MHT CET 2023 Mathematics Question Paper with Answer and Solution

(B) $1$. Identify the lines bounding the region: The lines are $3x + 4y = 12$ (intercepts $(4,0)$ and $(0,3)$),$4x + 7y = 28$ (intercepts $(7,0)$ and $(0,4)$),and $y = 1$.
$2$. Analyze the inequalities:
- For the line $3x + 4y = 12$,the shaded region is away from the origin,so the inequality is $3x + 4y \geq 12$.
- For the line $4x + 7y = 28$,the shaded region is towards the origin,so the inequality is $4x + 7y \leq 28$.
- For the line $y = 1$,the shaded region is above the line,so the inequality is $y \geq 1$.
- Since the region is in the first quadrant,we have $x \geq 0$ and $y \geq 0$.
$3$. Combining these,the system of inequalities is $3x + 4y \geq 12, 4x + 7y \leq 28, y \geq 1, x \geq 0, y \geq 0$.

(C) The feasible region is a polygon with vertices $(0,0)$,$(3,0)$,$(3,2)$,$(2,3)$,and $(0,3)$.
We evaluate the objective function $z=10x+25y$ at each vertex:
At $(0,0)$,$z=10(0)+25(0)=0$
At $(3,0)$,$z=10(3)+25(0)=30$
At $(3,2)$,$z=10(3)+25(2)=30+50=80$
At $(2,3)$,$z=10(2)+25(3)=20+75=95$
At $(0,3)$,$z=10(0)+25(3)=75$
Comparing these values,the maximum value of $z$ is $95$.

(A) To find the feasible region,we analyze the given inequalities:
$1$. $4x + 3y \leq 60$: This represents the region on or below the line passing through $(15, 0)$ and $(0, 20)$.
$2$. $y \geq 2x$: This represents the region on or above the line passing through $(0, 0)$ and $(3, 6)$.
$3$. $x \geq 3$: This represents the region to the right of the vertical line $x = 3$.
$4$. $x, y \geq 0$: This represents the first quadrant.
By testing a point $(4, 10)$ which lies in the $S_2$ region:
- $4(4) + 3(10) = 16 + 30 = 46 \leq 60$ (True)
- $10 \geq 2(4) = 8$ (True)
- $4 \geq 3$ (True)
- $4, 10 \geq 0$ (True)
Since all conditions are satisfied,the solution set is represented by the $S_2$ region.

(D) To determine the linear constraints,we analyze the boundary lines of the shaded region:
$1$. The line passing through $(3, 0)$ and $(0, 2)$ is $2x + 3y = 6$. Since the shaded region is away from the origin,the constraint is $2x + 3y \geq 6$.
$2$. The line passing through $(0, 3)$ and $(6, 0)$ is $3x + 6y = 18$. Since the shaded region is towards the origin,the constraint is $3x + 6y \leq 18$.
$3$. The line passing through $(3, 0)$ and $(0, -1)$ is $x - 3y = 3$. Since the shaded region is towards the origin,the constraint is $x - 3y \leq 3$.
$4$. The line passing through $(0, 1)$ and $(2, 2)$ is $-x + 2y = 2$. Since the shaded region is towards the origin,the constraint is $-x + 2y \leq 2$.
$5$. Since the region is in the first quadrant,$x \geq 0$ and $y \geq 0$.
Thus,the correct constraints are $2x + 3y \geq 6, 3x + 6y \leq 18, x - 3y \leq 3, -x + 2y \leq 2, x \geq 0, y \geq 0$.

(A) The corner points of the feasible region are $A, B, C, D$.
From the graph,the lines are $y = 3$,$x = 4$,$y = x + 3$,and $2x + 3y = 12$.
$1$. Point $A$ is the intersection of $y = 3$ and $2x + 3y = 12$:
$2x + 3(3) = 12 \implies 2x = 3 \implies x = 1.5$. So,$A = (1.5, 3)$.
$2$. Point $B$ is the intersection of $y = 3$ and $x = 4$. So,$B = (4, 3)$.
$3$. Point $C$ is the intersection of $x = 4$ and $y = x + 3$:
$y = 4 + 3 = 7$. So,$C = (4, 7)$.
$4$. Point $D$ is the intersection of $y = x + 3$ and $2x + 3y = 12$:
$2x + 3(x + 3) = 12 \implies 5x + 9 = 12 \implies 5x = 3 \implies x = 0.6$.
$y = 0.6 + 3 = 3.6$. So,$D = (0.6, 3.6)$.
Now,evaluate $Z = 3x + 5y$ at these points:
$Z(A) = 3(1.5) + 5(3) = 4.5 + 15 = 19.5$.
$Z(B) = 3(4) + 5(3) = 12 + 15 = 27$.
$Z(C) = 3(4) + 5(7) = 12 + 35 = 47$.
$Z(D) = 3(0.6) + 5(3.6) = 1.8 + 18 = 19.8$.
The minimum value of $Z$ is $19.5$.

(A) Given the matrix equation $AX = B$:
$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 6 \\ 11 \\ 0 \end{bmatrix}$
This corresponds to the system of linear equations:
$1) \quad a + b + c = 6$
$2) \quad b + 3c = 11$
$3) \quad a - 2b + c = 0$
From equation $(3)$,we have $a + c = 2b$.
Substitute $a + c = 2b$ into equation $(1)$:
$2b + b = 6 \implies 3b = 6 \implies b = 2$.
Now,substitute $b = 2$ into equation $(2)$:
$2 + 3c = 11 \implies 3c = 9 \implies c = 3$.
Finally,substitute $b = 2$ and $c = 3$ into equation $(1)$:
$a + 2 + 3 = 6 \implies a + 5 = 6 \implies a = 1$.
We need to find the value of $2a + b + 2c$:
$2(1) + 2 + 2(3) = 2 + 2 + 6 = 10$.

(C) Given $A = \begin{bmatrix} 2a & -3b \\ 3 & 2 \end{bmatrix}$.
We know that $A \cdot \operatorname{adj} A = |A| I = \begin{bmatrix} |A| & 0 \\ 0 & |A| \end{bmatrix}$.
Calculating the determinant $|A| = (2a)(2) - (-3b)(3) = 4a + 9b$.
So,$A \cdot \operatorname{adj} A = \begin{bmatrix} 4a + 9b & 0 \\ 0 & 4a + 9b \end{bmatrix}$.
Now,calculate $A A^{T}$:
$A A^{T} = \begin{bmatrix} 2a & -3b \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 2a & 3 \\ -3b & 2 \end{bmatrix} = \begin{bmatrix} 4a^2 + 9b^2 & 6a - 6b \\ 6a - 6b & 9 + 4 \end{bmatrix} = \begin{bmatrix} 4a^2 + 9b^2 & 6a - 6b \\ 6a - 6b & 13 \end{bmatrix}$.
Given $A \cdot \operatorname{adj} A = A A^{T}$,we equate the matrices:
$\begin{bmatrix} 4a + 9b & 0 \\ 0 & 4a + 9b \end{bmatrix} = \begin{bmatrix} 4a^2 + 9b^2 & 6a - 6b \\ 6a - 6b & 13 \end{bmatrix}$.
Comparing the elements:
$6a - 6b = 0 \implies a = b$.
$4a + 9b = 13$.
Substituting $a = b$ into the second equation: $4a + 9a = 13 \implies 13a = 13 \implies a = 1$.
Since $a = b$,we have $a = 1$ and $b = 1$.
Therefore,$2a + 3b = 2(1) + 3(1) = 5$.

(B) We know that for a square matrix $A$ of order $n$,$|\operatorname{Adj} A| = |A|^{n-1}$.
Given that $B = \operatorname{Adj}(A)$ and $A$ is a $3 \times 3$ matrix,we have $n = 3$.
Therefore,$|B| = |A|^{3-1} = |A|^2$.
Given $|A| = 5$,we have $|B| = 5^2 = 25$.
Now,calculate the determinant of $B$:
$|B| = \begin{vmatrix} 1 & \alpha & 2 \\ 1 & 2 & 2 \\ 2 & 3 & 3 \end{vmatrix} = 1(2 \times 3 - 2 \times 3) - \alpha(1 \times 3 - 2 \times 2) + 2(1 \times 3 - 2 \times 2)$.
$|B| = 1(6 - 6) - \alpha(3 - 4) + 2(3 - 4)$.
$|B| = 0 - \alpha(-1) + 2(-1) = \alpha - 2$.
Equating the two values of $|B|$:
$\alpha - 2 = 25$.
$\alpha = 27$.

(A) We know that for a square matrix $A$ of order $n$,$|\operatorname{adj} A| = |A|^{n-1}$.
Given that $B = \operatorname{adj} A$ and $n = 3$,we have $|B| = |A|^{3-1} = |A|^2$.
Given $|A| = 4$,so $|B| = 4^2 = 16$.
Now,calculate the determinant of $B$:
$|B| = \begin{vmatrix} 3 & \alpha & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{vmatrix} = 3(9-1) - \alpha(3+1) - 1(1+3) = 3(8) - 4\alpha - 4 = 24 - 4\alpha - 4 = 20 - 4\alpha$.
Equating the two values of $|B|$:
$20 - 4\alpha = 16$
$4\alpha = 4$
$\alpha = 1$.

(B) We are given that $P = \text{adj}(A)$ and $|A| = 4$.
We know the property of the adjoint of a matrix: $|\text{adj}(A)| = |A|^{n-1}$,where $n$ is the order of the matrix.
Here,$n = 3$,so $|P| = |\text{adj}(A)| = |A|^{3-1} = |A|^2$.
Given $|A| = 4$,we have $|P| = 4^2 = 16$.
Now,calculate the determinant of $P$:
$|P| = \begin{vmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{vmatrix}$
$= 1(3 \times 4 - 3 \times 4) - \alpha(1 \times 4 - 3 \times 2) + 3(1 \times 4 - 3 \times 2)$
$= 1(12 - 12) - \alpha(4 - 6) + 3(4 - 6)$
$= 0 - \alpha(-2) + 3(-2)$
$= 2\alpha - 6$.
Equating the two values of $|P|$:
$2\alpha - 6 = 16$
$2\alpha = 22$
$\alpha = 11$.

(D) The sum of the product of elements of any row (or column) with their corresponding cofactors is equal to the determinant of the matrix $A$,i.e.,$\sum_{j=1}^{3} a_{ij}A_{ij} = |A|$.
Given $A = \begin{bmatrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$.
The expression $a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23}$ represents the determinant of matrix $A$ expanded along the second row.
Calculating the cofactors:
$A_{21} = (-1)^{2+1} \begin{vmatrix} 2 & 3 \\ 2 & 4 \end{vmatrix} = -1(8 - 6) = -2$
$A_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} = 1(4 - 3) = 1$
$A_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} = -1(2 - 2) = 0$
Now,substituting the values:
$a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} = (-1)(-2) + (1)(1) + (2)(0)$
$= 2 + 1 + 0 = 3$.

(D) Given $B = I - {}^{3}C_{1}(\operatorname{adj} A) + {}^{3}C_{2}(\operatorname{adj} A)^{2} - {}^{3}C_{3}(\operatorname{adj} A)^{3}$.
Using the binomial expansion $(I - X)^{3} = I - {}^{3}C_{1}X + {}^{3}C_{2}X^{2} - {}^{3}C_{3}X^{3}$,we get $B = (I - \operatorname{adj} A)^{3}$.
For $A = \begin{bmatrix} 2 & -1 \\ 0 & 2 \end{bmatrix}$,the adjoint is $\operatorname{adj} A = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}$.
Then $I - \operatorname{adj} A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix}$.
Now,$B = \begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix}^{3} = \begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix}$.
Calculating the square: $\begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$.
Calculating the cube: $\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & -3 \\ 0 & -1 \end{bmatrix}$.
The sum of all elements of matrix $B$ is $(-1) + (-3) + 0 + (-1) = -5$.

(A) Given $A = \begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix}$.
First,calculate $A^2 = A \times A = \begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 4+1 & -2-3 \\ -2-3 & 1+9 \end{bmatrix} = \begin{bmatrix} 5 & -5 \\ -5 & 10 \end{bmatrix}$.
Now,calculate $2A^2 + 5A = 2 \begin{bmatrix} 5 & -5 \\ -5 & 10 \end{bmatrix} + 5 \begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 10 & -10 \\ -10 & 20 \end{bmatrix} + \begin{bmatrix} 10 & -5 \\ -5 & 15 \end{bmatrix} = \begin{bmatrix} 20 & -15 \\ -15 & 35 \end{bmatrix}$.
Let $M = 2A^2 + 5A = \begin{bmatrix} 20 & -15 \\ -15 & 35 \end{bmatrix}$.
The determinant $|M| = (20)(35) - (-15)(-15) = 700 - 225 = 475$.
The inverse $M^{-1} = \frac{1}{|M|} \text{adj}(M) = \frac{1}{475} \begin{bmatrix} 35 & 15 \\ 15 & 20 \end{bmatrix}$.
Dividing by $5$,we get $M^{-1} = \frac{1}{95} \begin{bmatrix} 7 & 3 \\ 3 & 4 \end{bmatrix}$.

(D) Given $A = \begin{bmatrix} i & 1 \\ 1 & 0 \end{bmatrix}$.
Calculate $A^2 = \begin{bmatrix} i & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} i & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} i^2+1 & i \\ i & 1 \end{bmatrix} = \begin{bmatrix} 0 & i \\ i & 1 \end{bmatrix}$.
Calculate $A^3 = A^2 \times A = \begin{bmatrix} 0 & i \\ i & 1 \end{bmatrix} \begin{bmatrix} i & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix} = iI$.
Then $A^6 = (A^3)^2 = (iI)^2 = i^2 I = -I$.
Now,$B = A^{2029} = A^{6 \times 338 + 1} = (A^6)^{338} \times A = (-I)^{338} \times A = I \times A = A$.
Since $B = A$,we need $B^{-1} = A^{-1}$.
$|A| = (i)(0) - (1)(1) = -1$.
$A^{-1} = \frac{1}{|A|} \operatorname{adj} A = \frac{1}{-1} \operatorname{adj} A = -\operatorname{adj} A$.
Therefore,$B^{-1} = -\operatorname{adj} A$.

(B) Given $A = \begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix}$.
First,we find the determinant $|A| = (1)(1) - (2)(-5) = 1 + 10 = 11$.
Next,we find the inverse $A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{11} \begin{bmatrix} 1 & -2 \\ 5 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{11} & \frac{-2}{11} \\ \frac{5}{11} & \frac{1}{11} \end{bmatrix}$.
Given $A^{-1} = xA + yI$,we have:
$\begin{bmatrix} \frac{1}{11} & \frac{-2}{11} \\ \frac{5}{11} & \frac{1}{11} \end{bmatrix} = x \begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix} + y \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} x+y & 2x \\ -5x & x+y \end{bmatrix}$.
Comparing the elements,we get $2x = \frac{-2}{11} \Rightarrow x = \frac{-1}{11}$.
Also,$x+y = \frac{1}{11} \Rightarrow \frac{-1}{11} + y = \frac{1}{11} \Rightarrow y = \frac{2}{11}$.
Finally,$2x + 3y = 2(\frac{-1}{11}) + 3(\frac{2}{11}) = \frac{-2}{11} + \frac{6}{11} = \frac{4}{11}$.

(B) Given $A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix}$.
The determinant $|A| = (1)(1) - (\tan x)(-\tan x) = 1 + \tan^2 x = \sec^2 x$.
The transpose $A^T = \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}$.
The inverse $A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{1 + \tan^2 x} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}$.
Now,$A^T \cdot A^{-1} = \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \cdot \frac{1}{1 + \tan^2 x} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}$
$= \frac{1}{1 + \tan^2 x} \begin{bmatrix} 1 - \tan^2 x & -\tan x - \tan x \\ \tan x + \tan x & -\tan^2 x + 1 \end{bmatrix}$
$= \frac{1}{1 + \tan^2 x} \begin{bmatrix} 1 - \tan^2 x & -2 \tan x \\ 2 \tan x & 1 - \tan^2 x \end{bmatrix}$
$= \begin{bmatrix} \frac{1 - \tan^2 x}{1 + \tan^2 x} & \frac{-2 \tan x}{1 + \tan^2 x} \\ \frac{2 \tan x}{1 + \tan^2 x} & \frac{1 - \tan^2 x}{1 + \tan^2 x} \end{bmatrix}$
Using trigonometric identities $\cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x}$ and $\sin 2x = \frac{2 \tan x}{1 + \tan^2 x}$,we get:
$A^T \cdot A^{-1} = \begin{bmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix}$.

(B) First,calculate the sum of matrices $A$ and $B$:
$A+B = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} + \begin{bmatrix} 1 & 1 \\ 4 & -1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 6 & -2 \end{bmatrix}$
Next,find the determinant of $(A+B)$:
$|A+B| = (2)(-2) - (0)(6) = -4 - 0 = -4$
Since $|A+B| \neq 0$,the inverse exists.
For a matrix $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$,$M^{-1} = \frac{1}{|M|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Applying this to $A+B$:
$(A+B)^{-1} = \frac{1}{-4} \begin{bmatrix} -2 & 0 \\ -6 & 2 \end{bmatrix} = \begin{bmatrix} \frac{-2}{-4} & \frac{0}{-4} \\ \frac{-6}{-4} & \frac{2}{-4} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 0 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}$

(C) Let $p$ be the probability of drawing a ten in a single draw. Since there are $4$ tens in a pack of $52$ cards,$p = \frac{4}{52} = \frac{1}{13}$.
The probability of not drawing a ten is $q = 1 - p = 1 - \frac{1}{13} = \frac{12}{13}$.
Since the cards are drawn with replacement,the number of tens $X$ follows a binomial distribution $B(n, p)$ with $n = 2$ and $p = \frac{1}{13}$.
The mean of a binomial distribution is given by $E(X) = np$.
Therefore,the mean number of tens is $E(X) = 2 \times \frac{1}{13} = \frac{2}{13}$.

(C) Let a step forward be a success $(p = 0.4)$ and a step backward be a failure $(q = 0.6)$.
To be one step away from the starting point after $11$ steps,the number of forward steps $(n_f)$ and backward steps $(n_b)$ must satisfy $n_f - n_b = 1$ or $n_b - n_f = 1$.
Since $n_f + n_b = 11$,the possible cases are:
Case $1$: $n_f = 6$ and $n_b = 5$.
Case $2$: $n_f = 5$ and $n_b = 6$.
The required probability is $P = { }^{11} C_6 p^6 q^5 + { }^{11} C_5 p^5 q^6$.
Since ${ }^{11} C_6 = { }^{11} C_5$,we have:
$P = { }^{11} C_6 (p^6 q^5 + p^5 q^6) = { }^{11} C_6 p^5 q^5 (p + q)$.
Given $p + q = 0.4 + 0.6 = 1$,we get:
$P = { }^{11} C_6 (0.4)^5 (0.6)^5 (1) = { }^{11} C_6 (0.24)^5$.

(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 ₹ $-0.5$.

(B) Let $p$ be the probability of getting an even number and $q = 1 - p$ be the probability of getting an odd number.
Let the random variable $X \sim B(n, p)$ where $n = 5$.
Given that $P(X = 3) = 2 P(X = 2)$.
Using the binomial probability formula $P(X = k) = { }^n C_k p^k q^{n-k}$:
${ }^5 C_3 p^3 q^2 = 2 \times { }^5 C_2 p^2 q^3$.
Since ${ }^5 C_3 = 10$ and ${ }^5 C_2 = 10$,we have $10 p^3 q^2 = 20 p^2 q^3$.
Dividing both sides by $10 p^2 q^2$ (assuming $p, q \neq 0$),we get $p = 2q$.
Since $p + q = 1$,substituting $p = 2q$ gives $3q = 1$,so $q = \frac{1}{3}$ and $p = \frac{2}{3}$.
The probability of getting no even number in $5$ throws is $P(X = 0) = { }^5 C_0 p^0 q^5 = q^5 = (\frac{1}{3})^5 = \frac{1}{243}$.
In $6804$ sets of $5$ throws,the expected number of times to get no even number is $6804 \times \frac{1}{243} = 28$.

(C) Let $X$ be the number of problems the candidate is unable to solve. The probability of being unable to solve a problem is $p = 1 - \frac{4}{5} = \frac{1}{5}$.
The probability of being able to solve a problem is $q = \frac{4}{5}$.
Given $n = 50$ problems.
We need to find the probability that he is unable to solve less than $2$ problems,i.e.,$P(X < 2) = P(X = 0) + P(X = 1)$.
Using the binomial distribution formula $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$:
$P(X = 0) = {}^{50}C_{0} \left(\frac{1}{5}\right)^{0} \left(\frac{4}{5}\right)^{50} = \left(\frac{4}{5}\right)^{50}$.
$P(X = 1) = {}^{50}C_{1} \left(\frac{1}{5}\right)^{1} \left(\frac{4}{5}\right)^{49} = 50 \times \frac{1}{5} \times \left(\frac{4}{5}\right)^{49} = 10 \times \left(\frac{4}{5}\right)^{49}$.
$P(X < 2) = \left(\frac{4}{5}\right) \left(\frac{4}{5}\right)^{49} + 10 \left(\frac{4}{5}\right)^{49} = \left(\frac{4}{5} + 10\right) \left(\frac{4}{5}\right)^{49} = \left(\frac{4 + 50}{5}\right) \left(\frac{4}{5}\right)^{49} = \frac{54}{5} \left(\frac{4}{5}\right)^{49}$.

(A) Let $X$ denote the number of queens obtained in two draws with replacement.
The probability of drawing a queen in a single draw is $p = \frac{4}{52} = \frac{1}{13}$.
The probability of not drawing a queen is $q = 1 - p = 1 - \frac{1}{13} = \frac{12}{13}$.
Since the draws are with replacement,the trials are independent and follow a binomial distribution $B(n, p)$ with $n = 2$ and $p = \frac{1}{13}$.
$P(X = 0) = ^2C_0 \times (\frac{12}{13})^2 = 1 \times \frac{144}{169} = \frac{144}{169}$.
$P(X = 1) = ^2C_1 \times (\frac{1}{13})^1 \times (\frac{12}{13})^1 = 2 \times \frac{12}{169} = \frac{24}{169}$.
$P(X = 2) = ^2C_2 \times (\frac{1}{13})^2 = 1 \times \frac{1}{169} = \frac{1}{169}$.
Thus,the probability distribution is as given in option $A$.

(D) Given $P(A) = \frac{1}{3}$,$P(B) = \frac{1}{5}$,and $P(A \cup B) = \frac{1}{3}$.
First,find $P(A^{\prime} \cap B^{\prime}) = 1 - P(A \cup B) = 1 - \frac{1}{3} = \frac{2}{3}$.
Also,$P(A^{\prime}) = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3}$ and $P(B^{\prime}) = 1 - P(B) = 1 - \frac{1}{5} = \frac{4}{5}$.
Now,$P(A^{\prime} | B^{\prime}) = \frac{P(A^{\prime} \cap B^{\prime})}{P(B^{\prime})} = \frac{2/3}{4/5} = \frac{2}{3} \times \frac{5}{4} = \frac{5}{6}$.
And $P(B^{\prime} | A^{\prime}) = \frac{P(A^{\prime} \cap B^{\prime})}{P(A^{\prime})} = \frac{2/3}{2/3} = 1$.
Therefore,$P(A^{\prime} | B^{\prime}) + P(B^{\prime} | A^{\prime}) = \frac{5}{6} + 1 = \frac{11}{6}$.

(A) Let $X$ denote the number of defective bulbs.
$p$ is the probability that a bulb is defective:
$p = \frac{10}{100} = \frac{1}{10}$
$q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10}$
Since the bulbs are selected from a large lot,we use the binomial distribution:
$P(X = r) = { }^5 C_r \left(\frac{1}{10}\right)^r \left(\frac{9}{10}\right)^{5-r}, r = 0, 1, \dots, 5$
We need to find the probability that the store receives at most one defective bulb,i.e.,$P(X \leq 1)$:
$P(X \leq 1) = P(X = 0) + P(X = 1)$
$P(X = 0) = { }^5 C_0 \left(\frac{1}{10}\right)^0 \left(\frac{9}{10}\right)^5 = \left(\frac{9}{10}\right)^5$
$P(X = 1) = { }^5 C_1 \left(\frac{1}{10}\right)^1 \left(\frac{9}{10}\right)^4 = 5 \times \frac{1}{10} \times \left(\frac{9}{10}\right)^4 = \frac{1}{2} \left(\frac{9}{10}\right)^4$
$P(X \leq 1) = \left(\frac{9}{10}\right)^5 + \frac{1}{2} \left(\frac{9}{10}\right)^4 = \left(\frac{9}{10}\right)^4 \left[ \frac{9}{10} + \frac{1}{2} \right] = \left(\frac{9}{10}\right)^4 \left[ \frac{9+5}{10} \right] = \left(\frac{9}{10}\right)^4 \left( \frac{14}{10} \right) = \frac{7}{5} \left(\frac{9}{10}\right)^4$

(D) Given that $A$ and $B$ are independent events,we have $P(A \cap B) = P(A) \cdot P(B)$.
From the given equation: $P(A \cup B) = 2P(B) - P(A)$.
Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$,we get:
$P(A) + P(B) - P(A \cap B) = 2P(B) - P(A)$.
Substituting $P(A \cap B) = P(A) \cdot P(B)$:
$P(A) + P(B) - P(A) \cdot P(B) = 2P(B) - P(A)$.
Rearranging the terms to solve for $P(B)$:
$2P(A) = P(B) + P(A) \cdot P(B)$.
$2P(A) = P(B)(1 + P(A))$.
$P(B) = \frac{2P(A)}{1 + P(A)}$.
Given $P(A) = \frac{1}{4}$,we substitute the value:
$P(B) = \frac{2 \times \frac{1}{4}}{1 + \frac{1}{4}} = \frac{\frac{1}{2}}{\frac{5}{4}} = \frac{1}{2} \times \frac{4}{5} = \frac{2}{5}$.

(B) An experiment succeeds twice as often as it fails.
Let $p$ be the probability of success and $q$ be the probability of failure.
Given $p = 2q$.
Since $p + q = 1$,we have $2q + q = 1$,which implies $3q = 1$,so $q = \frac{1}{3}$ and $p = \frac{2}{3}$.
We have $n = 6$ trials. Let $X$ be the number of successes,where $X \sim B(n, p)$.
The required probability is $P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6)$.
Using the binomial distribution formula $P(X = k) = {^nC_k} p^k q^{n-k}$:
$P(X = 4) = {^6C_4} (\frac{2}{3})^4 (\frac{1}{3})^2 = 15 \times \frac{16}{81} \times \frac{1}{9} = \frac{240}{729}$.
$P(X = 5) = {^6C_5} (\frac{2}{3})^5 (\frac{1}{3})^1 = 6 \times \frac{32}{243} \times \frac{1}{3} = \frac{192}{729}$.
$P(X = 6) = {^6C_6} (\frac{2}{3})^6 (\frac{1}{3})^0 = 1 \times \frac{64}{729} \times 1 = \frac{64}{729}$.
Adding these probabilities: $P(X \geq 4) = \frac{240 + 192 + 64}{729} = \frac{496}{729}$.

(B) The probability mass function for a Binomial distribution is given by $P(X=k) = { }^n C_k p^k (1-p)^{n-k}$.
Given $n=4$,we have:
$P(X=3) = { }^4 C_3 p^3 (1-p)^1 = 4p^3(1-p)$
$P(X=2) = { }^4 C_2 p^2 (1-p)^2 = 6p^2(1-p)^2$
Given the condition $2 P(X=3) = 3 P(X=2)$:
$2 \times [4p^3(1-p)] = 3 \times [6p^2(1-p)^2]$
$8p^3(1-p) = 18p^2(1-p)^2$
Dividing both sides by $2p^2(1-p)$ (assuming $p \neq 0, 1$):
$4p = 9(1-p)$
$4p = 9 - 9p$
$13p = 9 \implies p = \frac{9}{13}$
Then $q = 1 - p = 1 - \frac{9}{13} = \frac{4}{13}$.
The variance of a Binomial distribution is $npq$:
$\text{Variance} = 4 \times \frac{9}{13} \times \frac{4}{13} = \frac{144}{169}$.

(D) Given,$P(X=4) = \frac{135}{2^{12}}$.
Using the binomial probability formula $P(X=k) = {}^nC_k p^k q^{n-k}$,we have:
${}^6C_4 p^4 q^2 = \frac{135}{2^{12}}$.
Since ${}^6C_4 = 15$,we get $15 p^4 q^2 = \frac{135}{2^{12}}$.
Dividing by $15$,$p^4 q^2 = \frac{9}{2^{12}} = \frac{3^2}{(2^6)^2} = \left(\frac{3}{64}\right)^2$.
Taking the square root,$p^2 q = \frac{3}{64}$.
Substituting $q = 1-p$,we have $p^2(1-p) = \frac{3}{64}$.
By inspection,if $p = \frac{1}{4}$,then $p^2(1-p) = (\frac{1}{16})(\frac{3}{4}) = \frac{3}{64}$.
Thus,$p = \frac{1}{4}$ and $q = \frac{3}{4}$.
The variance of a binomial distribution is given by $npq$.
Variance $= 6 \times \frac{1}{4} \times \frac{3}{4} = \frac{18}{16} = \frac{9}{8}$.

(B) Given the Binomial distribution $B(n, p)$ with $n=7$,the probability mass function is $P(X=k) = {^nC_k} p^k q^{n-k}$,where $q = 1-p$.
Given $P(X=3) = 5 P(X=4)$.
Substituting the formula:
${^7C_3} p^3 q^4 = 5 \times {^7C_4} p^4 q^3$
Since ${^7C_3} = {^7C_4} = 35$,we can divide both sides by $35 p^3 q^3$:
$q = 5p$
Since $q = 1-p$,we have $1-p = 5p$,which implies $6p = 1$,so $p = \frac{1}{6}$.
Then $q = 1 - \frac{1}{6} = \frac{5}{6}$.
The variance of a Binomial distribution is given by $Var(X) = npq$.
$Var(X) = 7 \times \frac{1}{6} \times \frac{5}{6} = \frac{35}{36}$.

(D) For a Binomial distribution with $n$ trials,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $q = 1 - p$.
Given $n = 5$ and $\mu + \sigma^2 = 1.8$.
Substituting the values,we get:
$np + npq = 1.8$
$5p + 5p(1 - p) = 1.8$
$5p + 5p - 5p^2 = 1.8$
$10p - 5p^2 = 1.8$
$5p^2 - 10p + 1.8 = 0$
Multiplying by $10$ to clear the decimal:
$50p^2 - 100p + 18 = 0$
Dividing by $2$:
$25p^2 - 50p + 9 = 0$
Using the quadratic formula $p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$p = \frac{50 \pm \sqrt{2500 - 4(25)(9)}}{50} = \frac{50 \pm \sqrt{2500 - 900}}{50} = \frac{50 \pm \sqrt{1600}}{50} = \frac{50 \pm 40}{50}$
$p_1 = \frac{90}{50} = 1.8$ (Not possible since $0 \le p \le 1$)
$p_2 = \frac{10}{50} = 0.2$
Thus,$p = 0.2$.

(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:

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

(B) For a Binomial Distribution,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $n$ is the number of trials,$p$ is the probability of success,and $q = 1-p$ is the probability of failure.
Given $n = 10$ and $\text{mean} + \text{variance} = \frac{15}{2}$.
Substituting the formulas: $np + npq = \frac{15}{2}$.
Since $q = 1-p$,we have $np + np(1-p) = \frac{15}{2}$.
$10p + 10p(1-p) = 7.5$.
$10p + 10p - 10p^2 = 7.5$.
$20p - 10p^2 = 7.5$.
Divide by $2.5$: $8p - 4p^2 = 3$.
$4p^2 - 8p + 3 = 0$.
Solving the quadratic equation: $(2p-1)(2p-3) = 0$.
This gives $p = \frac{1}{2}$ or $p = \frac{3}{2}$.
Since $0 < p < 1$,we must have $p = \frac{1}{2}$.
Then $q = 1 - \frac{1}{2} = \frac{1}{2}$.
Therefore,the variance is $\sigma^2 = npq = 10 \times \frac{1}{2} \times \frac{1}{2} = 2.5$.

(A) The probability mass function for a binomial distribution is given by $P(X=k) = {^nC_k} p^k q^{n-k}$,where $q = 1-p$.
Given $n=6$,we have $P(X=4) = {^6C_4} p^4 q^2$ and $P(X=2) = {^6C_2} p^2 q^4$.
The condition is $9 \cdot P(X=4) = P(X=2)$.
Substituting the values: $9 \cdot {^6C_4} p^4 q^2 = {^6C_2} p^2 q^4$.
Since ${^6C_4} = \frac{6 \times 5}{2 \times 1} = 15$ and ${^6C_2} = \frac{6 \times 5}{2 \times 1} = 15$,the equation becomes:
$9 \cdot 15 \cdot p^4 q^2 = 15 \cdot p^2 q^4$.
Dividing both sides by $15 p^2 q^2$ (assuming $p, q \neq 0$):
$9 p^2 = q^2$.
Taking the square root of both sides: $3p = q$.
Since $q = 1-p$,we have $3p = 1-p$.
$4p = 1$,which gives $p = \frac{1}{4}$.

(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 $E(X) = \sum x_i P(x_i) = 0 \times P(X=0) + 1 \times P(X=1) + 2 \times P(X=2)$
$E(X) = 0 + \frac{168}{380} + 2 \times \frac{30}{380} = \frac{168 + 60}{380} = \frac{228}{380} = 0.6$.

(D) Let $X \sim B(n, p)$.
Given that the mean $np = 8$ and variance $npq = 4$.
Since $q = 1 - p$,we have $8q = 4$,which implies $q = \frac{1}{2}$ and $p = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $np = 8$,we get $n = 16$.
We need to find $P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$.
Using the formula $P(X=r) = {}^{n}C_{r} p^{r} q^{n-r}$,we have:
$P(X \leq 2) = {}^{16}C_{0} (\frac{1}{2})^{0} (\frac{1}{2})^{16} + {}^{16}C_{1} (\frac{1}{2})^{1} (\frac{1}{2})^{15} + {}^{16}C_{2} (\frac{1}{2})^{2} (\frac{1}{2})^{14}$.
$P(X \leq 2) = \frac{{}^{16}C_{0} + {}^{16}C_{1} + {}^{16}C_{2}}{2^{16}}$.
Calculating the combinations: ${}^{16}C_{0} = 1$,${}^{16}C_{1} = 16$,and ${}^{16}C_{2} = \frac{16 \times 15}{2} = 120$.
Thus,$P(X \leq 2) = \frac{1 + 16 + 120}{2^{16}} = \frac{137}{2^{16}}$.
Comparing this with $\frac{K}{2^{16}}$,we get $K = 137$.

(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:

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

(A) Let $X$ be the random variable representing the number of jacks drawn in $2$ trials with replacement.
The total number of cards is $52$,and the number of jacks is $4$.
The probability of drawing a jack in a single trial is $p = \frac{4}{52} = \frac{1}{13}$.
The probability of not drawing a jack is $q = 1 - p = 1 - \frac{1}{13} = \frac{12}{13}$.
Since the trials are independent (with replacement),$X$ follows a binomial distribution $B(n, p)$ where $n = 2$.
$P(X = x) = ^nC_x p^x q^{n-x}$
$P(X = 0) = ^2C_0 (\frac{1}{13})^0 (\frac{12}{13})^2 = 1 \times 1 \times \frac{144}{169} = \frac{144}{169}$
$P(X = 1) = ^2C_1 (\frac{1}{13})^1 (\frac{12}{13})^1 = 2 \times \frac{1}{13} \times \frac{12}{13} = \frac{24}{169}$
$P(X = 2) = ^2C_2 (\frac{1}{13})^2 (\frac{12}{13})^0 = 1 \times \frac{1}{169} \times 1 = \frac{1}{169}$
Thus,the probability distribution is given by option $(A)$.

(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) The prime numbers in the set of values for $X$ are $\{2, 3, 5, 7\}$.
$P(E) = P(X=2) + P(X=3) + P(X=5) + P(X=7) = 0.23 + 0.12 + 0.08 + 0.05 = 0.48$.
The event $F = \{X < 5\}$ corresponds to $X \in \{1, 2, 3, 4\}$.
$P(F) = P(X=1) + P(X=2) + P(X=3) + P(X=4) = 0.15 + 0.23 + 0.12 + 0.20 = 0.70$.
The event $E \cap F$ represents $X$ being a prime number less than $5$,which is $\{2, 3\}$.
$P(E \cap F) = P(X=2) + P(X=3) = 0.23 + 0.12 = 0.35$.
Using the addition theorem of probability:
$P(E \cup F) = P(E) + P(F) - P(E \cap F)$
$P(E \cup F) = 0.48 + 0.70 - 0.35 = 0.83$.

(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,$\sum_{x=1}^{n} x^2 = \frac{n(n+1)(2n+1)}{6}$.
$E(X) = \frac{2}{n(n+1)} \cdot \frac{n(n+1)(2n+1)}{6}$
$E(X) = \frac{2n+1}{3}$