MHT CET 2026 Mathematics Question Paper with Answer and Solution

(C) $1$. Identify the feasible region defined by the constraints:
$2x + 3y = 6$ passes through $(3, 0)$ and $(0, 2)$.
$x + y = 1$ passes through $(1, 0)$ and $(0, 1)$.
$2$. The corner points of the feasible region are $(1, 0)$, $(3, 0)$, and $(0, 1)$.
$3$. Evaluate $Z = 3x + y$ at each corner point:
At $(1, 0): Z = 3(1) + 0 = 3$.
At $(3, 0): Z = 3(3) + 0 = 9$.
At $(0, 1): Z = 3(0) + 1 = 1$.
$4$. Comparing the values, the minimum value is $1$.

(D) Step $1$: Identify the corner points of the feasible region defined by the constraints.
- $x + 3y = 60$ and $x - y = 0 \implies x=15, y=15$. Point: $(15, 15)$.
- $x + 3y = 60$ and $x = 0 \implies y=20$. Point: $(0, 20)$.
- $x + y = 10$ and $x - y = 0 \implies x=5, y=5$. Point: $(5, 5)$.
- $x + y = 10$ and $x = 0 \implies y=10$. Point: $(0, 10)$.
Step $2$: Evaluate $Z = 3x + 5y$ at each corner point.
- At $(15, 15): Z = 3(15) + 5(15) = 45 + 75 = 120$.
- At $(0, 20): Z = 3(0) + 5(20) = 100$.
- At $(5, 5): Z = 3(5) + 5(5) = 15 + 25 = 40$.
- At $(0, 10): Z = 3(0) + 5(10) = 50$.
Step $3$: Find the maximum and minimum values.
- Maximum value $Z_{max} = 120$.
- Minimum value $Z_{min} = 40$.
Step $4$: Calculate the difference.
- Difference $= Z_{max} - Z_{min} = 120 - 40 = 80$.

(C) $1$. Plot the lines: $y = 2x + 4$, $x + y = 5$, $x = 4$, and $y = 2$.
$2$. The region $y - 2x \leq 4$ is above the line $y = 2x + 4$.
$3$. The region $x + y \geq 5$ is above the line $x + y = 5$.
$4$. The region $x \leq 4$ is to the left of the line $x = 4$.
$5$. The region $y \geq 2$ is above the line $y = 2$.
$6$. Intersection points are: $(0, 4)$, $(1, 6)$, $(4, 6)$, $(4, 2)$, and $(3, 2)$.
$7$. Since all constraints form a closed polygon with $5$ vertices, the feasible region is a convex bounded region with $5$ corner points.

(C) Step $1$: Identify the feasible region defined by the constraints.
Step $2$: The intersection points of the lines $x=80$, $y=60$, $x+y=200$, and the axes are the vertices of the feasible region.
Step $3$: The vertices are found by solving the system of equations:
- Intersection of $y=60$ and $x=0$ gives $(0, 60)$.
- Intersection of $y=60$ and $x+y=200$ gives $(140, 60)$, but $x \leq 80$, so we use $(80, 60)$.
- Intersection of $x=80$ and $x+y=200$ gives $(80, 120)$.
- Intersection of $x=0$ and $x+y=200$ gives $(0, 200)$.
Step $4$: Evaluate $z = 3x + 5y$ at each vertex:
- At $(0, 60)$: $z = 3(0) + 5(60) = 300$.
- At $(80, 60)$: $z = 3(80) + 5(60) = 240 + 300 = 540$.
- At $(80, 120)$: $z = 3(80) + 5(120) = 240 + 600 = 840$.
- At $(0, 200)$: $z = 3(0) + 5(200) = 1000$.
Step $5$: The minimum value is $300$ at point $(0, 60)$.

(D) Step $1$: Identify the corner points of the feasible region defined by $x + 2y \leq 10$, $3x + y \leq 12$, $x \geq 0$, $y \geq 0$.
Step $2$: The intersection of $x + 2y = 10$ and $3x + y = 12$ is found by solving the system: $y = 12 - 3x$. Substituting into the first equation: $x + 2(12 - 3x) = 10 \implies x + 24 - 6x = 10 \implies -5x = -14 \implies x = \frac{14}{5}$. Then $y = 12 - 3(\frac{14}{5}) = \frac{60 - 42}{5} = \frac{18}{5}$. The intersection point is $(\frac{14}{5}, \frac{18}{5})$.
Step $3$: Evaluate $z = 5x + 10y$ at corner points: $(0, 0) \implies z = 0$; $(4, 0) \implies z = 20$; $(0, 5) \implies z = 50$; $(\frac{14}{5}, \frac{18}{5}) \implies z = 5(\frac{14}{5}) + 10(\frac{18}{5}) = 14 + 36 = 50$.
Step $4$: Since $z$ is maximum at $(0, 5)$ and $(\frac{14}{5}, \frac{18}{5})$ with value $50$, the maximum occurs at every point on the line segment joining these two points.

(D) $1$. The shaded region lies in the first quadrant, so the non-negativity constraints are $x \geq 0$ and $y \geq 0$.
$2$. The region is bounded by three lines:
$(i)$ The line passing through $(0, 12)$ and $(6, 0)$ has the equation $\frac{x}{6} + \frac{y}{12} = 1$, which simplifies to $2x + y = 12$. Since the shaded region is towards the origin, the constraint is $2x + y \leq 12$.
(ii) The line passing through $(0, 6)$ and $(12, 0)$ has the equation $\frac{x}{12} + \frac{y}{6} = 1$, which simplifies to $x + 2y = 12$. Since the shaded region is towards the origin, the constraint is $x + 2y \leq 12$.
(iii) The line passing through $(0, 4)$ and $(5, 0)$ has the equation $\frac{x}{5} + \frac{y}{4} = 1$, which simplifies to $4x + 5y = 20$, or $x + 1.25y = 5$. Since the shaded region is away from the origin, the constraint is $x + 1.25y \geq 5$.
$3$. Combining these, the system of constraints is $2x + y \leq 12, x + 2y \leq 12, x + 1.25y \geq 5, x \geq 0, y \geq 0$. This matches option $D$.

(A) Step $1$: Identify the corner points of the feasible region defined by the constraints.
Step $2$: The lines are $3x + y = 15$ and $3x + 4y = 24$.
Step $3$: Intersection point: Subtracting the first from the second gives $3y = 9$, so $y = 3$. Substituting $y = 3$ into $3x + y = 15$ gives $3x = 12$, so $x = 4$. The intersection point is $(4, 3)$.
Step $4$: The corner points of the feasible region are $(0, 0)$, $(5, 0)$, $(4, 3)$, and $(0, 6)$.
Step $5$: Evaluate $Z = 4x + 5y$ at each corner point:
At $(0, 0)$, $Z = 4(0) + 5(0) = 0$.
At $(5, 0)$, $Z = 4(5) + 5(0) = 20$.
At $(4, 3)$, $Z = 4(4) + 5(3) = 16 + 15 = 31$.
At $(0, 6)$, $Z = 4(0) + 5(6) = 30$.
Step $6$: The maximum value is $31$.

(D) Step $1$: Identify the corner points of the feasible region defined by the constraints.
Step $2$: The intersection points of the lines are:
$(i)$ $x+y=5$ and $2x+y=7$ gives $x=2, y=3$.
(ii) $2x+y=7$ and $3x+2y=11$ gives $x=3, y=1$.
(iii) The axes intercepts are $(0, 0), (5, 0), (0, 5), (0, 5.5), (3.66, 0)$.
Step $3$: Evaluating $z = 4x + y$ at the corner points of the feasible region:
At $(0, 0)$, $z = 0$.
At $(3.5, 0)$, $z = 4(3.5) + 0 = 14$.
At $(3, 1)$, $z = 4(3) + 1 = 13$.
At $(2, 3)$, $z = 4(2) + 3 = 11$.
At $(0, 5)$, $z = 4(0) + 5 = 5$.
Step $4$: The maximum value is $14$.

(D) Step $1$: Define the objective function. Profit is $1500$ per executive ticket $(x_1)$ and $900$ per economy ticket $(x_2)$. Thus, maximize $z = 1500x_1 + 900x_2$.
Step $2$: Identify constraints. Total capacity is $250$, so $x_1 + x_2 \leq 250$.
Step $3$: Executive class reservation is at least $30$, so $x_1 \geq 30$.
Step $4$: Economy class preference is at least $4$ times executive class, so $x_2 \geq 4x_1$.
Step $5$: Non-negativity constraints are $x_1 \geq 0, x_2 \geq 0$. Combining these, option $D$ is correct.

(C) Step $1$: Identify the corner points of the feasible region defined by the constraints.
Step $2$: The lines are $2x + 3y = 6$ and $x + y = 1$.
Step $3$: Intersection of $2x + 3y = 6$ and $x + y = 1$ gives $x = 3, y = -2$ (outside the first quadrant).
Step $4$: The vertices of the feasible region are $(0, 1)$, $(0, 2)$, $(1, 0)$, and $(3, 0)$.
Step $5$: Evaluate $z = 3x + y$ at each vertex:
At $(0, 1)$, $z = 3(0) + 1 = 1$.
At $(0, 2)$, $z = 3(0) + 2 = 2$.
At $(1, 0)$, $z = 3(1) + 0 = 3$.
At $(3, 0)$, $z = 3(3) + 0 = 9$.
Step $6$: Maximum value is $9$ and minimum value is $1$.
Step $7$: The difference is $9 - 1 = 8$.

(A) Step $1$: Analyze the constraints. The constraints are $x + 3y \leq 6$ and $2x + 6y \leq 18$. Note that $2x + 6y \leq 18$ simplifies to $x + 3y \leq 9$.
Step $2$: Since $x + 3y \leq 6$ is a stricter condition than $x + 3y \leq 9$, the feasible region is determined solely by $x + 3y \leq 6$ along with $x \geq 0$ and $y \geq 0$.
Step $3$: The vertices of the feasible region are $(0, 0)$, $(6, 0)$, and $(0, 2)$.
Step $4$: Evaluate $z = 2x + 5y$ at these vertices:
At $(0, 0)$, $z = 2(0) + 5(0) = 0$.
At $(6, 0)$, $z = 2(6) + 5(0) = 12$.
At $(0, 2)$, $z = 2(0) + 5(2) = 10$.
Step $5$: The maximum value is $12$ at the point $(6, 0)$. Since there is a single point where the maximum is attained, the solution is unique.

(D) Step $1$: Identify the outcomes for events $E$ and $F$.
$E = \{ X \text{ is a prime number} \} = \{2, 3, 5\}$.
$F = \{ X < 4 \} = \{1, 2, 3\}$.
Step $2$: Identify the union $E \cup F$.
$E \cup F = \{1, 2, 3, 5\}$.
Step $3$: Calculate the probability $P(E \cup F)$.
$P(E \cup F) = P(X=1) + P(X=2) + P(X=3) + P(X=5)$.
$P(E \cup F) = 0.15 + 0.20 + 0.25 + 0.10 = 0.70$.
Note: The provided options do not contain $0.70$. Re-evaluating the question logic, if $E \cup F$ is calculated as $P(E) + P(F) - P(E \cap F)$:
$P(E) = 0.20 + 0.25 + 0.10 = 0.55$.
$P(F) = 0.15 + 0.20 + 0.25 = 0.60$.
$E \cap F = \{2, 3\}$, so $P(E \cap F) = 0.20 + 0.25 = 0.45$.
$P(E \cup F) = 0.55 + 0.60 - 0.45 = 0.70$.

(D) Step $1$: Calculate the mean $E(X) = \sum x_i P(x_i) = (0 \times 0.3) + (1 \times 0.4) + (2 \times 0.3) = 0 + 0.4 + 0.6 = 1.0$.
Step $2$: Calculate $E(X^2) = \sum x_i^2 P(x_i) = (0^2 \times 0.3) + (1^2 \times 0.4) + (2^2 \times 0.3) = 0 + 0.4 + 1.2 = 1.6$.
Step $3$: Calculate the variance $Var(X) = E(X^2) - [E(X)]^2 = 1.6 - (1.0)^2 = 1.6 - 1.0 = 0.6$.
Step $4$: Calculate the standard deviation $\sigma = \sqrt{Var(X)} = \sqrt{0.6} \approx 0.77$. Since the closest option is $0.7$, we select $D$.

(A) The cumulative distribution function $F(x)$ is defined as $F(x_i) = \sum_{j=1}^{i} P(X = x_j)$.
For $x=1$: $F(1) = P(X=1) = 1/10$.
For $x=2$: $F(2) = P(X=1) + P(X=2) = 1/10 + 2/10 = 3/10$.
For $x=3$: $F(3) = P(X=1) + P(X=2) + P(X=3) = 3/10 + 3/10 = 6/10$.
For $x=4$: $F(4) = P(X=1) + P(X=2) + P(X=3) + P(X=4) = 6/10 + 4/10 = 10/10 = 1$.
Thus, the correct distribution is given in option $A$.

(C) Step $1$: Identify the sample space for event $E = \{ X \text{ is a prime number} \}$. The prime numbers in $\{1, 2, 3, 4, 5\}$ are $2$ and $3$. So, $E = \{2, 3\}$.
Step $2$: Identify the sample space for event $F = \{ X < 4 \}$. The values less than $4$ are $1, 2, 3$. So, $F = \{1, 2, 3\}$.
Step $3$: Find the intersection $E \cap F$. The common elements are $\{2, 3\}$.
Step $4$: Calculate $P(E \cap F) = P(2) + P(3) = 0.2 + 0.3 = 0.5$.

(A) Step $1$: The sum of all probabilities must be $1$. Thus, $\sum P(x) = k(1^2 + 2^2 + 3^2 + 4^2) = 1$.
Step $2$: Calculate the sum: $k(1 + 4 + 9 + 16) = 30k = 1$, so $k = \frac{1}{30}$.
Step $3$: The mean $E(X) = \sum x \cdot P(x) = \sum x \cdot kx^2 = k \sum x^3$.
Step $4$: Calculate $\sum x^3 = 1^3 + 2^3 + 3^3 + 4^3 = 1 + 8 + 27 + 64 = 100$.
Step $5$: $E(X) = \frac{1}{30} \cdot 100 = \frac{10}{3} \approx 3.33$.

(C) Total number of balls $= 8 + N$.
Number of balls drawn $= 2$.
The random variable $X$ represents the number of green balls, so $X \in \{0, 1, 2\}$.
The probability distribution is given by hypergeometric distribution:
$P(X=k) = \frac{\binom{N}{k} \binom{8}{2-k}}{\binom{N+8}{2}}$.
For a sample of size $n=2$ from a population of $N+8$ containing $N$ green balls, the expected value $E(X) = n \cdot \frac{N}{N+8}$.
Given $E(X) = 1.2$, we have $2 \cdot \frac{N}{N+8} = 1.2$.
$\frac{N}{N+8} = 0.6$.
$N = 0.6(N+8) \implies N = 0.6N + 4.8$.
$0.4N = 4.8 \implies N = \frac{4.8}{0.4} = 12$.

(D) Let $P(X = 1) = a, P(X = 2) = b, P(X = 3) = c, P(X = 4) = d$. Given $2a = 3b = c = 5d = k$.
Then $a = k/2, b = k/3, c = k, d = k/5$.
Since the sum of probabilities is $1$, we have $a + b + c + d = 1$.
Substituting the values: $k/2 + k/3 + k + k/5 = 1$.
Taking the $LCM$ of $2, 3, 1, 5$, which is $30$: $(15k + 10k + 30k + 6k) / 30 = 1$.
$61k / 30 = 1$, so $k = 30/61$.
Since $P(X = 4) = d = k/5$, we have $P(X = 4) = (30/61) / 5 = 6/61$.

(A) Given that $F(x)$ is the c.d.f. of $X$, we have $F(x) = P(X \le x)$.
We are given $F(b) = P(X \le b) = 0.7$.
We are given $P(X > a) = 0.4$. Since $P(X \le a) + P(X > a) = 1$, we have $P(X \le a) = 1 - 0.4 = 0.6$.
We need to find $P(a < X < b)$.
Since $X$ is a continuous random variable, $P(a < X < b) = P(X < b) - P(X < a) = P(X \le b) - P(X \le a)$.
Substituting the values, $P(a < X < b) = 0.7 - 0.6 = 0.1$.

(B) For a probability density function, the total area under the curve must be $1$, so $\int_{-1}^{4} f(x) \, dx = 1$.
$\int_{-1}^{4} k(9 + 8x - x^2) \, dx = 1$.
$k \left[ 9x + 4x^2 - \frac{x^3}{3} \right]_{-1}^{4} = 1$.
$k \left[ (9(4) + 4(4^2) - \frac{4^3}{3}) - (9(-1) + 4(-1)^2 - \frac{(-1)^3}{3}) \right] = 1$.
$k \left[ (36 + 64 - \frac{64}{3}) - (-9 + 4 + \frac{1}{3}) \right] = 1$.
$k \left[ (100 - \frac{64}{3}) - (-5 + \frac{1}{3}) \right] = 1$.
$k \left[ \frac{300 - 64}{3} - \frac{-15 + 1}{3} \right] = 1$.
$k \left[ \frac{236}{3} + \frac{14}{3} \right] = 1$.
$k \left[ \frac{250}{3} \right] = 1$.
$k = \frac{3}{250}$.

(A) Let $X$ be the random variable representing the winning amount. The probability of each face $k$ appearing is $P(k) = 1/6$ for $k \in \{1, 2, 3, 4, 5, 6\}$.
The winning amount for a face $k$ is $W(k) = \frac{36}{k} - 2k$.
The expected value $E[X]$ is given by $\sum_{k=1}^{6} W(k) \cdot P(k)$.
$E[X] = \frac{1}{6} \sum_{k=1}^{6} (\frac{36}{k} - 2k) = \frac{1}{6} [(\frac{36}{1} - 2) + (\frac{36}{2} - 4) + (\frac{36}{3} - 6) + (\frac{36}{4} - 8) + (\frac{36}{5} - 10) + (\frac{36}{6} - 12)]$.
$E[X] = \frac{1}{6} [34 + 14 + 6 + 1 + (7.2 - 10) + (6 - 12)]$.
$E[X] = \frac{1}{6} [55 - 2.8 - 6] = \frac{1}{6} [46.2] = 7.7$.
Since the provided options do not match the calculated value, the question is flawed. Assuming the sum was intended to be over $k \in \{1, 2, 3, 4, 5, 6\}$ and the expression was $\sum (\frac{36}{k} - 2k)$, the result is $7.7$.

(D) The probability $P(X < k)$ is given by $\int_{0}^{k} f(x) \, dx$.
For $P(X < a/2) = \int_{0}^{a/2} \frac{1}{2a} \, dx = \left[ \frac{x}{2a} \right]_{0}^{a/2} = \frac{a/2}{2a} = \frac{1}{4}$.
For $P(X > 3a/2) = \int_{3a/2}^{2a} \frac{1}{2a} \, dx = \left[ \frac{x}{2a} \right]_{3a/2}^{2a} = \frac{2a - 3a/2}{2a} = \frac{a/2}{2a} = \frac{1}{4}$.
Since both probabilities equal $\frac{1}{4}$, we have $P(X < a/2) = P(X > 3a/2)$.

(A) Total number of ways to draw $2$ cards from $5$ is $\binom{5}{2} = 10$.
The possible pairs and their sums are:
$(1,1) \rightarrow 2$
$(1,2) \rightarrow 3$
$(1,2) \rightarrow 3$
$(1,3) \rightarrow 4$
$(1,2) \rightarrow 3$
$(1,2) \rightarrow 3$
$(1,3) \rightarrow 4$
$(2,2) \rightarrow 4$
$(2,3) \rightarrow 5$
$(2,3) \rightarrow 5$
Sum of all possible sums $= 2+3+3+4+3+3+4+4+5+5 = 36$.
Expected value $E[X] = \frac{\text{Sum of all sums}}{\text{Total number of outcomes}} = \frac{36}{10} = 3.6$.
Note: Since $3.6$ is not in the options, the question is flawed. Assuming the intended answer based on the sum of individual cards: $E[X] = 2 \times E[\text{single card}] = 2 \times \frac{1+1+2+2+3}{5} = 2 \times \frac{9}{5} = 3.6$.

(C) Step $1$: Use the property of the p.d.f. that $\int_{-\infty}^{\infty} f(x) dx = 1$.
Step $2$: Calculate the integral: $\int_{0}^{1} kx dx + \int_{1}^{2} k dx + \int_{2}^{3} (-kx + 3k) dx = 1$.
Step $3$: Evaluate the integrals: $[k \frac{x^2}{2}]_{0}^{1} + [kx]_{1}^{2} + [-k \frac{x^2}{2} + 3kx]_{2}^{3} = 1$.
Step $4$: $\frac{k}{2} + k + [(- \frac{9k}{2} + 9k) - (- \frac{4k}{2} + 6k)] = 1$.
Step $5$: $\frac{k}{2} + k + [\frac{9k}{2} - 4k] = 1 \implies \frac{k}{2} + k + \frac{k}{2} = 1 \implies 2k = 1 \implies k = 1/2$.
Step $6$: Calculate $P(2 < X \leq 3) = \int_{2}^{3} (-kx + 3k) dx = [-\frac{kx^2}{2} + 3kx]_{2}^{3}$.
Step $7$: Substitute $k = 1/2$: $[-\frac{x^2}{4} + \frac{3x}{2}]_{2}^{3} = (-\frac{9}{4} + \frac{9}{2}) - (-1 + 3) = \frac{9}{4} - 2 = 1/4$.

(B) Step $1$: Calculate the Expected value $E(X) = \sum x_i P(x_i)$.
$E(X) = (1 \times 1/5) + (2 \times 2/5) + (3 \times 2/5) = 1/5 + 4/5 + 6/5 = 11/5$.
Step $2$: Calculate $E(X^2) = \sum x_i^2 P(x_i)$.
$E(X^2) = (1^2 \times 1/5) + (2^2 \times 2/5) + (3^2 \times 2/5) = 1/5 + 8/5 + 18/5 = 27/5$.
Step $3$: Calculate Variance $Var(X) = E(X^2) - [E(X)]^2$.
$Var(X) = 27/5 - (11/5)^2 = 27/5 - 121/25 = (135 - 121) / 25 = 14/25$.
Thus, the values are $11/5$ and $14/25$.

(C) For a binomial distribution $X \sim B(n, p)$, we have $E(X) = np = 2$ and $Var(X) = npq = 1$.
Dividing the variance by the mean: $q = \frac{npq}{np} = \frac{1}{2} = 0.5$.
Since $p + q = 1$, $p = 1 - 0.5 = 0.5$.
Substituting $p$ into $np = 2$: $n(0.5) = 2 \implies n = 4$.
We need to find $P(X \le 1) = P(X = 0) + P(X = 1)$.
$P(X = k) = \binom{n}{k} p^k q^{n-k}$.
$P(X = 0) = \binom{4}{0} (0.5)^0 (0.5)^4 = 1 \times 1 \times 0.0625 = 0.0625$.
$P(X = 1) = \binom{4}{1} (0.5)^1 (0.5)^3 = 4 \times 0.5 \times 0.125 = 0.25$.
$P(X \le 1) = 0.0625 + 0.25 = 0.3125$.

(A) The given probability mass function is $P(X = x) = \frac{^5C_x}{2^5}$.
This is a binomial distribution $B(n, p)$ with $n = 5$ and $p = 0.5$.
Calculate $P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) = \frac{^5C_0 + ^5C_1 + ^5C_2}{2^5} = \frac{1 + 5 + 10}{32} = \frac{16}{32} = 0.5$.
Calculate $P(X \geq 3) = P(X=3) + P(X=4) + P(X=5) = \frac{^5C_3 + ^5C_4 + ^5C_5}{2^5} = \frac{10 + 5 + 1}{32} = \frac{16}{32} = 0.5$.
Since $P(X \leq 2) = 0.5$ and $P(X \geq 3) = 0.5$, we have $P(X \leq 2) = P(X \geq 3)$.

(A) The probability mass function of a binomial distribution $X \sim B(n, p)$ is given by $P(X = k) = \binom{n}{k} p^k q^{n-k}$, where $q = 1 - p$.
We have $P(X = k) = \frac{n!}{k!(n-k)!} p^k q^{n-k}$ and $P(X = k-1) = \frac{n!}{(k-1)!(n-k+1)!} p^{k-1} q^{n-k+1}$.
Taking the ratio: $\frac{P(X = k)}{P(X = k - 1)} = \frac{n!}{k!(n-k)!} p^k q^{n-k} \times \frac{(k-1)!(n-k+1)!}{n! p^{k-1} q^{n-k+1}}$.
Simplifying the terms: $\frac{P(X = k)}{P(X = k - 1)} = \frac{(k-1)!}{k!} \times \frac{(n-k+1)!}{(n-k)!} \times \frac{p^k}{p^{k-1}} \times \frac{q^{n-k}}{q^{n-k+1}}$.
This results in $\frac{1}{k} \times (n-k+1) \times p \times \frac{1}{q} = \frac{n-k+1}{k} \cdot \frac{p}{q}$.

(C) For a binomial distribution $X \sim B(n, p)$, the probability mass function is given by $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$.
Here, $n = 6$ and $p = 1/2$.
Since $p = 1/2$, the distribution is symmetric.
For a symmetric binomial distribution, the maximum probability occurs at the mean, which is $E[X] = np = 6 \times (1/2) = 3$.
Alternatively, calculating for $k=3$: $P(X=3) = \binom{6}{3} (1/2)^3 (1/2)^3 = 20 \times (1/64) = 20/64$.
For $k=2$ or $k=4$: $P(X=2) = P(X=4) = \binom{6}{2} (1/2)^6 = 15 \times (1/64) = 15/64$.
Since $20/64 > 15/64$, the maximum probability occurs at $X = 3$.

(C) Let $x$ be the number of correct answers and $y$ be the number of incorrect answers.
Total questions $x + y = 10$.
Marks obtained $4x - y = 30$.
Adding the two equations: $5x = 40 \implies x = 8$.
Then $y = 10 - 8 = 2$.
The probability of a correct answer is $p = 1/4$ and an incorrect answer is $q = 3/4$.
The probability of getting $8$ correct and $2$ incorrect answers is given by the binomial distribution formula $P(X=8) = \binom{10}{8} p^8 q^2$.
$P(X=8) = \binom{10}{2} (1/4)^8 (3/4)^2 = 45 \times \frac{1}{4^8} \times \frac{9}{4^2} = \frac{45 \times 9}{4^{10}} = \frac{405}{4^{10}}$.

(D) The probability of $A$ winning a single game is $p = \frac{1}{1+2} = \frac{1}{3}$.
The probability of $A$ losing a single game is $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.
Let $X$ be the number of games won by $A$ in $n=3$ games. $X$ follows a binomial distribution $B(n, p) = B(3, 1/3)$.
The probability of $A$ winning at least two games is $P(X \ge 2) = P(X=2) + P(X=3)$.
$P(X=2) = \binom{3}{2} p^2 q^1 = 3 \times (\frac{1}{3})^2 \times (\frac{2}{3}) = 3 \times \frac{1}{9} \times \frac{2}{3} = \frac{6}{27}$.
$P(X=3) = \binom{3}{3} p^3 q^0 = 1 \times (\frac{1}{3})^3 \times 1 = \frac{1}{27}$.
$P(X \ge 2) = \frac{6}{27} + \frac{1}{27} = \frac{7}{27}$.

(B) This is a binomial distribution problem where $n = 8$, $p = 0.8$, and $q = 1 - p = 0.2$.
The probability of exactly $k$ successes is given by $P(X = k) = \binom{n}{k} p^k q^{n-k}$.
For $k = 4$:
$P(X = 4) = \binom{8}{4} (0.8)^4 (0.2)^{8-4}$
$P(X = 4) = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} \times (0.8)^4 \times (0.2)^4$
$P(X = 4) = 70 \times (0.8)^4 \times (0.2)^4$
Since $70 \times (0.2)^4 = 70 \times 0.0016 = 0.112$, the probability is $(0.112)(0.8)^4$.

(D) $1$. The total number of outcomes when two dice are thrown is $6 \times 6 = 36$.
$2$. $A$ doublet is $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$. There are $6$ such outcomes.
$3$. Probability of success $p = \frac{6}{36} = \frac{1}{6}$.
$4$. Probability of failure $q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.
$5$. The number of trials $n = 4$. The probability of getting at least $1$ success is $P(X \ge 1) = 1 - P(X = 0)$.
$6$. $P(X = 0) = ^nC_0 \cdot p^0 \cdot q^n = 1 \cdot 1 \cdot (\frac{5}{6})^4 = \frac{625}{1296}$.
$7$. $P(X \ge 1) = 1 - \frac{625}{1296} = \frac{1296 - 625}{1296} = \frac{671}{1296}$.

(B) For a fair coin, the probability of getting a head is $p = 1/2$ and the probability of getting a tail is $q = 1/2$. The number of trials is $n = 8$.
Using the binomial distribution formula $P(X = k) = \binom{n}{k} p^k q^{n-k}$, we need to find $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$.
$P(X=0) = \binom{8}{0} (1/2)^8 = 1 \times (1/256) = 1/256$.
$P(X=1) = \binom{8}{1} (1/2)^8 = 8 \times (1/256) = 8/256$.
$P(X=2) = \binom{8}{2} (1/2)^8 = 28 \times (1/256) = 28/256$.
Summing these probabilities: $P(X \le 2) = (1 + 8 + 28) / 256 = 37/256$.

(B) For a binomial distribution, mean $\mu = np = 1$ and variance $\sigma^2 = npq = 0.75$.
Dividing variance by mean: $\frac{npq}{np} = \frac{0.75}{1} \implies q = 0.75 = \frac{3}{4}$.
Since $p + q = 1$, $p = 1 - 0.75 = 0.25 = \frac{1}{4}$.
Using $np = 1$, $n(\frac{1}{4}) = 1 \implies n = 4$.
The probability mass function is $P(X = k) = \binom{4}{k} (\frac{1}{4})^k (\frac{3}{4})^{4-k}$.
For option $(C)$: $P(X = 3) = \binom{4}{3} (\frac{1}{4})^3 (\frac{3}{4})^1 = 4 \cdot \frac{1}{64} \cdot \frac{3}{4} = \frac{12}{256}$.
$P(X = 4) = \binom{4}{4} (\frac{1}{4})^4 (\frac{3}{4})^0 = 1 \cdot \frac{1}{256} \cdot 1 = \frac{1}{256}$.
Thus, $P(X = 3) = 12 \cdot P(X = 4)$. Checking option $(B)$: $P(X = 1) = \binom{4}{1} (\frac{1}{4})^1 (\frac{3}{4})^3 = 4 \cdot \frac{1}{4} \cdot \frac{27}{64} = \frac{27}{64}$.
$P(X = 2) = \binom{4}{2} (\frac{1}{4})^2 (\frac{3}{4})^2 = 6 \cdot \frac{1}{16} \cdot \frac{9}{16} = \frac{54}{256} = \frac{27}{128}$.
Since $P(X = 1) = 2 \cdot P(X = 2)$, option $(B)$ is correct.

(D) Let $n = 6$ be the number of trials and $p = 1/10$ be the probability of a person being busy. Then $q = 1 - p = 9/10$.
Using the binomial distribution formula $P(X = k) = \binom{n}{k} p^k q^{n-k}$:
For at least $5$ persons to be busy, we need $P(X \ge 5) = P(X = 5) + P(X = 6)$.
$P(X = 5) = \binom{6}{5} (1/10)^5 (9/10)^1 = 6 \times (1/10^5) \times (9/10) = 54/10^6$.
$P(X = 6) = \binom{6}{6} (1/10)^6 (9/10)^0 = 1 \times (1/10^6) \times 1 = 1/10^6$.
$P(X \ge 5) = 54/10^6 + 1/10^6 = 55/10^6$.

(C) Let $X$ be the number of successful predictions (heads). Since the coin is fair, the probability of heads is $p = 1/2$ and tails is $q = 1/2$. The number of tosses is $n = 9$.
$X$ follows a binomial distribution $B(9, 1/2)$.
We want the probability that the number of successful predictions is strictly greater than the number of unsuccessful predictions.
Let $S$ be the number of successes and $F$ be the number of failures. We have $S + F = 9$ and we want $S > F$.
Since $F = 9 - S$, the condition $S > 9 - S$ implies $2S > 9$, or $S > 4.5$.
Since $S$ must be an integer, we need $S \in \{5, 6, 7, 8, 9\}$.
Because the distribution is symmetric for $p = 1/2$, $P(S \ge 5) = P(S \le 4)$.
Since $P(S \le 9) = 1$, we have $P(S \le 4) + P(S \ge 5) = 1$.
Therefore, $2 \cdot P(S \ge 5) = 1$, which gives $P(S \ge 5) = 1/2$.

(A) $1$. The sample space of a fair die is $S = \{1, 2, 3, 4, 5, 6\}$, so the total outcomes $n(S) = 6$.
$2$. Prime numbers on a die are $\{2, 3, 5\}$, so the number of successful outcomes is $3$.
$3$. Probability of success in a single throw, $p = 3/6 = 1/2$.
$4$. Probability of failure in a single throw, $q = 1 - p = 1 - 1/2 = 1/2$.
$5$. The die is thrown $n = 4$ times. The probability of getting no success (i.e., $0$ successes) is given by the binomial distribution formula $P(X = k) = \binom{n}{k} p^k q^{n-k}$.
$6$. For $k = 0$, $P(X = 0) = \binom{4}{0} (1/2)^0 (1/2)^{4-0} = 1 \times 1 \times (1/2)^4 = 1/16$.

(D) For a binomial distribution, $P(X = k) = \binom{n}{k} p^k q^{n-k}$, where $n = 6$, $p$ is the probability of success, and $q = 1 - p$ is the probability of failure.
Given $9P(X = 4) = P(X = 2)$, we have $9 \binom{6}{4} p^4 q^2 = \binom{6}{2} p^2 q^4$.
Since $\binom{6}{4} = \binom{6}{2} = 15$, the equation simplifies to $9 \cdot 15 \cdot p^4 q^2 = 15 \cdot p^2 q^4$.
Dividing both sides by $15 p^2 q^2$ (assuming $p, q \neq 0$), we get $9p^2 = q^2$.
Taking the square root of both sides, $3p = q$.
Since $p + q = 1$, substitute $q = 3p$ into the equation: $p + 3p = 1 \implies 4p = 1 \implies p = 0.25$.
The probability of failure $q = 1 - p = 1 - 0.25 = 0.75$.

(D) Let $p$ be the probability that a bomb hits the target, so $p = 0.8 = 4/5$.
Let $q$ be the probability that a bomb misses the target, so $q = 1 - p = 1 - 0.8 = 0.2 = 1/5$.
We are dropping $n = 6$ bombs. Let $X$ be the number of bombs that miss the target.
We want to find the probability that at least $1$ bomb misses the target, which is $P(X \ge 1)$.
Using the complement rule, $P(X \ge 1) = 1 - P(X = 0)$.
$P(X = 0)$ is the probability that no bombs miss the target, meaning all $6$ bombs hit the target.
$P(X = 0) = p^6 = (4/5)^6$.
Therefore, $P(X \ge 1) = 1 - (4/5)^6$.

(C) Let $n = 4$ be the number of trials.
The probability of drawing a club card in a single draw is $p = \frac{13}{52} = \frac{1}{4}$.
The probability of not drawing a club card is $q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4}$.
Using the binomial distribution formula $P(X = k) = \binom{n}{k} p^k q^{n-k}$, where $k = 2$:
$P(X = 2) = \binom{4}{2} \left(\frac{1}{4}\right)^2 \left(\frac{3}{4}\right)^{4-2}$
$P(X = 2) = 6 \times \left(\frac{1}{16}\right) \times \left(\frac{9}{16}\right)$
$P(X = 2) = \frac{54}{256} = \frac{27}{128}$.

(A) The probability mass function for a binomial distribution is $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$.
Given $n = 6$, we have $P(X = 4) = \binom{6}{4} p^4 (1-p)^2 = 15 p^4 (1-p)^2$ and $P(X = 2) = \binom{6}{2} p^2 (1-p)^4 = 15 p^2 (1-p)^4$.
Given $9P(X = 4) = P(X = 2)$, substitute the expressions:
$9 \times 15 p^4 (1-p)^2 = 15 p^2 (1-p)^4$.
Divide both sides by $15 p^2 (1-p)^2$ (assuming $p \neq 0, 1$):
$9 p^2 = (1-p)^2$.
Taking the square root of both sides:
$3p = 1-p$ or $3p = -(1-p)$.
For $3p = 1-p$, we get $4p = 1$, so $p = 1/4$.
For $3p = p-1$, we get $2p = -1$, which gives $p = -1/2$ (not possible as $0 \le p \le 1$).
Thus, $p = 1/4$.

(C) Let $D$ be the event that the person has the disease and $T$ be the event that the test is positive.
Given: $P(D) = 0.01$, $P(D^c) = 0.99$.
Sensitivity $P(T|D) = 0.98$.
Specificity $P(T^c|D^c) = 0.95$, so $P(T|D^c) = 1 - 0.95 = 0.05$.
Using Bayes' Theorem, $P(D|T) = \frac{P(T|D)P(D)}{P(T|D)P(D) + P(T|D^c)P(D^c)}$.
$P(D|T) = \frac{0.98 \times 0.01}{(0.98 \times 0.01) + (0.05 \times 0.99)}$.
$P(D|T) = \frac{0.0098}{0.0098 + 0.0495} = \frac{0.0098}{0.0593} \approx 0.16526$.
Thus, the probability is approximately $16.5\%$.

(A) Given that $E_1$ and $E_2$ are equally likely, mutually exclusive, and exhaustive events, we have $P(E_1) = P(E_2) = 0.5$.
Using Bayes' Theorem, $P(E_1|A) = \frac{P(A|E_1)P(E_1)}{P(A|E_1)P(E_1) + P(A|E_2)P(E_2)}$.
Substitute the given values: $P(E_1|A) = \frac{0.2 \times 0.5}{(0.2 \times 0.5) + (0.3 \times 0.5)}$.
$P(E_1|A) = \frac{0.1}{0.1 + 0.15} = \frac{0.1}{0.25}$.
$P(E_1|A) = \frac{10}{25} = \frac{2}{5}$.

(B) Let the number of men be $5x$ and the number of women be $4x$. Total population $= 9x$.
Literate men $= 80\% \text{ of } 5x = 0.8 \times 5x = 4x$. Illiterate men $= 5x - 4x = x$.
Literate women $= 90\% \text{ of } 4x = 0.9 \times 4x = 3.6x$. Illiterate women $= 4x - 3.6x = 0.4x$.
Total illiterate people $= x + 0.4x = 1.4x$.
The probability that an illiterate person is a man is given by $P = \frac{\text{Illiterate men}}{\text{Total illiterate people}} = \frac{x}{1.4x} = \frac{1}{1.4} = \frac{10}{14} = \frac{5}{7}$.

(D) Let $P(A) = 0.4$ be the probability of event $A$ occurring in a single trial.
The probability that event $A$ does not occur in a single trial is $P(A') = 1 - P(A) = 1 - 0.4 = 0.6$.
Since there are $3$ independent trials, the probability that event $A$ does not occur in any of the $3$ trials is $(P(A'))^3 = (0.6)^3 = 0.216$.
The probability that event $A$ happens at least once is $1 - P(\text{event } A \text{ never happens}) = 1 - 0.216 = 0.784$.

(C) Given: $P(A) = \frac{1}{4}$, $P(A \cap B) = \frac{1}{10}$ and $P(A|B) = 2P(B|A)$.
Using the definition of conditional probability, $P(A|B) = \frac{P(A \cap B)}{P(B)}$ and $P(B|A) = \frac{P(A \cap B)}{P(A)}$.
Substitute these into the given equation: $\frac{P(A \cap B)}{P(B)} = 2 \times \frac{P(A \cap B)}{P(A)}$.
Since $P(A \cap B) \neq 0$, we can divide both sides by $P(A \cap B)$ to get: $\frac{1}{P(B)} = \frac{2}{P(A)}$.
Rearranging gives $P(B) = \frac{P(A)}{2}$.
Substitute $P(A) = \frac{1}{4}$: $P(B) = \frac{1/4}{2} = \frac{1}{8}$.

(B) Given $P(A) = \frac{2}{3}$, $P(B) = \frac{1}{2}$, and $P(A | B) = \frac{2}{3}$.
Since $P(A | B) = \frac{P(A \cap B)}{P(B)}$, we have $P(A \cap B) = P(A | B) \cdot P(B) = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3}$.
Note that $P(A \cap B) = P(A)$, which implies $A \subset B$.
We need to find $P(A' \cup B) + P(A \cup B')$.
Using $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$:
$P(A' \cup B) = P(A') + P(B) - P(A' \cap B) = (1 - P(A)) + P(B) - (P(B) - P(A \cap B)) = 1 - P(A) + P(A \cap B) = 1 - \frac{2}{3} + \frac{1}{3} = \frac{2}{3}$.
$P(A \cup B') = P(A) + P(B') - P(A \cap B') = P(A) + (1 - P(B)) - (P(A) - P(A \cap B)) = 1 - P(B) + P(A \cap B) = 1 - \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$.
Sum $= \frac{2}{3} + \frac{5}{6} = \frac{4+5}{6} = \frac{9}{6} = \frac{3}{2}$.
Wait, re-evaluating: $P(A' \cup B) = P(A' \cup B) = 1 - P(A \cap B') = 1 - (P(A) - P(A \cap B)) = 1 - (\frac{2}{3} - \frac{1}{3}) = 1 - \frac{1}{3} = \frac{2}{3}$.
$P(A \cup B') = 1 - P(A' \cap B) = 1 - (P(B) - P(A \cap B)) = 1 - (\frac{1}{2} - \frac{1}{3}) = 1 - \frac{1}{6} = \frac{5}{6}$.
Sum $= \frac{2}{3} + \frac{5}{6} = \frac{9}{6} = 1.5$. Given the options, let's re-check the expression. If $A \subset B$, then $P(A' \cup B) = P(S) = 1$ and $P(A \cup B') = P(B') = 1 - 1/2 = 1/2$. Sum $= 1.5$. The provided options seem inconsistent with the calculation.

(B) Let $K$ be the event that the student knows the answer and $G$ be the event that the student guesses the answer.
Let $C$ be the event that the student gets the correct answer.
Given: $P(K) = 0.9$, $P(G) = 1 - 0.9 = 0.1$.
If the student knows the answer, the probability of getting it correct is $P(C|K) = 1$.
If the student guesses, the probability of getting it correct is $P(C|G) = \frac{1}{4} = 0.25$.
Using Bayes' Theorem, the probability that the student was guessing given they got the answer correct is $P(G|C) = \frac{P(C|G)P(G)}{P(C|G)P(G) + P(C|K)P(K)}$.
$P(G|C) = \frac{0.25 \times 0.1}{(0.25 \times 0.1) + (1 \times 0.9)} = \frac{0.025}{0.025 + 0.9} = \frac{0.025}{0.925} = \frac{25}{925} = \frac{1}{37}$.

(C) Let $W_A$ be the event of drawing a white ball from bag $A$ and $B_A$ be the event of drawing a black ball from bag $A$.
$P(W_A) = \frac{3}{8}$ and $P(B_A) = \frac{5}{8}$.
If a white ball is transferred, bag $B$ now contains $5$ white and $3$ black balls (total $8$). The probability of drawing a white ball from $B$ is $P(W|W_A) = \frac{5}{8}$.
If a black ball is transferred, bag $B$ now contains $4$ white and $4$ black balls (total $8$). The probability of drawing a white ball from $B$ is $P(W|B_A) = \frac{4}{8}$.
Using the law of total probability:
$P(W) = P(W_A) \cdot P(W|W_A) + P(B_A) \cdot P(W|B_A)$
$P(W) = (\frac{3}{8} \cdot \frac{5}{8}) + (\frac{5}{8} \cdot \frac{4}{8}) = \frac{15}{64} + \frac{20}{64} = \frac{35}{64}$.