$
$k = \frac{1}{16}$

(D) The mean $E(X)$ is calculated as:
$E(X) = \sum x_i P_i = (-2)(0.2) + (-1)(0.3) + (0)(0.1) + (1)(0.15) + (2)(0.25)$
$E(X) = -0.4 - 0.3 + 0 + 0.15 + 0.5 = -0.05$
Next,we calculate $E(X^2)$:
$E(X^2) = \sum x_i^2 P_i = (-2)^2(0.2) + (-1)^2(0.3) + (0)^2(0.1) + (1)^2(0.15) + (2)^2(0.25)$
$E(X^2) = (4)(0.2) + (1)(0.3) + 0 + (1)(0.15) + (4)(0.25)$
$E(X^2) = 0.8 + 0.3 + 0.15 + 1.0 = 2.25$
The variance $Var(X)$ is given by:
$Var(X) = E(X^2) - [E(X)]^2$
$Var(X) = 2.25 - (-0.05)^2$
$Var(X) = 2.25 - 0.0025 = 2.2475$

(C) The given probability mass function is $P(X=x) = \frac{\binom{5}{x}}{2^5}$ for $x \in \{0, 1, 2, 3, 4, 5\}$.
Calculating the probabilities:
$P(X=0) = \frac{\binom{5}{0}}{32} = \frac{1}{32}$,$P(X=1) = \frac{\binom{5}{1}}{32} = \frac{5}{32}$,$P(X=2) = \frac{\binom{5}{2}}{32} = \frac{10}{32}$,$P(X=3) = \frac{\binom{5}{3}}{32} = \frac{10}{32}$,$P(X=4) = \frac{\binom{5}{4}}{32} = \frac{5}{32}$,$P(X=5) = \frac{\binom{5}{5}}{32} = \frac{1}{32}$.
Now evaluate the options:
$A$. $P(X \leq 1) = P(0) + P(1) = \frac{1+5}{32} = \frac{6}{32}$ and $P(X \geq 4) = P(4) + P(5) = \frac{5+1}{32} = \frac{6}{32}$. Thus,$P(X \leq 1) = P(X \geq 4)$ is true.
$B$. $P(X \leq 2) = P(0) + P(1) + P(2) = \frac{1+5+10}{32} = \frac{16}{32} = 0.5$. $P(X \geq 4) = \frac{6}{32} = 0.1875$. Since $0.5 \geq 0.1875$,this is true.
$C$. $P(X \leq 3) = P(0) + P(1) + P(2) + P(3) = \frac{1+5+10+10}{32} = \frac{26}{32} = 0.8125$. $P(X \geq 3) = P(3) + P(4) + P(5) = \frac{10+5+1}{32} = \frac{16}{32} = 0.5$. Since $0.8125 \leq 0.5$ is false,option $C$ is not true.
$D$. $P(X \leq 2) = \frac{16}{32} = 0.5$ and $P(X \geq 3) = \frac{16}{32} = 0.5$. Thus,$P(X \leq 2) = P(X \geq 3)$ is true.

(C) We are given the probability density function $f(x) = \frac{x+2}{18}$ for $-2 < x < 4$.
We need to find $P[|x| < 1]$.
The condition $|x| < 1$ is equivalent to $-1 < x < 1$.
Thus,$P[|x| < 1] = \int_{-1}^{1} f(x) \, dx = \int_{-1}^{1} \frac{x+2}{18} \, dx$.
Evaluating the integral:
$P = \frac{1}{18} \int_{-1}^{1} (x+2) \, dx = \frac{1}{18} \left[ \frac{x^2}{2} + 2x \right]_{-1}^{1}$.
Substituting the limits:
$P = \frac{1}{18} \left[ (\frac{1}{2} + 2) - (\frac{(-1)^2}{2} + 2(-1)) \right]$.
$P = \frac{1}{18} \left[ (\frac{1}{2} + 2) - (\frac{1}{2} - 2) \right]$.
$P = \frac{1}{18} [ \frac{1}{2} + 2 - \frac{1}{2} + 2 ] = \frac{1}{18} [4] = \frac{4}{18} = \frac{2}{9}$.
Therefore,the correct option is $C$.

(B) When a die is thrown,the possible outcomes are $S = \{1, 2, 3, 4, 5, 6\}$.
The probability of getting any number $x \in S$ is $P(x) = \frac{1}{6}$.
The expectation $E(X)$ is given by the formula $E(X) = \sum x \cdot P(x)$.
$E(X) = (1 \times \frac{1}{6}) + (2 \times \frac{1}{6}) + (3 \times \frac{1}{6}) + (4 \times \frac{1}{6}) + (5 \times \frac{1}{6}) + (6 \times \frac{1}{6})$.
$E(X) = \frac{1}{6} \times (1 + 2 + 3 + 4 + 5 + 6)$.
$E(X) = \frac{1}{6} \times 21 = 3.5$.

(C) To find $P(X \leq 2)$,we integrate the probability density function $f(x)$ from the lower bound $0$ to $2$.
$P(X \leq 2) = \int_{0}^{2} f(x) \, dx$
$P(X \leq 2) = \int_{0}^{2} \frac{x}{8} \, dx$
$P(X \leq 2) = \frac{1}{8} \left[ \frac{x^2}{2} \right]_{0}^{2}$
$P(X \leq 2) = \frac{1}{8} \left( \frac{2^2}{2} - \frac{0^2}{2} \right)$
$P(X \leq 2) = \frac{1}{8} \left( \frac{4}{2} \right) = \frac{1}{8} \times 2 = \frac{2}{8} = \frac{1}{4}$

(B) To find the standard deviation,we first calculate the mean $\mu = E(X)$ and the variance $\sigma^2 = E(X^2) - [E(X)]^2$.
The probability distribution is:

$x_i$	$p_i$	$p_i x_i$	$p_i x_i^2$
$0$	$q^2$	$0$	$0$
$1$	$2pq$	$2pq$	$2pq$
$2$	$p^2$	$2p^2$	$4p^2$
Total	$1$	$2pq + 2p^2$	$2pq + 4p^2$

Mean $\mu = E(X) = \sum p_i x_i = 2pq + 2p^2 = 2p(q+p)$.
Since $p+q=1$,we have $\mu = 2p(1) = 2p$.
Variance $\sigma^2 = E(X^2) - \mu^2 = (2pq + 4p^2) - (2p)^2 = 2pq + 4p^2 - 4p^2 = 2pq$.
Standard deviation $\sigma = \sqrt{\text{Variance}} = \sqrt{2pq}$.

(B) For a discrete random variable $X$,the probability mass function $P[X=x]$ is related to the cumulative distribution function $F(x)$ by the formula:
$P[X=x] = F(x) - F(x^-)$
where $F(x^-)$ is the value of the cumulative distribution function at the value immediately preceding $x$.
In this case,we want to find $P[X=3]$.
Looking at the table,the value of $X$ immediately preceding $3$ is $1$.
Therefore,$P[X=3] = F(3) - F(1)$.
From the table:
$F(3) = 0.75$
$F(1) = 0.65$
Substituting these values:
$P[X=3] = 0.75 - 0.65 = 0.10$
Thus,the correct option is $(B)$.

(B) The cumulative distribution function $F(x)$ is defined as $F(x) = P(X \leq x)$.
We need to calculate $F(1.5) - F(-0.5) = P(X \leq 1.5) - P(X \leq -0.5)$.
By the definition of probability for a discrete random variable,$P(X \leq x) = \sum_{x_i \leq x} P(X = x_i)$.
Therefore,$P(X \leq 1.5) = P(X = -1.5) + P(X = -0.5) + P(X = 0.5) + P(X = 1.5) = 0.05 + 0.2 + 0.15 + 0.25 = 0.65$.
And $P(X \leq -0.5) = P(X = -1.5) + P(X = -0.5) = 0.05 + 0.2 = 0.25$.
Thus,$F(1.5) - F(-0.5) = 0.65 - 0.25 = 0.4$.
Alternatively,$F(1.5) - F(-0.5) = P(-0.5 < X \leq 1.5) = P(X = 0.5) + P(X = 1.5) = 0.15 + 0.25 = 0.4$.
The correct option is $B$.

(A) The cumulative distribution function $F(x)$ is defined as the integral of the probability density function $f(x)$.
$F(x) = \int_{0}^{x} f(t) dt = \int_{0}^{x} 3(1 - 2t^2) dt$.
Evaluating the integral:
$F(x) = 3 \left[ t - \frac{2t^3}{3} \right]_{0}^{x} = 3 \left( x - \frac{2x^3}{3} \right)$.
Comparing this with the given form $F(x) = k(x - \frac{2x^3}{k})$,we observe that $k = 3$ is not directly matching the structure unless we rewrite the expression.
Wait,let us re-evaluate the expression: $F(x) = 3x - 2x^3$.
If $F(x) = k(x - \frac{2x^3}{k}) = kx - 2x^3$,then by comparing coefficients,$k = 3$ and $2 = 2$. Thus,$k = 3$.

(A) For a probability distribution,the sum of all probabilities must be $1$.
Therefore,$k + 3k + 3k + k = 1$,which implies $8k = 1$,so $k = \frac{1}{8}$.
The mean $E(X) = \sum x_i P(x_i) = (0 \times \frac{1}{8}) + (1 \times \frac{3}{8}) + (2 \times \frac{3}{8}) + (3 \times \frac{1}{8}) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{12}{8} = \frac{3}{2}$.
The expected value of the square,$E(X^2) = \sum x_i^2 P(x_i) = (0^2 \times \frac{1}{8}) + (1^2 \times \frac{3}{8}) + (2^2 \times \frac{3}{8}) + (3^2 \times \frac{1}{8}) = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = \frac{24}{8} = 3$.
The variance $Var(X) = E(X^2) - [E(X)]^2$.
$Var(X) = 3 - (\frac{3}{2})^2 = 3 - \frac{9}{4} = \frac{12 - 9}{4} = \frac{3}{4}$.

(A) The cumulative distribution function $F(x)$ is defined as $F(x) = \int_{-\infty}^{x} f(t) dt$.
For $0 < x < 4$,$F(x) = \int_{0}^{x} \frac{t}{8} dt$.
$F(x) = \frac{1}{8} \left[ \frac{t^2}{2} \right]_{0}^{x} = \frac{1}{16} (x^2 - 0) = \frac{x^2}{16}$.
Now,substitute $x = 0.5$ into the expression for $F(x)$:
$F(0.5) = \frac{(0.5)^2}{16} = \frac{0.25}{16} = \frac{1/4}{16} = \frac{1}{64}$.

(A) Given $X \sim B\left(8, \frac{1}{2}\right)$,we have $n=8, p=\frac{1}{2}, q=\frac{1}{2}$.
We need to find $P(|X-4| \leq 2)$.
This inequality $|X-4| \leq 2$ implies $-2 \leq X-4 \leq 2$,which simplifies to $2 \leq X \leq 6$.
Thus,we need to calculate $P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6)$.
The probability mass function is $P(X=k) = {}^{8}C_{k} \left(\frac{1}{2}\right)^{k} \left(\frac{1}{2}\right)^{8-k} = {}^{8}C_{k} \left(\frac{1}{2}\right)^{8}$.
Summing these probabilities:
$P(2 \leq X \leq 6) = \left(\frac{1}{2}\right)^{8} \left[ {}^{8}C_{2} + {}^{8}C_{3} + {}^{8}C_{4} + {}^{8}C_{5} + {}^{8}C_{6} \right]$.
Calculating the combinations:
${}^{8}C_{2} = 28, {}^{8}C_{3} = 56, {}^{8}C_{4} = 70, {}^{8}C_{5} = 56, {}^{8}C_{6} = 28$.
Sum $= 28 + 56 + 70 + 56 + 28 = 238$.
Therefore,$P(2 \leq X \leq 6) = \frac{238}{256} = \frac{119}{128}$.

(A) To find the variance of the random variable $X$,we use the formula $Var(X) = E(X^2) - [E(X)]^2$.

$x_{i}$	$p(x_{i})$	$p_{i} x_{i}$	$p_{i} x_{i}^{2}$
$0$	$\frac{1}{5}$	$0$	$0$
$1$	$\frac{2}{5}$	$\frac{2}{5}$	$\frac{2}{5}$
$2$	$\frac{2}{5}$	$\frac{4}{5}$	$\frac{8}{5}$

First,calculate the mean $E(X) = \sum p_{i} x_{i} = 0 + \frac{2}{5} + \frac{4}{5} = \frac{6}{5}$.
Next,calculate $E(X^2) = \sum p_{i} x_{i}^{2} = 0 + \frac{2}{5} + \frac{8}{5} = \frac{10}{5} = 2$.
Now,calculate the variance: $Var(X) = E(X^2) - [E(X)]^2 = 2 - (\frac{6}{5})^2 = 2 - \frac{36}{25} = \frac{50 - 36}{25} = \frac{14}{25}$.

(C) The random variable $X$ takes values $x_i \in \{0, 1, 2\}$.
Given the mean $E(X) = \sum x_i P(x_i) = 1.2$.
Substituting the values: $(0 \times P(X=0)) + (1 \times P(X=1)) + (2 \times P(X=2)) = 1.2$.
Since $P(X=0) = 0.3$,we have: $0 + P(X=1) + 2P(X=2) = 1.2 \implies P(X=1) + 2P(X=2) = 1.2$ (Equation $1$).
Also,the sum of probabilities is $1$: $P(X=0) + P(X=1) + P(X=2) = 1$.
$0.3 + P(X=1) + P(X=2) = 1 \implies P(X=1) + P(X=2) = 0.7$ (Equation $2$).
Subtracting Equation $2$ from Equation $1$: $(P(X=1) + 2P(X=2)) - (P(X=1) + P(X=2)) = 1.2 - 0.7$.
$P(X=2) = 0.5$.
Substituting $P(X=2) = 0.5$ into Equation $2$: $P(X=1) + 0.5 = 0.7 \implies P(X=1) = 0.2$.

(B) Let $X$ denote the number of successes (getting an even number) in $100$ trials.
Then $X$ follows a binomial distribution with $n = 100$,$p = \frac{1}{2}$,and $q = \frac{1}{2}$.
Variance of $X = npq = 100 \times \frac{1}{2} \times \frac{1}{2} = 25$.
Standard Deviation = $\sqrt{\text{Variance}} = \sqrt{25} = 5$.

(C) Let the direction cosines of the line be $l, m, n$. Since the line makes equal angles $\alpha$ with each of the coordinate axes,we have $l = \cos \alpha$,$m = \cos \alpha$,and $n = \cos \alpha$.
We know that for any line,$l^2 + m^2 + n^2 = 1$.
Substituting the values,we get $\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1$.
$3 \cos^2 \alpha = 1 \Rightarrow \cos^2 \alpha = \frac{1}{3}$.
Since the line is in the first octant $OXYZ$,the direction cosines must be positive,so $\cos \alpha = \frac{1}{\sqrt{3}}$.
Thus,$l = m = n = \frac{1}{\sqrt{3}}$.

(C) Let the direction angles of the line be $\alpha, \beta, \gamma$.
Since the line makes equal angles with the coordinate axes,we have $\alpha = \beta = \gamma$.
The direction cosines are given by $l = \cos \alpha, m = \cos \beta, n = \cos \gamma$.
We know the identity $l^2 + m^2 + n^2 = 1$,which implies $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Substituting $\alpha = \beta = \gamma$,we get $3 \cos^2 \alpha = 1$.
This gives $\cos^2 \alpha = \frac{1}{3}$,so $\cos \alpha = \pm \frac{1}{\sqrt{3}}$.
Since the angles are given to be acute,$\cos \alpha, \cos \beta, \cos \gamma$ must be positive.
Therefore,the direction cosines are $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$.

(D) Two planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ are parallel if their normal vectors are proportional,i.e.,$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Given planes are $2x - 5y + z - 8 = 0$ and $2\lambda x - 15y + \lambda z + 6 = 0$.
Comparing the coefficients,we get:
$\frac{2}{2\lambda} = \frac{-5}{-15} = \frac{1}{\lambda}$.
Simplifying the ratios:
$\frac{1}{\lambda} = \frac{1}{3} = \frac{1}{\lambda}$.
From this,we conclude that $\lambda = 3$.

(B) Direction cosines are the cosines of the angles which a line makes with the positive coordinate axes. They are represented by $\langle l, m, n \rangle$ where $l, m, n$ correspond to the $x$-axis,$y$-axis,and $z$-axis respectively. The sum of the squares of the direction cosines is unity,i.e.,$l^{2} + m^{2} + n^{2} = 1$.
Since the line lies in the $ZOX$ plane,it is perpendicular to the $y$-axis. Therefore,the angle made with the $y$-axis is $90^{\circ}$,so $m = \cos(90^{\circ}) = 0$.
The line makes an angle of $30^{\circ}$ with the $Z$-axis,so $n = \cos(30^{\circ}) = \frac{\sqrt{3}}{2}$.
Using the property $l^{2} + m^{2} + n^{2} = 1$,we substitute $m = 0$ and $n = \frac{\sqrt{3}}{2}$:
$l^{2} + 0^{2} + \left(\frac{\sqrt{3}}{2}\right)^{2} = 1$
$l^{2} + \frac{3}{4} = 1$
$l^{2} = 1 - \frac{3}{4} = \frac{1}{4}$
$l = \pm \frac{1}{2}$.
Thus,the direction cosines are $\langle \pm \frac{1}{2}, 0, \frac{\sqrt{3}}{2} \rangle$.

(A) Two vectors $\vec{a} = a_1 \hat{\imath} + a_2 \hat{\jmath} + a_3 \hat{k}$ and $\vec{b} = b_1 \hat{\imath} + b_2 \hat{\jmath} + b_3 \hat{k}$ are collinear if their corresponding components are proportional,i.e.,$\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3}$.
Given vectors are $(2, -q, 3)$ and $(4, -5, 6)$.
Setting the ratios equal,we get:
$\frac{2}{4} = \frac{-q}{-5} = \frac{3}{6}$
Simplifying the fractions:
$\frac{1}{2} = \frac{q}{5} = \frac{1}{2}$
From $\frac{1}{2} = \frac{q}{5}$,we find:
$q = \frac{5}{2}$

(C) The formula for the cosine of the angle $\theta$ between two lines with direction ratios $a_1, b_1, c_1$ and $a_2, b_2, c_2$ is given by $\cos \theta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \right|$.
Given $\theta = \frac{\pi}{3}$,$a_1=4, b_1=-3, c_1=5$ and $a_2=3, b_2=4, c_2=k$.
Substituting these values,we get $\cos \frac{\pi}{3} = \left| \frac{4(3) + (-3)(4) + 5k}{\sqrt{4^2 + (-3)^2 + 5^2} \sqrt{3^2 + 4^2 + k^2}} \right|$.
$\frac{1}{2} = \left| \frac{12 - 12 + 5k}{\sqrt{16 + 9 + 25} \sqrt{9 + 16 + k^2}} \right|$.
$\frac{1}{2} = \left| \frac{5k}{\sqrt{50} \sqrt{25 + k^2}} \right| = \left| \frac{5k}{5\sqrt{2} \sqrt{25 + k^2}} \right| = \left| \frac{k}{\sqrt{2} \sqrt{25 + k^2}} \right|$.
Squaring both sides,$\frac{1}{4} = \frac{k^2}{2(25 + k^2)}$.
$2(25 + k^2) = 4k^2$.
$50 + 2k^2 = 4k^2$.
$2k^2 = 50 \Rightarrow k^2 = 25$.
Therefore,$k = \pm 5$.

(C) The direction cosines of a line are denoted by $\ell, m, n$.
Given that $\ell = \frac{1}{c}$,$m = \frac{1}{c}$,and $n = \frac{1}{c}$.
We know the fundamental property of direction cosines is $\ell^{2} + m^{2} + n^{2} = 1$.
Substituting the given values: $(\frac{1}{c})^{2} + (\frac{1}{c})^{2} + (\frac{1}{c})^{2} = 1$.
This simplifies to $\frac{1}{c^{2}} + \frac{1}{c^{2}} + \frac{1}{c^{2}} = 1$,which is $\frac{3}{c^{2}} = 1$.
Therefore,$c^{2} = 3$,which gives $c = \pm \sqrt{3}$.

(B) We know that for a line making angles $\alpha, \beta, \gamma$ with the coordinate axes,the direction cosines are $\cos \alpha, \cos \beta, \cos \gamma$.
The sum of the squares of the direction cosines is given by $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Now,we need to evaluate $\cos 2\alpha + \cos 2\beta + \cos 2\gamma$.
Using the identity $\cos 2\theta = 2\cos^2 \theta - 1$,we get:
$\cos 2\alpha + \cos 2\beta + \cos 2\gamma = (2\cos^2 \alpha - 1) + (2\cos^2 \beta - 1) + (2\cos^2 \gamma - 1)$
$= 2(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma) - 3$
Substituting the value $1$ for the sum of squares:
$= 2(1) - 3 = 2 - 3 = -1$.

(B) Let $\vec{a} = 3\hat{i} - 2\hat{j} + 4\hat{k}$ and $\vec{b} = 1\hat{i} + 3\hat{j} - 2\hat{k}$.
Since the line is perpendicular to both lines,its direction vector is given by the cross product $\vec{n} = \vec{a} \times \vec{b}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -2 & 4 \\ 1 & 3 & -2 \end{vmatrix} = \hat{i}(4 - 12) - \hat{j}(-6 - 4) + \hat{k}(9 + 2) = -8\hat{i} + 10\hat{j} + 11\hat{k}$.
The magnitude of the vector is $|\vec{n}| = \sqrt{(-8)^2 + 10^2 + 11^2} = \sqrt{64 + 100 + 121} = \sqrt{285}$.
The direction cosines are the components of the unit vector,which are $\frac{-8}{\sqrt{285}}, \frac{10}{\sqrt{285}}, \frac{11}{\sqrt{285}}$.

(D) The direction ratios of the first line $\frac{x - 1}{2 \lambda} = \frac{y - 1}{-5} = \frac{z - 1}{2}$ are $(2 \lambda, -5, 2)$.
The direction ratios of the second line $\frac{x + 2}{\lambda} = \frac{y + 3}{\lambda} = \frac{z + 5}{1}$ are $(\lambda, \lambda, 1)$.
Since the two lines are parallel,their direction ratios must be proportional:
$\frac{2 \lambda}{\lambda} = \frac{-5}{\lambda} = \frac{2}{1}$.
From the ratio $\frac{-5}{\lambda} = \frac{2}{1}$,we get $2 \lambda = -5$,which implies $\lambda = \frac{-5}{2}$.
Checking the first ratio: $\frac{2 \lambda}{\lambda} = 2$,which is consistent with the third ratio $\frac{2}{1} = 2$ for $\lambda \neq 0$.

(C) The unit vectors along the positive $Y$ and $Z$ axes are $\hat{j} = (0, 1, 0)$ and $\hat{k} = (0, 0, 1)$ respectively.
Any vector along the bisector of the angle between these two axes is given by the sum of these unit vectors: $\vec{v} = \hat{j} + \hat{k} = (0, 1, 1)$.
The magnitude of this vector is $|\vec{v}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$.
The direction cosines are obtained by dividing the components of the vector by its magnitude:
$l = 0/\sqrt{2} = 0$,
$m = 1/\sqrt{2}$,
$n = 1/\sqrt{2}$.
Thus,the direction cosines are $(0, 1/\sqrt{2}, 1/\sqrt{2})$.

(B) We know that the sum of the squares of the direction cosines of a line is $1$,i.e.,$\cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1$.
Given that $\alpha + \beta = 90^{\circ}$,we have $\alpha = 90^{\circ} - \beta$.
Substituting this into the expression for $\cos \alpha$,we get $\cos \alpha = \cos(90^{\circ} - \beta) = \sin \beta$.
Therefore,$\cos^{2} \alpha = \sin^{2} \beta$.
Using the identity $\sin^{2} \beta = 1 - \cos^{2} \beta$,we get $\cos^{2} \alpha = 1 - \cos^{2} \beta$,which implies $\cos^{2} \alpha + \cos^{2} \beta = 1$.
Substituting this into the original equation: $(1) + \cos^{2} \gamma = 1$.
This simplifies to $\cos^{2} \gamma = 0$,which means $\cos \gamma = 0$.
Thus,$\gamma = 90^{\circ}$.

(B) Let the direction angles of the line be $\alpha = \frac{\pi}{6}$,$\beta = \frac{\pi}{3}$,and $\gamma$ be the angle with the $Z$ axis.
The sum of the squares of the direction cosines is always $1$,i.e.,$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Substituting the given values: $\cos^2(\frac{\pi}{6}) + \cos^2(\frac{\pi}{3}) + \cos^2 \gamma = 1$.
We know that $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\cos(\frac{\pi}{3}) = \frac{1}{2}$.
So,$(\frac{\sqrt{3}}{2})^2 + (\frac{1}{2})^2 + \cos^2 \gamma = 1$.
$\frac{3}{4} + \frac{1}{4} + \cos^2 \gamma = 1$.
$1 + \cos^2 \gamma = 1$.
$\cos^2 \gamma = 0$,which implies $\cos \gamma = 0$.
Therefore,$\gamma = \frac{\pi}{2}$.

(A) Let the angles made by the line with the $x$,$y$,and $z$-axes be $\alpha$,$\beta$,and $\gamma$ respectively.
Given $\alpha = 45^{\circ}$ and $\beta = \gamma = \theta$.
The direction cosines are $l = \cos \alpha$,$m = \cos \beta$,and $n = \cos \gamma$.
We know that $l^2 + m^2 + n^2 = 1$.
Substituting the values,we get $\cos^2 45^{\circ} + \cos^2 \theta + \cos^2 \theta = 1$.
$(\frac{1}{\sqrt{2}})^2 + 2 \cos^2 \theta = 1$.
$\frac{1}{2} + 2 \cos^2 \theta = 1$.
$2 \cos^2 \theta = 1 - \frac{1}{2} = \frac{1}{2}$.
$\cos^2 \theta = \frac{1}{4} \Rightarrow \cos \theta = \pm \frac{1}{2}$.
Thus,the direction cosines are $(\cos 45^{\circ}, \cos \theta, \cos \theta)$,which gives $(\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2})$ or $(\frac{1}{\sqrt{2}}, -\frac{1}{2}, -\frac{1}{2})$.

(A) The line passes through points $A(3, 4, -7)$ and $B(1, -1, 6)$.
First,find the direction vector $\vec{v} = B - A = (1 - 3, -1 - 4, 6 - (-7)) = (-2, -5, 13)$.
The parametric equations of a line passing through $(x_1, y_1, z_1)$ with direction vector $(a, b, c)$ are given by $x = x_1 + a\lambda, y = y_1 + b\lambda, z = z_1 + c\lambda$.
Using point $A(3, 4, -7)$ and direction vector $(-2, -5, 13)$,we get:
$x = 3 - 2\lambda$
$y = 4 - 5\lambda$
$z = -7 + 13\lambda$
Thus,the correct option is $A$.

(D) The given Cartesian equation is $x-1 = 2y+3 = 3-z$.
Rewrite the equation in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$:
$\frac{x-1}{1} = \frac{y - (-3/2)}{1/2} = \frac{z-3}{-1}$.
Thus,the point on the line is $A(1, -3/2, 3)$ and the direction ratios are $(1, 1/2, -1)$.
Multiplying the direction ratios by $2$,we get the proportional direction ratios $(2, 1, -2)$.
The vector equation of a line passing through point $\vec{a}$ and parallel to vector $\vec{b}$ is $\vec{r} = \vec{a} + \lambda \vec{b}$.
Substituting the values,we get $\vec{r} = (\hat{i} - \frac{3}{2}\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + \hat{j} - 2\hat{k})$.

(B) The given equation of the line is $\frac{x+3}{2}=\frac{2y-3}{5}; z=-1$.
First,rewrite the equation in the standard form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$.
$\frac{x+3}{2}=\frac{2(y-3/2)}{5}; z=-1$
$\frac{x-(-3)}{2}=\frac{y-3/2}{5/2}; z=-1$.
This line passes through the point $(-3, 3/2, -1)$ and has direction ratios proportional to $(2, 5/2, 0)$.
Multiplying the direction ratios by $2$,we get $(4, 5, 0)$.
The vector equation of a line passing through point $\vec{a}$ and parallel to vector $\vec{b}$ is $\vec{r}=\vec{a}+\lambda\vec{b}$.
Here,$\vec{a}=-3\hat{i}+\frac{3}{2}\hat{j}-\hat{k}$ and $\vec{b}=4\hat{i}+5\hat{j}$.
Thus,the vector equation is $\vec{r}=\left(-3\hat{i}+\frac{3}{2}\hat{j}-\hat{k}\right)+\lambda(4\hat{i}+5\hat{j})$.

(C) The given lines are $\frac{1-x}{2}=\frac{y-8}{\lambda}=\frac{z-5}{2}$ and $\frac{x-11}{5}=\frac{y-3}{3}=\frac{z-1}{1}$.
First,rewrite the first line in standard form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$:
$\frac{x-1}{-2}=\frac{y-8}{\lambda}=\frac{z-5}{2}$.
The direction ratios of the first line are $\vec{v_1} = (-2, \lambda, 2)$.
The direction ratios of the second line are $\vec{v_2} = (5, 3, 1)$.
Since the lines are perpendicular,the dot product of their direction vectors must be zero:
$\vec{v_1} \cdot \vec{v_2} = 0$
$(-2)(5) + (\lambda)(3) + (2)(1) = 0$
$-10 + 3\lambda + 2 = 0$
$3\lambda - 8 = 0$
$3\lambda = 8$
$\lambda = \frac{8}{3}$.

(D) The $XOZ$ plane is the plane where $y = 0$. The normal vector to the $XOZ$ plane is parallel to the $y$-axis,which is given by the vector $\vec{n} = (0, 1, 0)$.
Since the required line is perpendicular to the $XOZ$ plane,it must be parallel to the normal vector $\vec{n} = (0, 1, 0)$.
The line passes through the point $(2, 3, -4)$.
Using the symmetric form of the line equation,$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \lambda$,where $(x_1, y_1, z_1) = (2, 3, -4)$ and the direction ratios are $(a, b, c) = (0, 1, 0)$.
Thus,the equation is $\frac{x - 2}{0} = \frac{y - 3}{1} = \frac{z + 4}{0} = \lambda$.
This implies $x - 2 = 0 \implies x = 2$,$z + 4 = 0 \implies z = -4$,and $y - 3 = \lambda \implies y = 3 + \lambda$.
Therefore,the equation of the line is $x = 2, \quad y = 3 + \lambda, \quad z = -4$.