MHT CET 2020 Mathematics Question Paper with Answer and Solution

(A) The feasible region is determined by the constraints $2x + y \geq 16$,$x \geq 6$,and $y \geq 1$.
The corner points of the feasible region are found by the intersection of these lines:
$1$. Intersection of $x = 6$ and $y = 1$ is $(6, 1)$,but this point does not satisfy $2x + y \geq 16$ $(12 + 1 = 13 < 16)$.
$2$. Intersection of $2x + y = 16$ and $y = 1$: $2x + 1 = 16 \implies 2x = 15 \implies x = 7.5$. So,point $E = (7.5, 1)$.
$3$. Intersection of $2x + y = 16$ and $x = 6$: $2(6) + y = 16 \implies 12 + y = 16 \implies y = 4$. So,point $F = (6, 4)$.
Since the region is unbounded,we check the values of $Z$ at the corner points:
$Z(E) = Z(7.5, 1) = 6(7.5) + 2(1) = 45 + 2 = 47$.
$Z(F) = Z(6, 4) = 6(6) + 2(4) = 36 + 8 = 44$.
Comparing the values,the minimum value is $44$.

(D) The feasible region is determined by the constraints $x + y \leq 3$,$4x + y \leq 6$,$x \geq 0$,and $y \geq 0$.
The corner points of the feasible region are $O(0, 0)$,$A(1.5, 0)$,$B(1, 2)$,and $C(0, 3)$.
We evaluate the objective function $Z = 8x + 3y$ at each corner point:
$1$. At $O(0, 0)$,$Z = 8(0) + 3(0) = 0$.
$2$. At $A(1.5, 0)$,$Z = 8(1.5) + 3(0) = 12$.
$3$. At $B(1, 2)$,$Z = 8(1) + 3(2) = 8 + 6 = 14$.
$4$. At $C(0, 3)$,$Z = 8(0) + 3(3) = 9$.
The maximum value of $Z$ is $14$,which occurs at the point $(1, 2)$.
Therefore,the optimal solution is $x = 1, y = 2$.

(C) The constraints are $x + y \geq 5$,$0 \leq x \leq 4$,and $y \geq 2$.
From the graph,the feasible region is the triangle formed by the intersection of the lines $x + y = 5$,$x = 4$,and $y = 2$.
To find the vertices of the feasible region:
$1$. Intersection of $x + y = 5$ and $y = 2$: Substituting $y = 2$ into $x + y = 5$,we get $x = 3$. So,vertex $P = (3, 2)$.
$2$. Intersection of $x + y = 5$ and $x = 4$: Substituting $x = 4$ into $x + y = 5$,we get $y = 1$. However,the constraint is $y \geq 2$. Looking at the graph,the vertex is at $x = 4$ and $y = 2$,which is point $D = (4, 2)$.
$3$. The third vertex is the intersection of $x = 4$ and $x + y = 5$,which is $C = (4, 1)$. But since $y \geq 2$,the region is bounded by $P(3, 2)$,$D(4, 2)$,and the point where $x=4$ meets $x+y=5$ is $(4, 1)$,which is outside the feasible region $y \geq 2$.
Re-evaluating the vertices from the graph: The vertices of the shaded region are $P(3, 2)$,$D(4, 2)$,and the point where $x=4$ intersects $x+y=5$ is $(4, 1)$,but the region is bounded by $y \geq 2$. The vertices are $P(3, 2)$ and $D(4, 2)$. The line $x+y=5$ passes through $(3, 2)$ and $(4, 1)$. The feasible region is the triangle with vertices $(3, 2)$,$(4, 2)$,and $(4, 1)$ is incorrect. The region is bounded by $x+y \geq 5$,$x \leq 4$,$y \geq 2$. The vertices are $P(3, 2)$ and $D(4, 2)$. The line $x+y=5$ intersects $x=4$ at $(4, 1)$. The region is the triangle with vertices $(3, 2)$,$(4, 2)$,and $(4, 1)$ is not possible as $y \geq 2$. The vertices are $P(3, 2)$ and $D(4, 2)$ and the line $x+y=5$ segment. The vertices are $P(3, 2)$ and $D(4, 2)$. Evaluating $Z = 5x + 8y$ at these points:
$Z(P) = 5(3) + 8(2) = 15 + 16 = 31$.
$Z(D) = 5(4) + 8(2) = 20 + 16 = 36$.
The minimum value is $31$.

(C) The feasible region is determined by the constraints $3x+2y \leq 18$,$x \leq 4$,$y \leq 6$,and $x, y \geq 0$. The vertices of the feasible region are $O(0,0)$,$D(4,0)$,$Q(4,3)$,$P(2,6)$,and $C(0,6)$.
We evaluate the objective function $Z=3x+5y$ at each vertex:
At $O(0,0)$: $Z = 3(0) + 5(0) = 0$
At $D(4,0)$: $Z = 3(4) + 5(0) = 12$
At $Q(4,3)$: $Z = 3(4) + 5(3) = 12 + 15 = 27$
At $P(2,6)$: $Z = 3(2) + 5(6) = 6 + 30 = 36$
At $C(0,6)$: $Z = 3(0) + 5(6) = 30$
Comparing these values,the maximum value of $Z$ is $36$ at the point $(2,6)$.

(C) The constraints are $x \leq 3, y \leq 3, x+y \leq 5, x \geq 0, y \geq 0$.
We identify the vertices of the feasible region by finding the intersection points of the boundary lines:
$1$. $x=0, y=0 \Rightarrow O(0,0)$
$2$. $x=3, y=0 \Rightarrow A(3,0)$
$3$. $x=3, x+y=5 \Rightarrow P(3,2)$
$4$. $x+y=5, y=3 \Rightarrow Q(2,3)$
$5$. $x=0, y=3 \Rightarrow D(0,3)$
Now,we evaluate the objective function $Z=10x+25y$ at each vertex:
- At $O(0,0): Z = 10(0) + 25(0) = 0$
- At $A(3,0): Z = 10(3) + 25(0) = 30$
- At $P(3,2): Z = 10(3) + 25(2) = 30 + 50 = 80$
- At $Q(2,3): Z = 10(2) + 25(3) = 20 + 75 = 95$
- At $D(0,3): Z = 10(0) + 25(3) = 0 + 75 = 75$
The maximum value of $Z$ is $95$,which occurs at the point $(2,3)$.
Therefore,the correct option is $C$.

(A) The given constraints are:
$1$) $x + y \leq 1$
$2$) $2x + 2y \geq 6 \implies x + y \geq 3$
$3$) $x \geq 0, y \geq 0$
From the first constraint,the region is towards the origin for the line $x + y = 1$.
From the second constraint,the region is away from the origin for the line $x + y = 3$.
Since there is no point $(x, y)$ that satisfies both $x + y \leq 1$ and $x + y \geq 3$ simultaneously,there is no common feasible region.
Therefore,the given $L$.$P$.$P$. has no solution.

(D) The constraints are $x + y \geq 5$,$x \leq 4$,$y \leq 2$,$x \geq 0$,and $y \geq 0$.
To find the feasible region,we identify the intersection points of the lines:
$1$. $x + y = 5$ and $x = 4$ gives $4 + y = 5 \implies y = 1$. Point: $(4, 1)$.
$2$. $x + y = 5$ and $y = 2$ gives $x + 2 = 5 \implies x = 3$. Point: $(3, 2)$.
$3$. $x = 4$ and $y = 2$ gives the point $(4, 2)$.
The vertices of the feasible region are $(4, 1)$,$(4, 2)$,and $(3, 2)$.
Now,evaluate the objective function $Z = 5x + 8y$ at these vertices:
- At $(4, 1)$: $Z = 5(4) + 8(1) = 20 + 8 = 28$.
- At $(4, 2)$: $Z = 5(4) + 8(2) = 20 + 16 = 36$.
- At $(3, 2)$: $Z = 5(3) + 8(2) = 15 + 16 = 31$.
The minimum value is $28$,which occurs at the point $(4, 1)$.

(A) In a Linear Programming Problem $(L.P.P.)$,the objective function is a linear function.
If the objective function attains the same optimal value at two distinct corner points of the feasible region,then it will also attain the same optimal value at every point on the line segment joining these two points.
Since a line segment contains an infinite number of points,the $L.P.P.$ will have infinite solutions.

(C) First,we calculate $A^{2}$:
$A^{2} = A \times A = \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix} = \begin{bmatrix} (1)(1) + (0)(-1) & (1)(0) + (0)(7) \\ (-1)(1) + (7)(-1) & (-1)(0) + (7)(7) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -8 & 49 \end{bmatrix}$
Given the equation $A^{2} = 8A + kI$,we substitute the matrices:
$\begin{bmatrix} 1 & 0 \\ -8 & 49 \end{bmatrix} = 8 \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix} + k \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$\begin{bmatrix} 1 & 0 \\ -8 & 49 \end{bmatrix} = \begin{bmatrix} 8 & 0 \\ -8 & 56 \end{bmatrix} + \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}$
$\begin{bmatrix} 1 & 0 \\ -8 & 49 \end{bmatrix} = \begin{bmatrix} 8 + k & 0 \\ -8 & 56 + k \end{bmatrix}$
By comparing the corresponding elements,we get:
$8 + k = 1 \Rightarrow k = 1 - 8 = -7$
Also,$56 + k = 49 \Rightarrow k = 49 - 56 = -7$
Thus,the value of $k$ is $-7$.

(C) We know that for a square matrix $A$ of order $n$,the property of the adjoint of an adjoint is given by $\operatorname{adj}(\operatorname{adj} A) = |A|^{n-2} A$.
Here,the order of matrix $A$ is $n = 3$.
First,calculate the determinant $|A|$:
$|A| = \begin{vmatrix} 1 & 2 & i \\ 1 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix} = 1(0 - 1) - 2(0 - 1) + i(1 - 1) = -1 + 2 + 0 = 1$.
Now,substitute $n = 3$ and $|A| = 1$ into the formula:
$\operatorname{adj}(\operatorname{adj} A) = (1)^{3-2} A = (1)^1 A = A$.
Therefore,$[\operatorname{adj}(\operatorname{adj} A)]^{-1} = (A)^{-1} = A^{-1}$.

(A) For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$,the adjoint is given by $\operatorname{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Given $A = \begin{bmatrix} 2 & -3 \\ 3 & 5 \end{bmatrix}$,we have $a = 2, b = -3, c = 3, d = 5$.
Substituting these values into the formula,we get $\operatorname{adj}(A) = \begin{bmatrix} 5 & -(-3) \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 3 \\ -3 & 2 \end{bmatrix}$.

(B) Let the matrix be $A = \begin{bmatrix} 1 & 3 & 2 \\ -2 & 0 & 1 \\ 5 & 2 & 1 \end{bmatrix}$.
The cofactors of the elements of the second row $(A_{21}, A_{22}, A_{23})$ are calculated as follows:
$A_{21} = (-1)^{2+1} \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} = (-1)(3(1) - 2(2)) = (-1)(3 - 4) = (-1)(-1) = 1$.
$A_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 2 \\ 5 & 1 \end{vmatrix} = (1)(1(1) - 5(2)) = (1)(1 - 10) = -9$.
$A_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 3 \\ 5 & 2 \end{vmatrix} = (-1)(1(2) - 5(3)) = (-1)(2 - 15) = (-1)(-13) = 13$.
The sum of the cofactors is $A_{21} + A_{22} + A_{23} = 1 + (-9) + 13 = 5$.
Therefore,the correct option is $B$.

(B) The cofactor $C_{ij}$ of an element $a_{ij}$ is given by $C_{ij} = (-1)^{i+j} M_{ij}$,where $M_{ij}$ is the minor of the element $a_{ij}$.
For the first column,we need to find $C_{11}, C_{21},$ and $C_{31}$.
$C_{11} = (-1)^{1+1} \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} = (1)(2 - 2) = 0$.
$C_{21} = (-1)^{2+1} \begin{vmatrix} 0 & -1 \\ 1 & 2 \end{vmatrix} = (-1)(0 - (-1)) = (-1)(1) = -1$.
$C_{31} = (-1)^{3+1} \begin{vmatrix} 0 & -1 \\ 1 & 2 \end{vmatrix} = (1)(0 - (-1)) = (1)(1) = 1$.
Thus,the cofactors are $0, -1, 1$.

(D) Given the matrix $A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$.
The determinant of $A$ is $|A| = (2)(2) - (-1)(-1) = 4 - 1 = 3$.
The adjoint of $A$ is $\text{adj}(A) = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$.
The inverse of a matrix is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
Substituting the values,we get $A^{-1} = \frac{1}{3} \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$.
Alternatively,using the characteristic equation $A^{2} - 4A + 3I = 0$,multiply by $A^{-1}$:
$A - 4I + 3A^{-1} = 0 \Rightarrow 3A^{-1} = 4I - A$.
$3A^{-1} = 4 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$.
Thus,$A^{-1} = \frac{1}{3} \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$.

(C) square matrix $A$ is invertible if and only if its determinant $|A| \neq 0$.
Let us calculate the determinant for each matrix:
$1$. For $A_{1} = \begin{bmatrix} 4 & 2 \\ 2 & 1 \end{bmatrix}$,$|A_{1}| = (4 \times 1) - (2 \times 2) = 4 - 4 = 0$. Thus,$A_{1}$ is not invertible.
$2$. For $A_{2} = \begin{bmatrix} -1 & -2 & 3 \\ 4 & 5 & 7 \\ 2 & 4 & -6 \end{bmatrix}$,notice that row $3$ is $-2$ times row $1$ $(R_{3} = -2R_{1})$. Since two rows are proportional,$|A_{2}| = 0$. Thus,$A_{2}$ is not invertible.
$3$. For $A_{3} = \begin{bmatrix} 1 & 0 & 0 \\ 5 & 2 & 1 \\ 7 & 2 & 1 \end{bmatrix}$,notice that row $2$ and row $3$ are not proportional,but let's calculate: $|A_{3}| = 1(2-2) - 0 + 0 = 0$. Thus,$A_{3}$ is not invertible.
$4$. For $A_{4} = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{bmatrix}$,$|A_{4}| = 1(2-6) - 0(0-3) + 1(0-2) = 1(-4) + 1(-2) = -4 - 2 = -6$.
Since $|A_{4}| = -6 \neq 0$,the matrix $A_{4}$ is invertible.

(A) First,calculate the product $AB$:
$AB = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} (2 \times 1 + 3 \times 3) & (2 \times 0 + 3 \times 1) \\ (1 \times 1 + 2 \times 3) & (1 \times 0 + 2 \times 1) \end{bmatrix} = \begin{bmatrix} 11 & 3 \\ 7 & 2 \end{bmatrix}$
Next,find the determinant $|AB|$:
$|AB| = (11 \times 2) - (3 \times 7) = 22 - 21 = 1$
Now,find the adjoint of $AB$:
For a matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$,the adjoint is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Thus,$\text{adj}(AB) = \begin{bmatrix} 2 & -3 \\ -7 & 11 \end{bmatrix}$
Finally,$(AB)^{-1} = \frac{1}{|AB|} \text{adj}(AB) = \frac{1}{1} \begin{bmatrix} 2 & -3 \\ -7 & 11 \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ -7 & 11 \end{bmatrix}$

(D) Given $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta \end{bmatrix}$.
First,we find the determinant of $A$:
$|A| = (\cos \theta)(-\cos \theta) - (-\sin \theta)(-\sin \theta) = -\cos^2 \theta - \sin^2 \theta = -(\cos^2 \theta + \sin^2 \theta) = -1$.
Next,we find the adjoint of $A$ by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$\text{adj}(A) = \begin{bmatrix} -\cos \theta & \sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$.
Finally,the inverse is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A)$:
$A^{-1} = \frac{1}{-1} \begin{bmatrix} -\cos \theta & \sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta \end{bmatrix}$.
Thus,$A^{-1} = A$.

(D) We know that for any invertible matrices $A$ and $B$,the property $(B^{-1} A^{-1})^{-1} = (A^{-1})^{-1} (B^{-1})^{-1}$ holds true.
Since $(A^{-1})^{-1} = A$ and $(B^{-1})^{-1} = B$,the expression simplifies to $AB$.
Now,we calculate the product $AB$:
$AB = \left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right] \left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]$
$AB = \left[\begin{array}{cc} (2)(2) + (3)(-1) & (2)(-3) + (3)(2) \\ (1)(2) + (2)(-1) & (1)(-3) + (2)(2) \end{array}\right]$
$AB = \left[\begin{array}{cc} 4 - 3 & -6 + 6 \\ 2 - 2 & -3 + 4 \end{array}\right]$
$AB = \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$
Thus,the correct option is $D$.

(A) We know that $A^4 A^{-1} = A^{4-1} = A^3$.
Since $A$ is a diagonal matrix,$A^n = \begin{bmatrix} a_{11}^n & 0 & 0 \\ 0 & a_{22}^n & 0 \\ 0 & 0 & a_{33}^n \end{bmatrix}$.
Given $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -1 \end{bmatrix}$,we calculate $A^3$:
$A^3 = \begin{bmatrix} 2^3 & 0 & 0 \\ 0 & (-2)^3 & 0 \\ 0 & 0 & (-1)^3 \end{bmatrix} = \begin{bmatrix} 8 & 0 & 0 \\ 0 & -8 & 0 \\ 0 & 0 & -1 \end{bmatrix}$.
Thus,$A^4 A^{-1} = \begin{bmatrix} 8 & 0 & 0 \\ 0 & -8 & 0 \\ 0 & 0 & -1 \end{bmatrix}$.

(D) Given the matrix $M = \left[\begin{array}{ccc}1 & \omega & \omega^{2} \\ \omega & \omega^{2} & 1 \\ \omega^{2} & 1 & \omega\end{array}\right]$.
The matrix $A$ consists of the reciprocals of the elements of $M$,so $A = \left[\begin{array}{ccc}1 & \frac{1}{\omega} & \frac{1}{\omega^2} \\ \frac{1}{\omega} & \frac{1}{\omega^2} & 1 \\ \frac{1}{\omega^2} & 1 & \frac{1}{\omega}\end{array}\right]$.
Using the property $\omega^3 = 1$,we can write $\frac{1}{\omega} = \omega^2$ and $\frac{1}{\omega^2} = \omega$.
Thus,$A = \left[\begin{array}{ccc}1 & \omega^2 & \omega \\ \omega^2 & \omega & 1 \\ \omega & 1 & \omega^2\end{array}\right]$.
Now,calculate the determinant $|A|$:
$|A| = 1(\omega^3 - 1) - \omega^2(\omega^4 - \omega) + \omega(\omega^2 - \omega^2)$
$|A| = 1(1 - 1) - \omega^2(\omega - \omega) + \omega(0)$
$|A| = 0 - 0 + 0 = 0$.
Since the determinant of matrix $A$ is $0$,the matrix $A$ is singular,and therefore $A^{-1}$ does not exist.

(D) matrix is not invertible if its determinant is equal to $0$.
We set the determinant of the given matrix to $0$:
$\begin{vmatrix} x & 2 & 3 \\ 4 & 5 & 6 \\ 2 & 3 & 5 \end{vmatrix} = 0$
Expanding along the first row:
$x(5 \times 5 - 6 \times 3) - 2(4 \times 5 - 6 \times 2) + 3(4 \times 3 - 5 \times 2) = 0$
$x(25 - 18) - 2(20 - 12) + 3(12 - 10) = 0$
$x(7) - 2(8) + 3(2) = 0$
$7x - 16 + 6 = 0$
$7x - 10 = 0$
$7x = 10$
$x = \frac{10}{7}$

(B) Given $A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$.
First,we find the determinant $|A| = 0(0) - 0(0) - 1(0 - 1) = -1(-1) = 1$.
Since $|A| \neq 0$,the matrix $A$ is invertible.
Now,we calculate $A^2 = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$.
Since $A^2 = I$,multiplying both sides by $A^{-1}$ gives $A = A^{-1}$.

(C) First,calculate the sum of matrices $A$ and $B$:
$A+B = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} + \begin{bmatrix} 4 & 1 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 2 \\ 4 & 3 \end{bmatrix}$
Next,find the determinant of $(A+B)$:
$|A+B| = (5 \times 3) - (2 \times 4) = 15 - 8 = 7$
Now,find the adjoint of $(A+B)$ by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$\text{adj}(A+B) = \begin{bmatrix} 3 & -2 \\ -4 & 5 \end{bmatrix}$
Finally,use the formula $(A+B)^{-1} = \frac{1}{|A+B|} \text{adj}(A+B)$:
$(A+B)^{-1} = \frac{1}{7} \begin{bmatrix} 3 & -2 \\ -4 & 5 \end{bmatrix}$

(B) First,calculate the product $AB$:
$AB = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} (1)(1)+(2)(2)+(1)(0) & (1)(2)+(2)(1)+(1)(1) \\ (2)(1)+(1)(2)+(0)(0) & (2)(2)+(1)(1)+(0)(1) \end{bmatrix} = \begin{bmatrix} 5 & 5 \\ 4 & 5 \end{bmatrix}$
Now,find the determinant $|AB|$:
$|AB| = (5)(5) - (5)(4) = 25 - 20 = 5$
Next,find the adjoint of $AB$:
$adj(AB) = \begin{bmatrix} 5 & -5 \\ -4 & 5 \end{bmatrix}$
Finally,calculate the inverse $(AB)^{-1} = \frac{1}{|AB|} adj(AB)$:
$(AB)^{-1} = \frac{1}{5} \begin{bmatrix} 5 & -5 \\ -4 & 5 \end{bmatrix}$
Thus,the correct option is $B$.

(B) We know that $A \cdot A^{-1} = I$,where $I$ is the identity matrix of order $3 \times 3$.
$\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix} \begin{bmatrix} 3 & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Multiplying the matrices on the left side:
Row $1$,Column $1$: $(2)(3) + (0)(\alpha) + (-1)(\beta) = 6 - \beta$
Since the result must be $I$,$6 - \beta = 1$,which gives $\beta = 5$.
Row $2$,Column $1$: $(5)(3) + (1)(\alpha) + (0)(\beta) = 15 + \alpha$
Since the result must be $I$,$15 + \alpha = 0$,which gives $\alpha = -15$.
Thus,the values are $\alpha = -15$ and $\beta = 5$.

(A) matrix $A$ is not invertible if and only if its determinant is zero,i.e.,$|A| = 0$.
Given the matrix $A = \begin{bmatrix} a & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2 \end{bmatrix}$.
Calculating the determinant $|A|$:
$|A| = a(0 \times 2 - 1 \times 1) - (-1)(-3 \times 2 - 1 \times (-1)) + 4(-3 \times 1 - 0 \times (-1)) = 0$
$|A| = a(0 - 1) + 1(-6 + 1) + 4(-3 - 0) = 0$
$|A| = a(-1) + 1(-5) + 4(-3) = 0$
$-a - 5 - 12 = 0$
$-a - 17 = 0$
$a = -17$.
Thus,the matrix is not invertible when $a = -17$.

(A) We know that $(AB)^{-1} = B^{-1} A^{-1}$.
First,we find the product $AB$:
$AB = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} (2)(1) + (3)(3) & (2)(0) + (3)(1) \\ (1)(1) + (2)(3) & (1)(0) + (2)(1) \end{bmatrix} = \begin{bmatrix} 11 & 3 \\ 7 & 2 \end{bmatrix}$.
Now,we find $(AB)^{-1} = \frac{1}{|AB|} \text{adj}(AB)$.
$|AB| = (11)(2) - (3)(7) = 22 - 21 = 1$.
$\text{adj}(AB) = \begin{bmatrix} 2 & -3 \\ -7 & 11 \end{bmatrix}$.
Therefore,$B^{-1} A^{-1} = (AB)^{-1} = \frac{1}{1} \begin{bmatrix} 2 & -3 \\ -7 & 11 \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ -7 & 11 \end{bmatrix}$.

(A) Given $A = \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix}$.
We know that $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
First,calculate the determinant $|A| = \omega \cdot \omega - 0 \cdot 0 = \omega^{2}$.
Next,find the adjoint of $A$ for a diagonal matrix $\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}$ is $\begin{bmatrix} b & 0 \\ 0 & a \end{bmatrix}$.
Thus,$\text{adj}(A) = \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix}$.
Therefore,$A^{-1} = \frac{1}{\omega^{2}} \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix} = \begin{bmatrix} \frac{\omega}{\omega^{2}} & 0 \\ 0 & \frac{\omega}{\omega^{2}} \end{bmatrix} = \begin{bmatrix} \frac{1}{\omega} & 0 \\ 0 & \frac{1}{\omega} \end{bmatrix}$.
Since $\omega^{3} = 1$,we have $\frac{1}{\omega} = \omega^{2}$.
So,$A^{-1} = \begin{bmatrix} \omega^{2} & 0 \\ 0 & \omega^{2} \end{bmatrix}$.
Now,check the options:
$A^{2} = \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix} \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix} = \begin{bmatrix} \omega^{2} & 0 \\ 0 & \omega^{2} \end{bmatrix}$.
Thus,$A^{-1} = A^{2}$.

(C) Given the characteristic equation $A^{2}-5A-6I=0$.
Multiplying by $A^{-1}$ on both sides,we get:
$A^{-1}(A^{2}-5A-6I) = A^{-1}(0)$
$A - 5I - 6A^{-1} = 0$
$6A^{-1} = A - 5I$
Substituting the matrices:
$6A^{-1} = \begin{bmatrix} 4 & 5 \\ 2 & 1 \end{bmatrix} - 5\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$6A^{-1} = \begin{bmatrix} 4-5 & 5-0 \\ 2-0 & 1-5 \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 2 & -4 \end{bmatrix}$
$A^{-1} = \frac{1}{6}\begin{bmatrix} -1 & 5 \\ 2 & -4 \end{bmatrix}$
Thus,the correct option is $C$.

(C) Let $A = \left[\begin{array}{ccc}1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0\end{array}\right]$.
First,we calculate the determinant $|A|$:
$|A| = 1(0 - 1) - (-3)(0 - (-2)) + 2(3 - 6)$
$|A| = 1(-1) + 3(2) + 2(-3) = -1 + 6 - 6 = -1$.
The element in the third row and first column of $A^{-1}$ is given by $\frac{C_{13}}{|A|}$,where $C_{13}$ is the cofactor of the element in the first row and third column of $A$.
$C_{13} = (-1)^{1+3} \left|\begin{array}{cc}-3 & 3 \\ 2 & -1\end{array}\right| = 1(3 - 6) = -3$.
Therefore,the element in the third row and first column of $A^{-1}$ is $\frac{-3}{-1} = 3$.

(B) First,find the determinant of $A$: $|A| = (1)(1) - (2)(-5) = 1 + 10 = 11$.
Next,find the adjoint of $A$: $\text{adj } A = \left[\begin{array}{cc}1 & -2 \\ 5 & 1\end{array}\right]$.
Therefore,$A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{11} \left[\begin{array}{cc}1 & -2 \\ 5 & 1\end{array}\right]$.
Given the equation $A^{-1} = xA + yI$,we substitute the matrices:
$\frac{1}{11} \left[\begin{array}{cc}1 & -2 \\ 5 & 1\end{array}\right] = x \left[\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right] + y \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$.
$\left[\begin{array}{cc}1/11 & -2/11 \\ 5/11 & 1/11\end{array}\right] = \left[\begin{array}{cc}x+y & 2x \\ -5x & x+y\end{array}\right]$.
Comparing the elements,we get $2x = -2/11$,which implies $x = -1/11$.
Also,$x+y = 1/11$. Substituting $x = -1/11$,we get $-1/11 + y = 1/11$,so $y = 2/11$.
Thus,the values are $x = -1/11$ and $y = 2/11$.

(C) Given $A = \begin{bmatrix} 2 & -3 \\ 5 & -7 \end{bmatrix}$.
First,we find the determinant $|A| = (2)(-7) - (-3)(5) = -14 + 15 = 1$.
Since $|A| \neq 0$,$A^{-1}$ exists.
The inverse is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{1} \begin{bmatrix} -7 & 3 \\ -5 & 2 \end{bmatrix} = \begin{bmatrix} -7 & 3 \\ -5 & 2 \end{bmatrix}$.
Now,calculate $2A = 2 \begin{bmatrix} 2 & -3 \\ 5 & -7 \end{bmatrix} = \begin{bmatrix} 4 & -6 \\ 10 & -14 \end{bmatrix}$.
Next,calculate $3A^{-1} = 3 \begin{bmatrix} -7 & 3 \\ -5 & 2 \end{bmatrix} = \begin{bmatrix} -21 & 9 \\ -15 & 6 \end{bmatrix}$.
Finally,$2A - 3A^{-1} = \begin{bmatrix} 4 & -6 \\ 10 & -14 \end{bmatrix} - \begin{bmatrix} -21 & 9 \\ -15 & 6 \end{bmatrix} = \begin{bmatrix} 4 - (-21) & -6 - 9 \\ 10 - (-15) & -14 - 6 \end{bmatrix} = \begin{bmatrix} 25 & -15 \\ 25 & -20 \end{bmatrix}$.

(D) Given $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $AX = I$.
Since $AX = I$,it follows that $X = A^{-1}$.
First,find the determinant of $A$: $|A| = (1)(4) - (2)(3) = 4 - 6 = -2$.
The inverse of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $\frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Therefore,$A^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$.
$X = \begin{bmatrix} \frac{4}{-2} & \frac{-2}{-2} \\ \frac{-3}{-2} & \frac{1}{-2} \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}$.

(A) The total number of outcomes when throwing three dice is $n(S) = 6 \times 6 \times 6 = 216$.
The sums less than $5$ are obtained from the outcomes: $(1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1)$.
The number of outcomes with a sum less than $5$ is $n(E') = 4$.
The probability of getting a sum less than $5$ is $P(< 5) = \frac{4}{216} = \frac{1}{54}$.
The probability of getting a sum at least $5$ is $P(\geq 5) = 1 - P(< 5)$.
$P(\geq 5) = 1 - \frac{1}{54} = \frac{53}{54}$.

(C) Let $n = 5$ be the number of lottery tickets purchased.
Let $p$ be the probability of winning a prize on a single ticket,so $p = \frac{1}{4}$.
Let $q$ be the probability of not winning a prize on a single ticket,so $q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4}$.
We want to find the probability of winning at least one prize,which is $P(X \ge 1)$.
Using the complement rule,$P(X \ge 1) = 1 - P(X = 0)$.
The probability of winning zero prizes in $5$ trials is given by the binomial distribution formula $P(X = k) = \binom{n}{k} p^k q^{n-k}$.
For $k = 0$,$P(X = 0) = \binom{5}{0} (\frac{1}{4})^0 (\frac{3}{4})^5 = 1 \times 1 \times \frac{243}{1024} = \frac{243}{1024}$.
Therefore,$P(X \ge 1) = 1 - \frac{243}{1024} = \frac{1024 - 243}{1024} = \frac{781}{1024}$.

(A) There are $52$ cards in a deck,and there are $4$ queens in total.
When the first card is drawn,the probability of getting a queen is $P(Q_1) = \frac{4}{52} = \frac{1}{13}$.
Since the card is drawn without replacement,there are now $51$ cards left,and $3$ queens remaining.
The probability of drawing a second queen given that the first was a queen is $P(Q_2|Q_1) = \frac{3}{51} = \frac{1}{17}$.
The probability that both cards are queens is $P(Q_1 \cap Q_2) = P(Q_1) \times P(Q_2|Q_1) = \frac{4}{52} \times \frac{3}{51} = \frac{1}{13} \times \frac{1}{17} = \frac{1}{221}$.

(D) There are $5$ questions and each question has $3$ options of which one is correct.
Probability of getting a correct answer for any question is $p = \frac{1}{3}$.
Probability of getting an incorrect answer is $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.
We need to find the probability of getting at least one correct answer.
$P(\text{at least one correct}) = 1 - P(\text{none correct})$.
The probability that none of the $5$ questions are answered correctly is given by $P(X=0) = {}^{5}C_{0} \times p^{0} \times q^{5}$.
$P(X=0) = 1 \times 1 \times (\frac{2}{3})^{5} = \frac{32}{243}$.
Therefore,$P(\text{at least one correct}) = 1 - \frac{32}{243} = \frac{243 - 32}{243} = \frac{211}{243}$.

(B) Total number of red balls $= 4$.
Total number of white balls $= 5$.
Total number of balls $= 4 + 5 = 9$.
Let $R_1$ be the event that the first ball drawn is red and $R_2$ be the event that the second ball drawn is red.
The probability of drawing the first red ball is $P(R_1) = \frac{4}{9}$.
Since the balls are drawn without replacement,if the first ball is red,there are now $3$ red balls left out of $8$ total balls.
The probability of drawing the second red ball given that the first was red is $P(R_2|R_1) = \frac{3}{8}$.
The probability that both balls are red is $P(R_1 \cap R_2) = P(R_1) \times P(R_2|R_1) = \frac{4}{9} \times \frac{3}{8} = \frac{12}{72} = \frac{1}{6}$.

(B) Given $P(A') = 0.6$,we have $P(A) = 1 - 0.6 = 0.4$.
We are given $P(B) = 0.8$ and $P(B/A) = 0.3$.
Using the multiplication theorem,$P(A \cap B) = P(A) \cdot P(B/A) = 0.4 \times 0.3 = 0.12$.
We know that $P(A/B) = \frac{P(A \cap B)}{P(B)}$.
Substituting the values,$P(A/B) = \frac{0.12}{0.8} = \frac{12}{80} = \frac{3}{20}$.

(D) This is a binomial distribution problem where $n = 5$ and the probability of success $p = 0.2$.
The probability of failure is $q = 1 - p = 1 - 0.2 = 0.8$.
We need to find the probability that exactly $x = 4$ patients survive.
The formula for binomial probability is $P(X = x) = {}^{n}C_{x} p^{x} q^{n-x}$.
Substituting the values: $P(X = 4) = {}^{5}C_{4} (0.2)^{4} (0.8)^{5-4}$.
$P(X = 4) = 5 \times (0.0016) \times (0.8)$.
$P(X = 4) = 5 \times 0.00128 = 0.0064$.

(D) Given that $A$ and $B$ are independent events,$P(A) = \frac{2}{3}$ and $P(B) = \frac{3}{5}$.
Since $A$ and $B$ are independent,$A^{\prime}$ and $B$ are also independent.
Therefore,$P(A^{\prime} \cap B) = P(A^{\prime}) \times P(B)$.
We know that $P(A^{\prime}) = 1 - P(A) = 1 - \frac{2}{3} = \frac{1}{3}$.
Substituting the values,we get $P(A^{\prime} \cap B) = \frac{1}{3} \times \frac{3}{5} = \frac{1}{5}$.
Thus,the correct option is $D$.

(B) Given that $A$ and $B$ are independent events,$P(A) = \frac{3}{5}$ and $P(B) = \frac{2}{3}$.
We need to find $P(A' \cap B')$.
Since $A$ and $B$ are independent,$A'$ and $B'$ are also independent.
Therefore,$P(A' \cap B') = P(A') \cdot P(B')$.
First,calculate $P(A')$ and $P(B')$:
$P(A') = 1 - P(A) = 1 - \frac{3}{5} = \frac{2}{5}$.
$P(B') = 1 - P(B) = 1 - \frac{2}{3} = \frac{1}{3}$.
Now,calculate the product:
$P(A' \cap B') = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}$.
Thus,the correct option is $B$.

(A) Let $M$ be the event that the selected person is a man,$W$ be the event that the selected person is a woman,and $G$ be the event that the selected person has gray hair.
Given that the number of males and females is equal,we have $P(M) = P(W) = \frac{1}{2}$.
The probability of a man having gray hair is $P(G|M) = \frac{5}{100}$.
The probability of a woman having gray hair is $P(G|W) = \frac{0.25}{100} = \frac{1}{400}$.
We need to find the probability that the person is a man,given that they have gray hair,which is $P(M|G)$.
Using Bayes' Theorem:
$P(M|G) = \frac{P(M) \times P(G|M)}{P(M) \times P(G|M) + P(W) \times P(G|W)}$
$P(M|G) = \frac{\frac{1}{2} \times \frac{5}{100}}{\frac{1}{2} \times \frac{5}{100} + \frac{1}{2} \times \frac{0.25}{100}}$
$P(M|G) = \frac{5}{5 + 0.25} = \frac{5}{5.25} = \frac{500}{525} = \frac{20}{21}$.

(A) Given $X \sim B(4, p)$,where $n=4$. The probability mass function is $P(X=k) = {}^{n}C_{k} p^{k} q^{n-k}$.
Given the condition $2 P(X=3) = 3 P(X=2)$.
Substituting the values: $2 \times ({}^{4}C_{3} p^{3} q^{1}) = 3 \times ({}^{4}C_{2} p^{2} q^{2})$.
Calculating combinations: $2 \times (4 p^{3} q) = 3 \times (6 p^{2} q^{2})$.
Simplifying: $8 p^{3} q = 18 p^{2} q^{2}$.
Dividing both sides by $2 p^{2} q$ (assuming $p, q \neq 0$): $4 p = 9 q$.
Since $q = 1 - p$,we have $4 p = 9(1 - p)$.
$4 p = 9 - 9 p$.
$13 p = 9$.
Therefore,$p = \frac{9}{13}$.

(D) Let $X$ denote the number of defective bulbs out of $20$ bulbs selected.
Given that the total number of bulbs is $100$ and $10$ are defective,the probability $p$ of selecting a defective bulb is $p = \frac{10}{100} = \frac{1}{10}$.
Consequently,the probability $q$ of selecting a non-defective bulb is $q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10}$.
We are selecting $n = 20$ bulbs. The probability of getting no defective bulb is given by the binomial distribution formula $P(X = k) = {}^{n}C_{k} \cdot p^{k} \cdot q^{n-k}$.
For $k = 0$,$P(X = 0) = {}^{20}C_{0} \cdot \left(\frac{1}{10}\right)^{0} \cdot \left(\frac{9}{10}\right)^{20-0}$.
$P(X = 0) = 1 \cdot 1 \cdot \left(\frac{9}{10}\right)^{20} = \left(\frac{9}{10}\right)^{20}$.

(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 $₹ 2.50$.

(C) Let $X$ be the number of heads in $8$ tosses of a fair coin. Here,$n=8$,$p=\frac{1}{2}$,and $q=\frac{1}{2}$.
We want to find the probability that heads show more than tails,which means $X > 4$.
Since the total number of outcomes is $2^8 = 256$,and the distribution is symmetric,we have $P(X < 4) = P(X > 4)$.
We know that $\sum_{k=0}^{8} P(X=k) = 1$,so $P(X < 4) + P(X=4) + P(X > 4) = 1$.
Thus,$2P(X > 4) + P(X=4) = 1$,which implies $P(X > 4) = \frac{1 - P(X=4)}{2}$.
Calculating $P(X=4) = {}^{8}C_{4} \left(\frac{1}{2}\right)^{8} = \frac{70}{256}$.
Therefore,$P(X > 4) = \frac{1 - \frac{70}{256}}{2} = \frac{\frac{186}{256}}{2} = \frac{93}{256}$.

(B) The probability of missing the target is $q = 0.2 = \frac{1}{5}$.
So,the probability of hitting the target is $p = 1 - q = 1 - 0.2 = 0.8 = \frac{4}{5}$.
Given $n = 10$ bombs are dropped,and we want exactly $r = 2$ hits.
Using the binomial distribution formula $P(X = r) = {}^{n}C_{r} p^{r} q^{n-r}$:
$P(X = 2) = {}^{10}C_{2} \times (0.8)^{2} \times (0.2)^{8}$
$P(X = 2) = \frac{10 \times 9}{2 \times 1} \times \left(\frac{4}{5}\right)^{2} \times \left(\frac{1}{5}\right)^{8}$
$P(X = 2) = 45 \times \frac{16}{25} \times \frac{1}{5^{8}}$
$P(X = 2) = 45 \times \frac{16}{5^{2} \times 5^{8}} = 45 \times \frac{16}{5^{10}}$
$P(X = 2) = (9 \times 5) \times \frac{16}{5^{10}} = \frac{9 \times 16}{5^{9}} = \frac{144}{5^{9}}$
Thus,the correct option is $B$.

(B) Let the probability of a person having a common cold be $p = \frac{10}{100} = \frac{1}{10}$.
Then,the probability of a person not having a common cold is $q = 1 - p = \frac{9}{10}$.
We are selecting $n = 5$ persons. Let $X$ be the number of persons having a common cold. $X$ follows a binomial distribution $B(n, p) = B(5, 0.1)$.
We need to find the probability that at most one person has a common cold,which is $P(X \le 1) = P(X = 0) + P(X = 1)$.
Using the binomial probability formula $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$:
$P(X = 0) = {}^{5}C_{0} \left(\frac{1}{10}\right)^{0} \left(\frac{9}{10}\right)^{5} = 1 \times 1 \times \frac{59049}{100000} = 0.59049$.
$P(X = 1) = {}^{5}C_{1} \left(\frac{1}{10}\right)^{1} \left(\frac{9}{10}\right)^{4} = 5 \times \frac{1}{10} \times \frac{6561}{10000} = \frac{32805}{100000} = 0.32805$.
Therefore,$P(X \le 1) = 0.59049 + 0.32805 = 0.91854$.
Rounding to four decimal places,the probability is $0.9185$.

(D) For a binomial distribution $X \sim B(n, p)$,the mean is given by $E(X) = np$ and the variance is given by $\operatorname{Var}(X) = npq$,where $q = 1 - p$.
Given $E(X) = np = 4$.
Given $\operatorname{Var}(X) = npq = 2.4$.
Substituting $np = 4$ into the variance equation: $4q = 2.4$.
Solving for $q$: $q = \frac{2.4}{4} = 0.6 = \frac{3}{5}$.
Since $p = 1 - q$,we have $p = 1 - 0.6 = 0.4 = \frac{2}{5}$.
Now,substitute $p$ back into the mean equation: $n \times \frac{2}{5} = 4$.
$n = 4 \times \frac{5}{2} = 10$.
Thus,the value of $n$ is $10$.