Question 1

   Question: Three distinct numbers are selected randomly from the set $\{ 1 , 2 , 3 , \ldots , 4 0 \}$. If the probability, that the selected numbers are in an increasing G.P., is $\textstyle { \frac { m } { n } }$, $\operatorname { g c d } ( m , n ) = 1$, then $m + n$ is equal to.

   Options: Numerical Answer Type (No options provided in the source).

   Correct Answer: 2477.

   Year: JEE Main 2025 (Online) 2nd April Morning Shift.

   Solution (Source): Total choices $= { } ^ { 4 0 } C _ { 3 } = 9 8 8 0$. Required probability $= { \frac { 1 8 } { 9 8 8 0 } } = { \frac { 9 } { 4 9 4 0 } } = { \frac { m } { n } }$. (Note: The source text shows a calculation error in the number of favorable cases, which should be 28 to reach the provided answer of 2477).

   Step Solution:

    1.  Total Outcomes: $n(S) = {}^{40}C_3 = \frac{40 \times 39 \times 38}{3 \times 2 \times 1} = 9880$.

    2.  Favorable Cases: Find sets $(a, ar, ar^2)$ where $1 \le a < ar^2 \le 40$. The common ratio $r$ can be $x/y$ where $x > y$.

    3.  Count G.P.s: By testing $r \in \{2, 3, 4, 5, 6, 3/2, 5/2, 4/3, 5/3, 5/4, 6/5\}$, we find cases: $10+4+2+1+1+4+1+2+1+1+1 = 28$.

    4.  Probability: $P = \frac{28}{9880} = \frac{7}{2470}$.

    5.  Final Sum: Here $m = 7$ and $n = 2470$. Therefore, $m + n = 7 + 2470 = 2477$.

   Difficulty Level: Hard.

   Concept Name: Classical Probability / Geometric Progression.

   Short cut solution: List G.P.s by the square of the common ratio's denominator ($y^2$). For a ratio $x/y$, the first term $a$ must be a multiple of $y^2$, and the third term $a(x/y)^2$ must be $\le 40$.

 Question 6

   Question: One die has two faces marked 1, two faces marked 2, one face marked 3 and one face marked 4. Another die has one face marked 1, two faces marked 2, two faces marked 3 and one face marked 4. The probability of getting the sum of numbers to be 4 or 5, when both the dice are thrown together, is.

   Options: 

       A. 2/3 

       B. 3/5 

       C. 4/9 

       D. 1/2.

   Correct Answer: D.

   Year: JEE Main 2025 (Online) 23rd January Morning Shift.

   Solution (Source): Required probability $= \frac{2}{6} \times \frac{2}{6} + \frac{1}{6} \times \frac{1}{6} + \frac{2}{6} \times \frac{2}{6} + \frac{2}{6} \times \frac{2}{6} + \frac{1}{6} \times \frac{2}{6} + \frac{2}{6} \times \frac{1}{6} + \frac{2}{6} \times \frac{1}{6} + \frac{1}{6} \times \frac{2}{6} + \frac{2}{6} \times \frac{1}{6} + \frac{1}{6} \times \frac{2}{6} = \frac{18}{36} = \frac{1}{2}$.

   Step Solution:

    1.  Probability Distributions: Die 1: $P(1)=2/6, P(2)=2/6, P(3)=1/6, P(4)=1/6$. Die 2: $P(1)=1/6, P(2)=2/6, P(3)=2/6, P(4)=1/6$.

    2.  Sum 4 Pairs: $(1,3), (3,1), (2,2)$. $P(\text{Sum 4}) = (2/6)(2/6) + (1/6)(1/6) + (2/6)(2/6) = 9/36$.

    3.  Sum 5 Pairs: $(1,4), (4,1), (2,3), (3,2)$. $P(\text{Sum 5}) = (2/6)(1/6) + (1/6)(1/6) + (2/6)(2/6) + (1/6)(2/6) = 9/36$.

    4.  Total Probability: Sum the results: $9/36 + 9/36 = 18/36$.

    5.  Simplify: $18/36 = 1/2$.

   Difficulty Level: Medium.

   Concept Name: Classical Probability / Probability Distribution.

   Short cut solution: Use a frequency table. Total ways = 36. Sum up product of occurrences: Sum 4 has $4+1+4=9$ ways. Sum 5 has $2+1+4+2=9$ ways. Total $= 18/36 = 1/2$.

 Question 7

   Question: A board has 16 squares as shown in the figure: Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is. 

   Options: 

       A. 3/5 

       B. 4/5 

          C. 23/30 

       D. 7/10.

   Correct Answer: B.

   Year: JEE Main 2025 (Online) 23rd January Evening Shift.

   Solution (Source): Total ways for selecting any two squares $= {}^{16}C_2 = 120$. Total ways for selecting common side squares $= 24$. So required probability $= 1 - \frac{24}{120} = \frac{4}{5}$.

   Step Solution:

    1.  Total Ways: $n(S) = {}^{16}C_2 = \frac{16 \times 15}{2} = 120$.

    2.  Calculate Opposite Event: Find pairs with a common side (adjacent squares).

    3.  Horizontal Adjacents: In a $4 \times 4$ grid, there are 3 adjacent pairs per row $\times 4$ rows $= 12$.

    4.  Vertical Adjacents: There are 3 adjacent pairs per column $\times 4$ columns $= 12$.

    5.  Probability: Favorable pairs $= 120 - (12 + 12) = 96$. $P = 96/120 = 4/5$.

   Difficulty Level: Medium.

   Concept Name: Complementary Probability / Combinatorics.

   Short cut solution: For an $n \times n$ grid, adjacent pairs are $2n(n-1)$. Here $2 \times 4 \times 3 = 24$. Probability $= 1 - \frac{24}{120} = 4/5$.

Question 9

   Question: Let $A = [a_{ij}]$ be a square matrix of order 2 with entries either 0 or 1. Let E be the event that A is an invertible matrix. Then the probability $P(E)$ is.

   Options: 

       A. 3/8

       B. 1/8

       C. 3/16

       D. 5/8.

   Correct Answer: A.

   Year: JEE Main 2025 (Online) 24th January Evening Shift.

   Solution (Source): $C-I \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} 4$ ways; $C-II \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} \& \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} \to 2$ ways. $P = \frac{\text{favourable}}{\text{total}} = \frac{6}{16} = \frac{3}{8}$.

   Step Solution: 

    1.  Total Outcomes: A $2 \times 2$ matrix has 4 entries. Since each can be 0 or 1, total matrices $n(S) = 2^4 = 16$.

    2.  Invertibility Condition: A matrix is invertible if its determinant $|A| = ad - bc \neq 0$.

    3.  Case 1 ($ad - bc = 1$): This requires $ad = 1$ and $bc = 0$. Thus $a=1, d=1$ and $(b,c) \in \{(0,0), (0,1), (1,0)\}$. This gives 3 ways.

    4.  Case 2 ($ad - bc = -1$): This requires $ad = 0$ and $bc = 1$. Thus $b=1, c=1$ and $(a,d) \in \{(0,0), (0,1), (1,0)\}$. This gives 3 ways.

    5.  Calculate Probability: Total favorable cases $= 3 + 3 = 6$. $P(E) = \frac{6}{16} = \frac{3}{8}$.

   Difficulty Level: Medium.

   Concept Name: Classical Probability / Determinants.

   Short cut solution: In a binary $2 \times 2$ matrix, the only way to be singular is if rows are identical or one is zero. There are 10 such cases. $P = \frac{16-10}{16} = \frac{6}{16} = \frac{3}{8}$.

 Question 11

   Question: Two number $k_1$ and $k_2$ are randomly chosen from the set of natural numbers. Then, the probability that the value of $i^{k_1} + i^{k_2}$, $(i = \sqrt{-1})$ is non-zero, equals.

   Options: 

       A. 3/4

       B. 1/2

       C. 1/4

       D. 2/3.

   Correct Answer: A.

   Year: JEE Main 2025 (Online) 28th January Morning Shift.

   Solution (Source): $i^{k_1} + i^{k_2} \neq 0$. $i^{k_1}$ has 4 options: $i, -1, -i, 1$. Total cases $\implies 4 \times 4 = 16$. Unfavourable cases $\implies i^{k_1} + i^{k_2} = 0$: $\{1, -1\}, \{-1, 1\}, \{i, -i\}, \{-i, i\}$. 4 Cases $\implies$ Probability $= \frac{16 - 4}{16} = \frac{3}{4}$.

   Step Solution: 

    1.  Identify Possible Values: For any $k \in \mathbb{N}$, $i^k$ cyclically results in $\{i, -1, -i, 1\}$.

    2.  Total Outcomes: Selecting two values results in $4 \times 4 = 16$ possible pairs.

    3.  Find Unfavorable Outcomes: The sum is zero if $i^{k_1} = -i^{k_2}$. This happens for pairs $(1, -1), (-1, 1), (i, -i), (-i, i)$.

    4.  Count Non-Zero Cases: Total cases - zero-sum cases $= 16 - 4 = 12$.

    5.  Final Probability: $P = \frac{12}{16} = \frac{3}{4}$.

   Difficulty Level: Easy.

   Concept Name: Powers of $i$ / Classical Probability.

   Short cut solution: For any chosen $i^{k_1}$, there is exactly 1 value out of 4 possible for $i^{k_2}$ that makes the sum zero. Thus, $P(\text{sum}=0) = 1/4$. $P(\text{non-zero}) = 1 - 1/4 = 3/4$.

 Question 12

   Question: Let S be the set of all the words that can be formed by arranging all the letters of the word GARDEN. From the set S, one word is selected at random. The probability that the selected word will NOT have vowels in alphabetical order is.

   Options: 

       A. 1/4

       B. 1/2

       C. 1/3

       D. (Not provided in source).

   Correct Answer: B.

   Year: JEE Main 2025 (Online) 28th January Evening Shift.

   Solution (Source): Vowels: A, E. Consonants: G, R, D, N. Total cases $= 6!$. Favourable cases (when A & E are in order) $= {}^6C_2 \cdot 4!$. $P = \frac{(15)4!}{(30)4!} = 1/2$. Probability when not in order $= 1 - 1/2 = 1/2$.

   Step Solution: 

    1.  Total Outcomes: The word GARDEN has 6 distinct letters. Total arrangements $n(S) = 6! = 720$.

    2.  Identify Vowels: The vowels are A and E.

    3.  Calculate Alphabetical Order Case: To have A and E in alphabetical order (A before E), we choose 2 spots out of 6 for them (${}^6C_2$) and arrange the other 4 letters ($4!$). Favorable $= 15 \times 24 = 360$.

    4.  Probability of Order: $P(\text{Order}) = \frac{360}{720} = \frac{1}{2}$.

    5.  Probability of NOT in Order: $P(\text{Not Order}) = 1 - P(\text{Order}) = 1 - 1/2 = 1/2$.

   Difficulty Level: Medium.

   Concept Name: Permutations / Relative Order.

   Short cut solution: In any arrangement of distinct letters, any two specific letters (like A and E) can only appear in two relative orders: A before E or E before A. By symmetry, both are equally likely. Thus, the probability is $1/2$.

 Question 18

   Question: The probability, of forming a 12 persons committee from 4 engineers, 2 doctors and 10 professors containing at least 3 engineers and at least 1 doctor, is.

   Options: 

       A. 129/182

       B. 17/26

       C. 19/26

       D. 103/182.

   Correct Answer: A.

   Year: JEE Main 2025 (Online) 4th April Morning Shift.

   Solution (Source): The committee can be formed in the following ways: (3E, 1D, 8P), (3E, 2D, 7P), (4E, 1D, 7P), (4E, 2D, 6P). The probability $P = \frac{{}^4C_3 \cdot {}^2C_1 \cdot {}^{10}C_8 + {}^4C_3 \cdot {}^2C_2 \cdot {}^{10}C_7 + {}^4C_4 \cdot {}^2C_1 \cdot {}^{10}C_7 + {}^4C_4 \cdot {}^2C_2 \cdot {}^{10}C_6}{{}^{16}C_{12}} = \frac{129}{182}$.

   Step Solution:

    1.  Total Outcomes: The total number of ways to select 12 people from 16 (4E + 2D + 10P) is $n(S) = {}^{16}C_{12} = {}^{16}C_4 = \frac{16 \times 15 \times 14 \times 13}{4 \times 3 \times 2 \times 1} = 1820$.

    2.  Case 1 (3E, 1D, 8P): ${}^4C_3 \times {}^2C_1 \times {}^{10}C_8 = 4 \times 2 \times 45 = 360$.

    3.  Case 2 & 3 (3E, 2D, 7P & 4E, 1D, 7P): $(4 \times 1 \times 120) + (1 \times 2 \times 120) = 480 + 240 = 720$.

    4.  Case 4 (4E, 2D, 6P): ${}^4C_4 \times {}^2C_2 \times {}^{10}C_6 = 1 \times 1 \times 210 = 210$.

    5.  Probability: $P = \frac{360 + 720 + 210}{1820} = \frac{1290}{1820} = \frac{129}{182}$.

   Difficulty Level: Hard.

   Concept Name: Combinatorics / Case-based Probability.

   Short cut solution: Focus on calculating ${}^{16}C_4$ as the denominator first to eliminate options that don't have 182 or 26 as factors.

 Question 25

   Question: An integer is chosen at random from the integers 1, 2, 3,...,50. The probability that the chosen integer is a multiple of atleast one of 4, 6 and 7 is.

   Options: 

       A. 8/25

       B. 21/50

       C. 9/50

       D. 14/25.

   Correct Answer: B.

   Year: 29th January 2024 Shift 2.

   Solution (Source): $P(A) = 12/50$ (mult. of 4), $P(B) = 8/50$ (mult. of 6), $P(C) = 7/50$ (mult. of 7). $P(A \cap B) = 4/50$, $P(B \cap C) = 1/50$, $P(A \cap C) = 1/50$, $P(A \cap B \cap C) = 0$. $P(A \cup B \cup C) = \frac{12+8+7-4-1-1+0}{50} = \frac{21}{50}$.

   Step Solution:

    1.  Count Single Multiples: Multiples of 4 are 12; multiples of 6 are 8; multiples of 7 are 7.

    2.  Count Double Multiples: Multiples of $\text{lcm}(4,6)=12$ are 4; multiples of $\text{lcm}(6,7)=42$ is 1; multiples of $\text{lcm}(4,7)=28$ is 1.

    3.  Count Triple Multiples: Multiples of $\text{lcm}(4,6,7)=84$ is 0.

    4.  Inclusion-Exclusion: Total multiples $= (12 + 8 + 7) - (4 + 1 + 1) + 0 = 21$.

    5.  Probability: $P = \frac{21}{50}$.

   Difficulty Level: Medium.

   Concept Name: Principle of Inclusion-Exclusion.

   Short cut solution: List the multiples of the smallest sets first and subtract overlaps manually to avoid large PIE calculations.

 Question 26

   Question: Two integers x and y are chosen with replacement from the set {0, 1, 2, 3, ...., 10}. Then the probability that $|x - y| > 5$ is.

   Options: 

       A. 30/121

       B. 62/121

       C. 60/121

       D. 31/121.

   Correct Answer: A.

   Year: 30th January 2024 Shift 1.

   Solution (Source): Total possible ways $= (5+4+3+2+1) \times 2 = 30$. Required probability $= \frac{30}{11 \times 11} = \frac{30}{121}$.

   Step Solution:

    1.  Total Sample Space: With replacement from 11 numbers $\{0, \ldots, 10\}$, $n(S) = 11 \times 11 = 121$.

    2.  Analyze $|x-y| > 5$: This means $x - y \ge 6$ or $y - x \ge 6$.

    3.  Count for $x - y \ge 6$: If $x=6, y=0$; $x=7, y=0,1$; $x=8, y=0,1,2$; $x=9, y=0,1,2,3$; $x=10, y=0,1,2,3,4$.

    4.  Sum Favorable Cases: $1 + 2 + 3 + 4 + 5 = 15$ ways.

    5.  Total Favorable: Due to symmetry $(y > x)$, total favorable $= 15 \times 2 = 30$. $P = 30/121$.

   Difficulty Level: Medium.

   Concept Name: Enumeration / Absolute Difference.

   Short cut solution: For a set $\{0, \dots, n\}$, the number of pairs such that $|x-y| > k$ is $(n-k)(n-k+1)$. Here $n=10$ and $k=5$, so $5 \times 6 = 30$. Probability $= 30/121$.

 Question 33

   Question: Let $N$ denote the number that turns up when a fair die is rolled. If the probability that the system of equations $x + y + z = 1$, $2x + Ny + 2z = 2$, $3x + 3y + Nz = 3$ has unique solution is $k/6$, then the sum of value of $k$ and all possible values of $N$ is.

   Options: 

       A. 18

       B. 19

       C. 20

       D. 21

   Correct Answer: C.

   Year: JEE Main 2023, 24th January Shift 1.

   Solution (Source): $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 2 & N & 2 \\ 3 & 3 & N \end{vmatrix} = (N - 2)(N - 3)$. For unique solution $\Delta \neq 0 \implies N \neq 2, 3$. $P(\text{unique solution}) = 4/6 \implies k = 4$. Sum $= 4 + (1 + 4 + 5 + 6) = 20$.

   Step Solution:

    1.  Set up Determinant: For a unique solution, the determinant of the coefficient matrix ($\Delta$) must be non-zero.

    2.  Calculate $\Delta$: Expanding the matrix $\begin{vmatrix} 1 & 1 & 1 \\ 2 & N & 2 \\ 3 & 3 & N \end{vmatrix}$ gives $(N - 2)(N - 3)$.

    3.  Find Possible $N$: For $\Delta \neq 0$, $N$ cannot be 2 or 3. Since $N \in \{1, 2, 3, 4, 5, 6\}$, the possible values for $N$ are $\{1, 4, 5, 6\}$.

    4.  Find $k$: There are 4 favorable values out of 6, so the probability is $4/6$. Comparing this to $k/6$, we find $k = 4$.

    5.  Calculate Final Sum: The sum of $k$ and all valid values of $N$ is $4 + 1 + 4 + 5 + 6 = 20$.

   Difficulty Level: Medium.

   Concept Name: System of Linear Equations / Cramer's Rule.

   Short cut solution: Quickly identify $N=2$ and $N=3$ make rows proportional/identical, rendering the determinant zero; the remaining 4 faces of the die provide the unique solution.

 Question 35

   Question: Let $M$ be the maximum value of the product of two positive integers when their sum is 66. Let the sample space $S = \{x \in Z : x(66 - x) \geq \frac{5}{9} M\}$ and the event $A = \{x \in S : x \text{ is a multiple of 3}\}$. Then $P(A)$ is equal to.

   Options: 

       A. 15/44

       B. 1/3

       C. 1/5

          D. 7/22

   Correct Answer: B.

   Year: JEE Main 2023, 25th January Shift 1.

   Solution (Source): $M = 33 \times 33$. $x(66 - x) \geq \frac{5}{9} \times 33 \times 33 \implies 11 \leq x \leq 55$. $A : \{12, 15, 18, \ldots, 54\}$. $P(A) = 15/45 = 1/3$.

   Step Solution:

    1.  Find $M$: For a fixed sum, the product of two numbers is maximized when they are equal. $M = 33 \times 33 = 1089$.

    2.  Solve Inequality: $66x - x^2 \geq \frac{5}{9}(1089) \implies x^2 - 66x + 605 \leq 0$.

    3.  Determine Sample Space Range: Factoring $(x - 11)(x - 55) \leq 0$ gives $11 \leq x \leq 55$. The total number of integers in $S$ is $55 - 11 + 1 = 45$.

    4.  Identify Favorable Outcomes: List multiples of 3 in $$, which are $\{12, 15, \ldots, 54\}$. Number of terms $= \frac{54 - 12}{3} + 1 = 15$.

    5.  Calculate Probability: $P(A) = 15/45 = 1/3$.

   Difficulty Level: Medium.

   Concept Name: Maxima and Minima / Quadratic Inequalities.

   Short cut solution: The quadratic $x(66-x)$ is symmetric about 33. Solving $x(66-x) = \frac{5}{9}(33^2)$ quickly gives the bounds 11 and 55.

 Question 36

   Question: Let $x$ and $y$ be distinct integers where $1 \leq x \leq 25$ and $1 \leq y \leq 25$. Then, the number of ways of choosing $x$ and $y$, such that $x + y$ is divisible by 5, is.

   Options: (No options provided in the source; solution specifies a numerical answer).

   Correct Answer: 120.

   Year: JEE Main 2023, 25th January Shift 1.

   Solution (Source): $x + y = 5\lambda$. Cases: Total $= 120$.

   Step Solution:

    1.  Categorize by Remainder: Divide 1–25 into 5 sets based on remainder mod 5: $R_0, R_1, R_2, R_3, R_4$. Each set has 5 elements.

    2.  Identify Valid Pair Combinations: $x+y$ is divisible by 5 if remainders are $(0,0), (1,4), (4,1), (2,3),$ or $(3,2)$.

    3.  Calculate Case $(0,0)$: Choosing two distinct numbers from $R_0 = \{5, 10, 15, 20, 25\}$ gives $5 \times 4 = 20$ ways.

    4.  Calculate Other Cases: For $(1,4)$, $(4,1)$, $(2,3)$, and $(3,2)$, choose one from each set: $5 \times 5 = 25$ ways per case.

    5.  Sum Total Ways: Total ways $= 20 + 25 + 25 + 25 + 25 = 120$.

   Difficulty Level: Medium.

   Concept Name: Modular Arithmetic / Combinatorics.

   Short cut solution: Use the remainder sets. Total pairs $(x,y)$ with $x \neq y$ where $x+y$ is a multiple of 5 is simply $(5 \times 4) + 4(5 \times 5) = 20 + 100 = 120$.

Question 37

   Question: Let $N$ be the sum of the numbers appeared when two fair dice are rolled and let the probability that $N - 2, \sqrt{3N}, N + 2$ are in geometric progression be $\frac{k}{48}$. Then the value of $k$ is.

   Options: 

       A. 2 

       B. 4 

       C. 16 

       D. 8

   Correct Answer: B.

   Year: JEE Main 2023, 25th January Shift 2.

   Solution (Source): $n(s) = 36$. Given: $N - 2, \sqrt{3N}, N + 2$ are in G.P. $3N = (N - 2)(N + 2) \Rightarrow 3N = N^2 - 4 \Rightarrow N^2 - 3N - 4 = 0 \Rightarrow (N - 4)(N + 1) = 0 \Rightarrow N = 4$ or $N = -1$ (rejected). $(\text{Sum} = 4) \equiv \{(1, 3), (3, 1), (2, 2)\}$. $n(A) = 3$. $P(A) = \frac{3}{36} = \frac{1}{12} = \frac{4}{48} \Rightarrow k = 4$.

   Step Solution:

    1.  G.P. Condition: For $a, b, c$ in G.P., $b^2 = ac$. Thus, $(\sqrt{3N})^2 = (N-2)(N+2)$.

    2.  Solve for N: $3N = N^2 - 4 \Rightarrow N^2 - 3N - 4 = 0$. Factoring gives $(N-4)(N+1) = 0$.

    3.  Identify Valid Sum: $N=4$ (since a sum of dice $N$ must be $\ge 2$).

    4.  Count Outcomes: Possible pairs for sum 4 are $(1,3), (3,1), (2,2)$ (3 ways).

    5.  Calculate k: $P = \frac{3}{36} = \frac{1}{12}$. Equating $\frac{1}{12} = \frac{k}{48}$ gives $k=4$.

   Difficulty Level: Medium.

   Concept Name: Geometric Progression / Classical Probability.

   Short cut solution: Quickly recognize that for the expression $N^2 - 3N - 4 = 0$, only $N=4$ is a valid sum for two dice. Probability of sum 4 is always $3/36$.

 Question 41

   Question: If an unbiased die, marked with $-2, -1, 0, 1, 2, 3$ on its faces, is through five times, then the probability that the product of the outcomes is positive, is.

   Options: 

       A. 881/2592 

       B. 521/2592 

       C. 440/2592 

       D. 27/288

   Correct Answer: B.

   Year: JEE Main 2023, 30th January Shift 1.

   Solution (Source): Either all outcomes are positive or any two are negative. Now, $p = P(\text{positive}) = 3/6 = 1/2, q = P(\text{negative}) = 2/6 = 1/3$. Required probability $= {}^5C_5 (1/2)^5 + {}^5C_2 (1/3)^2 (1/2)^3 + {}^5C_4 (1/3)^4 (1/2)^1 = \frac{521}{2592}$.

   Step Solution:

    1.  Define Probabilities: $P(\text{pos}) = 3/6 = 1/2$; $P(\text{neg}) = 2/6 = 1/3$; $P(\text{zero}) = 1/6$.

    2.  Condition for Positive Product: Requires an even number of negative outcomes and NO zero outcomes.

    3.  Case 1 (0 Negative): All 5 positive: ${}^5C_5(1/2)^5 = 1/32$.

    4.  Case 2 (2 Negative): 3 positive and 2 negative: ${}^5C_2(1/2)^3(1/3)^2 = 10 \times \frac{1}{8} \times \frac{1}{9} = \frac{10}{72}$.

    5.  Case 3 (4 Negative): 1 positive and 4 negative: ${}^5C_4(1/2)^1(1/3)^4 = 5 \times \frac{1}{2} \times \frac{1}{81} = \frac{5}{162}$. Summing these gives $\frac{81+360+80}{2592} = \frac{521}{2592}$.

   Difficulty Level: Hard.

   Concept Name: Binomial Distribution / Properties of Products.

   Short cut solution: Focus only on even counts of negative faces. Since we need a positive product, zero must be excluded entirely from all 5 trials.

 Question 48

   Question: Three dice are rolled. If the probability of getting different numbers on the three dice is $\frac{p}{q}$, where $p$ and $q$ are co-prime, then $q - p$ is equal to.

   Options: 

       A. 1 

       B. 2 

       C. 4 

       D. 3

   Correct Answer: C.

   Year: JEE Main 2023, 6th April Shift 2.

   Solution (Source): $\text{Fav.} = \frac{({}^6C_3)(3!)}{6 \times 6 \times 6} = \frac{(20)(6)}{6 \cdot 6 \cdot 6} = \frac{20}{36} = \frac{5}{9} = \frac{p}{q}$. $p = 5, q = 9 \Rightarrow q - p = 4$.

   Step Solution:

    1.  Total Outcomes: $6 \times 6 \times 6 = 216$.

    2.  Favorable Outcomes: Selecting 3 distinct numbers out of 6 and arranging them: ${}^6C_3 \times 3!$.

    3.  Calculate Favorable: $20 \times 6 = 120$.

    4.  Find Simplified Probability: $P = \frac{120}{216} = \frac{5}{9}$.

    5.  Final Calculation: $p = 5, q = 9$. Thus, $q - p = 9 - 5 = 4$.

   Difficulty Level: Easy.

   Concept Name: Classical Probability / Permutations.

   Short cut solution: Use the direct arrangement formula for distinct faces: $\frac{6 \times 5 \times 4}{6 \times 6 \times 6} = \frac{5 \times 4}{6 \times 6} = \frac{20}{36} = \frac{5}{9}$.

 Question 51

   Question: Let $N$ denote the sum of the numbers obtained when two dice are rolled. If the probability that $2^N < N!$ is $\frac{m}{n}$, where $m$ and $n$ are coprime, then $4m - 3n$ equal to.

   Options: 

       A. 12 

       B. 8 

       C. 10 

       D. 6

   Correct Answer: B.

   Year: JEE Main 2023, 10th April Shift 1.

   Solution (Source): $2^N < N!$ is satisfied for $N \geq 4$. Required probability $P(N \geq 4) = 1 - P(N < 4)$. $N=1$ (Not possible). $N=2 (1, 1) \Rightarrow P(N=2) = 1/36$. $N=3 (1, 2), (2, 1) \Rightarrow P(N=3) = 2/36$. $P(N < 4) = 1/36 + 2/36 = 3/36$. $\therefore P(N \geq 4) = 1 - 3/36 = 33/36 = 11/12 = \frac{m}{n} \Rightarrow m=11, n=12$. $\therefore 4m - 3n = 4(11) - 3(12) = 8$.

   Step Solution:

    1.  Test Condition: Verify the inequality $2^N < N!$. For $N=2$, $4 < 2$ (False). For $N=3$, $8 < 6$ (False). For $N=4$, $16 < 24$ (True). The inequality holds for $N \in \{4, 5, \dots, 12\}$.

    2.  Total Outcomes: For two fair dice, $n(S) = 6 \times 6 = 36$.

    3.  Find Complementary Event: $N < 4$ includes sum 2 [(1,1)] and sum 3 [(1,2), (2,1)]. Total ways $= 1 + 2 = 3$.

    4.  Calculate Probability: $P(N \geq 4) = \frac{36 - 3}{36} = \frac{33}{36} = \frac{11}{12}$. Thus, $m = 11$ and $n = 12$.

    5.  Final Calculation: $4(11) - 3(12) = 44 - 36 = 8$.

   Difficulty Level: Medium.

   Concept Name: Factorial Inequalities / Classical Probability.

   Short cut solution: It is much faster to calculate the probability of the complement ($N < 4$) because it only involves three simple outcomes [(1,1), (1,2), (2,1)].

 Question 53

   Question: Let $S = \{ M = [a_{ij}], a_{ij} \in \{0, 1, 2\}, 1 \le i, j \le 2 \}$ be a sample space and $A = \{ M \in S : M \text{ is invertible} \}$ be an event. Then $P(A)$ is equal to.

   Options: 

       A. 16/27 

       B. 50/81 

       C. 47/81 

       D. 49/81

   Correct Answer: B.

   Year: JEE Main 2023, 11th April Shift 1.

   Solution (Source): $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, a, b, c, d \in \{0, 1, 2\}$. $n(S) = 3^4 = 81$. $|M| = 0 \Rightarrow ad = bc$. $ad = bc = 0 \Rightarrow 25$ ways. $ad = bc = 1 \Rightarrow 1^2 = 1$. $ad = bc = 2 \Rightarrow 2^2 = 4$. $ad = bc = 4 \Rightarrow 1^2 = 1$. Total $\bar{A} = 31 \Rightarrow P(A) = 50/81$.

   Step Solution:

    1.  Total Sample Space: Each of the 4 entries in the $2 \times 2$ matrix has 3 choices. $n(S) = 3^4 = 81$.

    2.  Singularity Condition: A matrix is not invertible if its determinant $ad - bc = 0$, or $ad = bc$.

    3.  Count $ad = bc = 0$: This occurs if at least one of $(a, d)$ is 0 and at least one of $(b, c)$ is 0. For $(a, d)$, choices are (0,0), (0,1), (0,2), (1,0), (2,0) = 5 ways. Thus, $5 \times 5 = 25$ ways.

    4.  Count Non-Zero products: $ad=1$ (1,1); $bc=1$ (1,1) $\rightarrow 1$ way. $ad=2$ (1,2), (2,1); $bc=2$ (1,2), (2,1) $\rightarrow 2 \times 2 = 4$ ways. $ad=4$ (2,2); $bc=4$ (2,2) $\rightarrow 1$ way.

    5.  Probability: Total singular cases $= 25 + 1 + 4 + 1 = 31$. $P(A) = 1 - \frac{31}{81} = \frac{50}{81}$.

   Difficulty Level: Hard.

   Concept Name: Singular Matrices / Counting Principle.

   Short cut solution: Systematically listing the products of $\{0, 1, 2\}$ as $0, 1, 2, 4$ allows you to quickly count the $ad=bc$ pairs using symmetry.

 Question 55

   Question: Two dice A and B are rolled. Let numbers obtained on A and B be $\alpha$ and $\beta$ respectively. If the variance of $\alpha - \beta$ is $\frac{p}{q}$, where $p$ and $q$ are co-prime, then the sum of the positive divisor of $p$ is equal to.

   Options: 

       A. 36 

       B. 31 

       C. 48 

       D. 72

   Correct Answer: C.

   Year: JEE Main 2023, 12th April Shift 1.

   Solution (Source): $\mu = \Sigma(x) = 0$ as data is symmetric. Sum of divisors $= (5^0 + 5^1)(7^0 + 7^1) = 6 \times 8 = 48$.

   Step Solution:

    1.  Individual Variance: For a fair die, the variance $Var(X) = \frac{n^2 - 1}{12}$. For $n=6$, $Var(X) = \frac{36-1}{12} = \frac{35}{12}$.

    2.  Variance of Difference: Since dice A and B are independent, $Var(\alpha - \beta) = Var(\alpha) + Var(\beta)$.

    3.  Calculation: $Var(\alpha - \beta) = \frac{35}{12} + \frac{35}{12} = \frac{70}{12} = \frac{35}{6}$.

    4.  Identify $p$: Since 35 and 6 are co-prime, $p = 35$.

    5.  Sum of Divisors: Divisors of $35$ (which is $5 \times 7$) are 1, 5, 7, and 35. Sum $= 1 + 5 + 7 + 35 = 48$.

   Difficulty Level: Medium.

   Concept Name: Variance of Independent Variables / Sum of Divisors.

   Short cut solution: Use the property that for independent $X$ and $Y$, $Var(X - Y) = Var(X) + Var(Y)$. Since both dice are identical, the result is simply $2 \times Var(\text{one die})$.

 Question 64

   Question: Five numbers $x_1, x_2, x_3, x_4, x_5$ are randomly selected from the numbers 1, 2, 3, ......., 18 and are arranged in the increasing order $(x_1 < x_2 < x_3 < x_4 < x_5)$. The probability that $x_2 = 7$ and $x_4 = 11$ is :

   Options: 

       A. 1/136

       B. 1/72

       C. 1/68

       D. 1/34

   Correct Answer: C

   Year: JEE Main 27-Jun-2022 Shift-1

   Solution (Source): No. of ways to select and arrange $x_1, x_2, x_3, x_4, x_5$ from 1, 2, 3... ....... $n(s) = {}^{18}C_5$. $n(E) = {}^6C_1 \times {}^3C_1 \times {}^7C_1$. $P(E) = \frac{6 \times 3 \times 7}{18} \dots \frac{1}{17 \times 4} = \frac{1}{68}$.

   Step Solution:

    1.  Total Outcomes: Selecting 5 distinct numbers from 18 and arranging them in increasing order is equivalent to choosing 5 numbers: $n(S) = {}^{18}C_5 = \frac{18 \times 17 \times 16 \times 15 \times 14}{120} = 8568$.

    2.  Define Constraints: Since $x_1 < x_2 < x_3 < x_4 < x_5$ and given $x_2 = 7$ and $x_4 = 11$, the numbers must satisfy: $x_1 < 7 < x_3 < 11 < x_5$.

    3.  Count Favorable Choices:

           $x_1$ must be from $\{1, 2, 3, 4, 5, 6\}$ (6 choices).

           $x_3$ must be from $\{8, 9, 10\}$ (3 choices).

           $x_5$ must be from $\{12, 13, 14, 15, 16, 17, 18\}$ (7 choices).

    4.  Calculate Total Favorable: $n(E) = 6 \times 3 \times 7 = 126$.

    5.  Final Probability: $P(E) = \frac{126}{8568} = \frac{1}{68}$.

   Difficulty Level: Medium

   Concept Name: Classical Probability / Combinations

   Short cut solution: Use the gap counting method: $P = \frac{(\text{choices for } x_1) \times (\text{choices for } x_3) \times (\text{choices for } x_5)}{{}^{18}C_5}$.

 Question 67

   Question: Let $S = \{E_1, E_2, \dots, E_8\}$ be a sample space of a random experiment such that $P(E_n) = \frac{n}{36}$ for every $n = 1, 2, \dots, 8$. Then the number of elements in the set $\{A \subseteq S : P(A) \ge \frac{4}{5}\}$ is _____.

   Options: Numerical Answer Type.

   Correct Answer: 19

   Year: JEE Main 27-Jun-2022 Shift-2

   Solution (Source): Here $P(E_n) = \frac{n}{36}$ for $n = 1, 2, \dots, 8$. $P(A) = \frac{\text{sum of indices}}{36} \ge \frac{4}{5} \Rightarrow \text{sum} \ge 29$. If one number is left, sum $\ge 29$ by 3 ways. Similarly leaving more terms gives 16 combinations. Total $= 16 + 3 = 19$.

   Step Solution:

    1.  Calculate Required Sum: For $P(A) = \frac{\sum n_i}{36} \ge \frac{4}{5}$, we need $\sum n_i \ge \frac{4}{5} \times 36 = 28.8$. Thus, the sum of indices in subset $A$ must be $\ge 29$.

    2.  Determine Total Sum: The sum of all indices in $S$ is $\frac{8 \times 9}{2} = 36$.

    3.  Use Complementary Sums: It is easier to count subsets where the sum of indices NOT in $A$ is $\le 36 - 29 = 7$.

    4.  List Subsets with Sum $\le 7$: 

           Sum 0: $\emptyset$ (1 way)

           Sum 1-7 (1-element): $\{1\}, \{2\}, \{3\}, \{4\}, \{5\}, \{6\}, \{7\}$ (7 ways)

           Sum 3-7 (2-elements): $\{1,2\}, \{1,3\}, \{1,4\}, \{1,5\}, \{1,6\}, \{2,3\}, \{2,4\}, \{2,5\}, \{3,4\}$ (9 ways)

           Sum 6-7 (3-elements): $\{1,2,3\}, \{1,2,4\}$ (2 ways).

    5.  Calculate Total: $1 + 7 + 9 + 2 = 19$.

   Difficulty Level: Hard

   Concept Name: Power Sets / Sum of Integers

   Short cut solution: Focus on the sum of elements in the complement $A^c$. Since total sum is 36, $P(A) \ge 0.8$ means $P(A^c) \le 0.2$. $0.2 \times 36 = 7.2$. Count subsets with sum $\le 7$.

 Question 68

   Question: The probability, that in a randomly selected 3-digit number at least two digits are odd, is

   Options: 

       A. 19/36

       B. 15/36

       C. 13/36

       D. 23/36

   Correct Answer: A

   Year: JEE Main 28-Jun-2022 Shift-1

   Solution (Source): Atleast two digits are odd = Exactly 2 digits odd + exactly 3 digits odd. For exactly three: $5 \times 5 \times 5 = 125$. For exactly two: (If 0 used) $2 \times 5 \times 5 = 50$; (If 0 not used) ${}^3C_1 \times 4 \times 5 \times 5 = 300$. Total = 475. Probability $= 475/900 = 19/36$.

   Step Solution:

    1.  Total 3-digit Numbers: $9 \times 10 \times 10 = 900$.

    2.  Case 1 (3 Odd Digits): Choices for each position are $\{1, 3, 5, 7, 9\}$. Total $= 5 \times 5 \times 5 = 125$.

    3.  Case 2 (Exactly 2 Odd Digits):

           Odd-Odd-Even: $5 \times 5 \times 5 = 125$ (Even includes $\{0, 2, 4, 6, 8\}$).

           Odd-Even-Odd: $5 \times 5 \times 5 = 125$.

           Even-Odd-Odd: $4 \times 5 \times 5 = 100$ (First digit cannot be 0, so 4 choices: $\{2, 4, 6, 8\}$).

    4.  Sum Favorable Cases: $125 (\text{3 odd}) + [125 + 125 + 100] (\text{2 odd}) = 475$.

    5.  Probability: $P = \frac{475}{900}$. Dividing both by 25 gives $\frac{19}{36}$.

   Difficulty Level: Medium

   Concept Name: Counting Principles / Parity

   Short cut solution: Calculate probability of each digit being odd/even. $P(\text{1st is O}) = 5/9, P(\text{others are O}) = 1/2$. Use Binomial-like expansion while noting the first digit's unique constraint.

 Question 81

   Question: Let a die be rolled n times. Let the probability of getting odd numbers seven times be equal to the probability of getting odd numbers nine times. If the probability of getting even numbers twice is $\textstyle { \frac { k } { 2 ^ { 1 5 } } }$, then $k$ is equal to.

   Options: 

       A. 60

       B. 30

       C. 90

       D. 15.

   Correct Answer: A.

   Year: JEE Main 2023, 10th April Shift 2.

   Solution (Source): $P(\text{odd } 7 \text{ times}) = P(\text{odd } 9 \text{ times})$. ${}^nC_7 (1/2)^7 (1/2)^{n-7} = {}^nC_9 (1/2)^9 (1/2)^{n-9} \Rightarrow {}^nC_7 = {}^nC_9 \Rightarrow n = 16$. Required $P = {}^{16}C_2 \times (1/2)^{16} = \frac{16 \cdot 15}{2} \times \frac{1}{2^{16}} = \frac{15}{2^{13}} \Rightarrow \frac{60}{2^{15}} \Rightarrow k = 60$.

   Step Solution:

    1.  Define success probability: For a fair die, $P(\text{odd}) = 1/2$ and $P(\text{even}) = 1/2$.

    2.  Equate probabilities: Using Binomial Distribution, ${}^nC_7 (1/2)^n = {}^nC_9 (1/2)^n$.

    3.  Find n: The coefficients must be equal, so ${}^nC_7 = {}^nC_9$, which implies $n = 7 + 9 = 16$.

    4.  Calculate even probability: For even numbers twice ($r=2$), $P = {}^{16}C_2 (1/2)^{16} = \frac{16 \times 15}{2} \times \frac{1}{2^{16}} = \frac{120}{2^{16}}$.

    5.  Solve for k: $\frac{120}{2^{16}} = \frac{60}{2^{15}}$. Thus, $k = 60$.

   Difficulty Level: Medium.

   Concept Name: Binomial Distribution.

   Short cut solution: Use the property ${}^nC_x = {}^nC_y \Rightarrow n = x + y$. Here $n = 7 + 9 = 16$. Then calculate ${}^{16}C_2$ and adjust the power of 2.

 Question 84

   Question: A seven digit number is formed using digits 3, 3, 4, 4, 4, 5, 5. The probability, that number so formed is divisible by 2, is.

   Options: 

       A. 6/7

       B. 1/7

       C. 3/7

       D. 4/7.

   Correct Answer: C.

   Year: JEE Main 2021, 26th February Shift-II.

   Solution (Source): Total 7 digit number $= \frac{7!}{3!2!2!} = 210$. If divisible by 2, last digit must be 4. Number divisible by 2 will be $= \frac{6!}{2!2!2!} = 90$. Required probability $= 90/210 = 3/7$.

   Step Solution:

    1.  Total Outcomes: Arrange the digits {3,3, 4,4,4, 5,5}. $n(S) = \frac{7!}{2!3!2!} = \frac{5040}{2 \times 6 \times 2} = 210$.

    2.  Condition for Divisibility: A number is divisible by 2 if its last digit is even. Here, the only even digit is 4.

    3.  Favorable Arrangements: Fix one '4' at the unit place. Arrange the remaining 6 digits: {3, 3, 4, 4, 5, 5}.

    4.  Calculate Favorable: $n(E) = \frac{6!}{2!2!2!} = \frac{720}{8} = 90$.

    5.  Final Probability: $P = \frac{90}{210} = \frac{3}{7}$.

   Difficulty Level: Easy.

   Concept Name: Permutations of Multisets / Classical Probability.

   Short cut solution: Since every digit is equally likely to occupy the last position, the probability is simply the number of even digits divided by the total number of digits $= 3/7$.

 Question 90

   Question: Let A be a set of all 4-digit natural numbers whose exactly one digit is 7. Then, the probability that a randomly chosen element of A leaves remainder 2 when divided by 5 is.

   Options: 

       A. 1/5

       B. 2/9

       C. 97/297

       D. 297/122 (Source text error in labels, but calculation provided).

   Correct Answer: C.

   Year: JEE Main 2021, 25th February Shift-II.

   Solution (Source): Total cases $= ({}^4C_1 \times 9 \times 9 \times 9) - ({}^3C_1 \times 9 \times 9) = 2673$. Remainder 2 when divided by 5 means unit digit is 2 or 7. Unit digit 2 cases $= 225$. Unit digit 7 cases $= 648$. Total $= 873/2673 = 97/297$.

   Step Solution:

    1.  Calculate Total Sample Space: Count 4-digit numbers with exactly one 7. Total $= (1 \times 9^3) + 3 \times (8 \times 1 \times 9^2) = 729 + 1944 = 2673$.

    2.  Remainder Condition: A number leaves remainder 2 mod 5 if it ends in 2 or 7.

    3.  Case 1 (Ends in 7): The 7 is fixed at the end. Thousands place has 8 choices (1-9, no 7, no 0). Others have 9 choices. Total $= 8 \times 9 \times 9 = 648$.

    4.  Case 2 (Ends in 2): Exactly one 7 is in the first three places. This gives $(1 \times 9 \times 9) + (8 \times 1 \times 9) + (8 \times 9 \times 1) = 81 + 72 + 72 = 225$.

    5.  Calculate Probability: $P = \frac{648 + 225}{2673} = \frac{873}{2673} = \frac{97}{297}$.

   Difficulty Level: Hard.

   Concept Name: Permutations with Constraints / Modular Arithmetic.

   Short cut solution: Systematically place the "7" in each of the four positions and subtract cases where "0" is at the start to find the denominator.

 Question 92

   Question: The probability that two randomly selected subsets of the set { 1, 2, 3, 4, 5 } have exactly two elements in their intersection, is.

   Options: 

       A. $\frac{65}{2^7}$ 

       B. $\frac{65}{2^8}$ 

       C. $\frac{135}{2^9}$ 

       D. $\frac{35}{2^7}$.

   Correct Answer: C.

   Year: JEE Main 2021, 24th Feb. Shift-II.

   Solution (Source): Given, set $= \{ 1 , 2 , 3 , 4 , 5 \}$. Let the two subsets be A and B. Then, $n(A \cap B) = 2$ (as given in question). $\therefore$ Required probability $= \frac{{}^5C_2 \times 3^3}{4^5} = \frac{10 \times 27}{2^{10}} = \frac{135}{2^9}$.

   Step Solution:

    1.  Total Outcomes: For each element in a 5-element set, there are 4 possibilities when picking two subsets $A$ and $B$: (in $A$ only, in $B$ only, in both, or in neither). Thus, total pairs $(A, B) = 4^5 = 2^{10}$.

    2.  Select Intersection Elements: Choose exactly 2 elements to be in both $A$ and $B$ from the 5 available: ${}^5C_2 = 10$ ways.

    3.  Arrange Remaining Elements: Each of the remaining 3 elements has 3 choices (in $A$ only, in $B$ only, or in neither): $3^3 = 27$ ways.

    4.  Calculate Favorable Cases: Multiply the selection ways by the remaining arrangements: $10 \times 27 = 270$.

    5.  Final Probability: $P = \frac{270}{2^{10}} = \frac{270}{1024} = \frac{135}{512} = \frac{135}{2^9}$.

   Difficulty Level: Hard.

   Concept Name: Intersection of Subsets / Multiplication Principle.

   Short cut solution: Use the general formula for a set of size $n$ having exactly $r$ elements in the intersection of two subsets: $P = \frac{{}^nC_r \cdot 3^{n-r}}{4^n}$. Here, $n=5, r=2 \Rightarrow \frac{{}^5C_2 \cdot 3^3}{4^5} = \frac{135}{2^9}$.

 Question 93

   Question: Two dices are rolled. If both dices have six faces numbered 1, 2, 3, 5, 7 and 11, then the probability that the sum of the numbers on the top faces is less than or equal to 8 is.

   Options: 

       A. 4/9 

       B. 17/36 

       C. 5/12 

       D. 1/2.

   Correct Answer: B.

   Year: JEE Main 2021, 17 March Shift-1.

   Solution (Source): Six faces have numbers $\{ 1 , 2 , 3 , 5 , 7 , 1 1 \}$. Sum of the numbers on top faces of both the sides is less than or equal to 8 $= ( 1 , 1 ) , ( 1 , 2 ) , ( 1 , 3 ) , ( 1 , 5 ) , ( 1 , 7 ) \to 5$; $( 2 , 1 ) , ( 2 , 2 ) , ( 2 , 3 ) , ( 2 , 5 ) \to 4$; $( 3 , 1 ) , ( 3 , 2 ) , ( 3 , 3 ) , ( 3 , 5 ) \to 4$; $( 5 , 1 ) , ( 5 , 2 ) , ( 5 , 3 ) \to 3$; $( 7 , 1 ) \to 1$. $n(S) = 5 + 4 + 4 + 3 + 1 = 17$. $n(T) = 6 \times 6 = 36$. $\therefore$ Probability $= 17/36$.

   Step Solution:

    1.  Total Sample Space: Each die has 6 faces, so total outcomes $= 6 \times 6 = 36$.

    2.  Enumerate Sums $\le 8$ (Die 1: 1, 2, 3): Pairs for 1 are (1,1), (1,2), (1,3), (1,5), (1,7); for 2 are (2,1), (2,2), (2,3), (2,5); for 3 are (3,1), (3,2), (3,3), (3,5).

    3.  Enumerate Sums $\le 8$ (Die 1: 5, 7, 11): Pairs for 5 are (5,1), (5,2), (5,3); for 7 is (7,1); for 11 is .

    4.  Total Favorable Cases: Sum the counts: $5 + 4 + 4 + 3 + 1 = 17$.

    5.  Probability Calculation: $P = \frac{\text{favorable}}{\text{total}} = \frac{17}{36}$.

   Difficulty Level: Easy.

   Concept Name: Classical Probability / Enumeration.

   Short cut solution: Create a $6 \times 6$ table with faces $\{1, 2, 3, 5, 7, 11\}$ and count entries $\le 8$; systematic counting is the most reliable method here.

 Question 95

   Question: Let A denote the event that a 6-digit integer formed by 0, 1, 2, 3, 4, 5, 6 without repetitions, be divisible by 3. Then, probability of event A is equal to.

   Options: 

       A. 9/56 

       B. 4/9 

       C. 3/7 

       D. 11/27.

   Correct Answer: B.

   Year: JEE Main 2021, 16 March Shift-II.

   Solution (Source): Total 6 digit numbers $= 6 \cdot 6!$. Sum of all 7 digits $= 21$. Divisible by 3 implies digit sum is divisible by 3. Remove multiple of 3 ($0, 3, 6$). Case I $\{1, 2, 3, 4, 5, 6\} \to 6!$; Case II $\{0, 1, 2, 4, 5, 6\} \to 5 \cdot 5!$; Case III $\{0, 1, 2, 3, 4, 5\} \to 5 \cdot 5!$. Probability $= \frac{6! + 5 \cdot 5! + 5 \cdot 5!}{6 \cdot 6!} = 16/36 = 4/9$.

   Step Solution:

    1.  Total Sample Space: To form a 6-digit number from 7 digits $\{0..6\}$, the first digit cannot be 0 (6 choices), and the remaining 5 spots are $6 \times 5 \times 4 \times 3 \times 2$. $n(S) = 6 \times 6! = 4320$.

    2.  Apply Divisibility Rule: A number is divisible by 3 if the sum of its digits is a multiple of 3. The sum of $\{0, 1, 2, 3, 4, 5, 6\} = 21$.

    3.  Find Favorable Digit Sets: We must exclude one digit such that the remaining 6 sum to a multiple of 3. This only happens if we exclude $0, 3,$ or $6$.

    4.  Count Case Outcomes: Excluding 0 (set $\{1..6\}$) gives $6! = 720$ ways. Excluding 3 or 6 (sets including 0) gives $5 \times 5! = 600$ ways each. Total favorable $= 720 + 600 + 600 = 1920$.

    5.  Calculate Probability: $P = \frac{1920}{4320} = \frac{192}{432} = \frac{4}{9}$.

   Difficulty Level: Hard.

   Concept Name: Divisibility Rules / Permutations with Constraints.

   Short cut solution: The probability is $\frac{6! + 2(5 \cdot 5!)}{6 \cdot 6!}$. Factor out $5!$: $\frac{6 + 10}{6 \times 6} = \frac{16}{36} = \frac{4}{9}$.

 Question 98

   Question: Let 9 distinct balls be distributed among 4 boxes, $B_1, B_2, B_3$ and $B_4$. If the probability that $B_3$ contains exactly 3 balls is $k \left( \frac{3}{4} \right)^9$, then $k$ lies in the set.

   Options: 

       A. $\{ x \in R : | x - 3 | < 1 \}$

       B. $\{ x \in R : | x - 2 | \le 1 \}$

       C. $\{ x \in R : | x - 13 | < 1 \}$

       D. $\{ x \in R : | x - 5 | \le 1 \}$.

   Correct Answer: A.

   Year: JEE Main 2021, 25 July Shift-1.

   Solution (Source): $B_3$ contains exactly 3 balls among 9 balls. So, number of ways $= {}^9C_3$. And the rest of the 6 balls can be placed in $3^6$ ways. $n(E) = {}^9C_3 \times 3^6$. And $n(s) = 4^9$. Probability $= \frac{{}^9C_3 \times 3^6}{4^9} = k \frac{3^9}{4^9}$. $k = \frac{{}^9C_3}{3^3} = \frac{9 \times 8 \times 7}{1 \times 2 \times 3 \times 27} = \frac{28}{9}$. Which satisfies $|x - 3| < 1$.

   Step Solution: 

    1.  Total Outcomes: Since each of the 9 distinct balls has 4 box choices, $n(S) = 4^9$.

    2.  Favorable Arrangements: Choose 3 specific balls for Box 3 in ${}^9C_3$ ways.

    3.  Distribute Remainder: Place the remaining 6 balls into the other 3 boxes ($B_1, B_2, B_4$) in $3^6$ ways.

    4.  Solve for k: Equate $\frac{{}^9C_3 \cdot 3^6}{4^9} = k \frac{3^9}{4^9}$, leading to $k = \frac{84}{3^3} = \frac{28}{9}$.

    5.  Set Verification: $28/9 \approx 3.11$, which satisfies $|3.11 - 3| = 0.11 < 1$.

   Difficulty Level: Medium.

   Concept Name: Combinatorics / Classical Probability.

   Short cut solution: Treat each ball placement as a Bernoulli trial where success is Box 3 ($p=1/4$). The probability of 3 successes is ${}^9C_3 (1/4)^3 (3/4)^6$. Factoring out $(3/4)^9$ gives $k = {}^9C_3 / 3^3$.

 Question 99

   Question: The probability that a randomly selected 2-digit number belongs to the set $\{ n \in N : (2^n - 2) \text{ is a multiple of 3} \}$ is equal to.

   Options: 

       A. 1/6 

       B. 2/3 

       C. 1/2 

       D. 1/3.

   Correct Answer: C.

   Year: JEE Main 2021, 27 July Shift-1.

   Solution (Source): Sample set $S = \{ 10, 11, 12, \dots, 99 \}$. $n(S) = 90$. As we can see, for all the odd values of n, $2^n - 2$ is divisible by 3. So, the event set will be all the two digit odd numbers. $E = \{ 11, 13, 15, \dots, 99 \}$.

   Step Solution: 

    1.  Define Sample Space: The 2-digit numbers are $\{10, 11, \dots, 99\}$, so $n(S) = 90$.

    2.  Test the Property: Using modular arithmetic, $2^n \equiv (-1)^n \pmod 3$.

    3.  Evaluate Condition: $2^n - 2 \equiv (-1)^n - 2 \pmod 3$.

    4.  Identify valid n: If $n$ is odd, $(-1) - 2 = -3 \equiv 0 \pmod 3$. If $n$ is even, $1 - 2 = -1 \equiv 2 \pmod 3$. Thus $n$ must be odd.

    5.  Count and Calculate: There are 45 odd 2-digit numbers in the set. $P = 45/90 = 1/2$.

   Difficulty Level: Easy.

   Concept Name: Modular Arithmetic / Power of Integers.

   Short cut solution: In any large consecutive range of integers, odd and even numbers appear with equal frequency ($1/2$ probability). Since the condition depends only on the parity of $n$, the answer is $1/2$.

 Question 101

   Question: Four dice are thrown simultaneously and the numbers shown on these dice are recorded in $2 \times 2$ matrices. The probability that such formed matrices have all different entries and are non singular, is.

   Options: 

       A. 45/162 

       B. 23/81 

       C. 22/81 

       D. 43/162.

   Correct Answer: D.

   Year: JEE Main 2021, 22 July Shift-II.

   Solution (Source): $n(s) = 6^4$. For non-singular matrix $|A| = ad - bc \neq 0$. If $ad = bc$ with different entries: sets are $\{1,2,3,6\}$ and $\{2,3,4,6\}$. Total arrangements for these two sets where $ad=bc$ is 16. Total arrangements with distinct entries is ${}^6C_4 \times 4! = 360$. $P = \frac{360 - 16}{6^4} = \frac{43}{162}$.

   Step Solution: 

    1.  Total Outcomes: Every matrix entry is a die face, so $n(S) = 6^4 = 1296$.

    2.  All Different Requirement: Number of ways to fill the matrix with 4 distinct digits is $P(6,4) = 6 \times 5 \times 4 \times 3 = 360$.

    3.  Identify Singular Cases: A matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is singular if $ad = bc$. We need sets of 4 distinct numbers $\{a,b,c,d\}$ where this is possible.

    4.  Count arrangements: The sets are $\{1,2,3,6\}$ (8 ways to have $ad=bc$) and $\{2,3,4,6\}$ (8 ways to have $ad=bc$). Total $= 16$.

    5.  Final Probability: Favorable $= 360 - 16 = 344$. $P = 344/1296 = 43/162$.

   Difficulty Level: Hard.

   Concept Name: Determinants of Matrices / Permutations.

   Short cut solution: Focus on the products. The only distinct digit sets from $\{1..6\}$ that have equal products are $\{1,6,2,3\}$ and $\{2,6,3,4\}$. Since each set has 24 arrangements and 8 are singular, subtract $8+8$ from the 360 total distinct arrangements.

 Question 103

   Question: Let $S = \{ 1 , 2 , 3 , 4 , 5 , 6 \}$. Then, the probability that a randomly chosen onto function $g$ from $S$ to $S$ satisfies $g(3) = 2g(1)$ is.

   Options: 

       A. 1/10 

       B. 1/15 

       C. 1/5 

       D. 1/30.

   Correct Answer: A.

   Year: JEE Main 2021, 31 Aug. Shift-II.

   Solution (Source): $g : S \longrightarrow S$ where $S = \{ 1 , 2 , 3 , 4 , 5 , 6 \}$. Total number of onto function $= 6!$. Given condition $g(3) = 2g(1)$. If $g(1) = 1$, then $g(3) = 2$; $g(1) = 2$, then $g(3) = 4$; and $g(1) = 3$, then $g(3) = 6$. There are only 3 possible cases. Number onto functions $= 3 \times 4!$. Required probability $= \frac{3 \times 4!}{6!} = \frac{1}{10}$.

   Step Solution:

    1.  Total Outcomes: For a set of 6 elements, the total number of onto functions from $S$ to itself is $6! = 720$.

    2.  Define Constraint: We need $g(3) = 2g(1)$. Since the codomain is $\{1, 2, 3, 4, 5, 6\}$, $g(1)$ can only be 1, 2, or 3.

    3.  Count Valid Pairs: This gives three possible pairs for $(g(1), g(3))$: $(1, 2), (2, 4),$ or $(3, 6)$.

    4.  Arrange Remaining Elements: For each of these 3 pairs, the remaining 4 elements in the domain must be mapped to the remaining 4 elements in the codomain in $4!$ ways.

    5.  Calculate Probability: $P = \frac{3 \times 4!}{6!} = \frac{3 \times 24}{720} = \frac{72}{720} = \frac{1}{10}$.

   Difficulty Level: Medium.

   Concept Name: Onto Functions / Permutations.

   Short cut solution: In a permutation of $n$ elements, the probability that two specific elements take specific values is $\frac{1}{n(n-1)}$. Since there are 3 valid pairs here, $P = 3 \times \frac{1}{6 \times 5} = \frac{3}{30} = \frac{1}{10}$.

 Question 108

   Question: Two squares are chosen at random on a chessboard (see figure). The probability that they have a side in common is.

   Options: 

       A. 2/7 

       B. 1/18 

       C. 1/7 

       D. 1/9.

   Correct Answer: B.

   Year: JEE Main (Standard Problem included in 2020-2021 collections).

   Solution (Source): Total ways $= {}^{64}C_2 = 32 \times 63$. Favourable ways $= 2 \times 7 \times 8$. Required probability $= \frac{2 \times 7 \times 8}{32 \times 63} = \frac{1}{18}$.

   Step Solution:

    1.  Total Sample Space: A chessboard has 64 squares. Choosing 2 squares is ${}^{64}C_2 = \frac{64 \times 63}{2} = 2016$.

    2.  Identify Adjacency: Squares have a side in common if they are horizontally or vertically adjacent.

    3.  Horizontal Pairs: In each of the 8 rows, there are 7 adjacent pairs. Total $= 8 \times 7 = 56$.

    4.  Vertical Pairs: In each of the 8 columns, there are 7 adjacent pairs. Total $= 8 \times 7 = 56$.

    5.  Calculate Probability: Total favorable $= 56 + 56 = 112$. $P = \frac{112}{2016} = \frac{1}{18}$.

   Difficulty Level: Easy.

   Concept Name: Combinatorics / Grid Adjacency.

   Short cut solution: For an $n \times n$ grid, the number of adjacent pairs is $2n(n-1)$. Here, $2 \times 8 \times 7 = 112$. Probability $= \frac{112}{2016} = \frac{1}{18}$.

 Question 114

   Question: If 10 different balls are to be placed in 4 distinct boxes at random, then the probability that two of these boxes contain exactly 2 and 3 balls is :.

   Options: 

       A. $\frac{965}{2^{11}}$ 

       B. $\frac{965}{2^{10}}$ 

       C. $\frac{945}{2^{10}}$ 

       D. $\frac{17 \times 945}{2^{15}}$ (Note: Option D is partially obscured in source text but the solution derives this value).

   Correct Answer: D.

   Year: Jan. 9, 2020 (II).

   Solution (Source): Total ways $= 4^{10}$. Cases for boxes containing 2 and 3 balls: Case-1 (2,3,0,5), Case-2 (2,3,1,4), Case-3 (2,3,2,3). Summing the multinomial distributions multiplied by box arrangements gives $2^5 \times 17 \times 945$. Probability $= \frac{2^5 \times 17 \times 945}{4^{10}} = \frac{17 \times 945}{2^{15}}$.

   Step Solution:

    1.  Total Sample Space: Since each of the 10 distinct balls has 4 box choices, $n(S) = 4^{10} = 2^{20}$.

    2.  Case 1 (2, 3, 0, 5 distribution): Select boxes and balls: $\frac{10!}{2!3!0!5!} \times 4! = 2520 \times 24 = 60480$.

    3.  Case 2 (2, 3, 1, 4 distribution): Select boxes and balls: $\frac{10!}{2!3!1!4!} \times 4! = 12600 \times 24 = 302400$.

    4.  Case 3 (2, 3, 2, 3 distribution): Note box sizes repeat. Ways: $\frac{10!}{2!2!3!3!} \times \frac{4!}{2!2!} = 12600 \times 6 = 75600$.

    5.  Total Probability: Sum the ways and divide by $4^{10}$ to reach the simplified result $\frac{17 \times 945}{2^{15}}$.

   Difficulty Level: Hard.

   Concept Name: Multinomial Theorem / Distribution of Distinct Objects.

   Short cut solution: Use the multinomial coefficient for each case of partitions of 10 into 4 parts where two parts are 2 and 3. Sum these and divide by $4^{10}$.

Question 118

   Question: The probability that a randomly chosen 5-digit number is made from exactly two digits is:

   Options: 

       A. $135 / 10^4$

       B. $121 / 10^4$

       C. $150 / 10^4$

       D. $134 / 10^4$

   Correct Answer: A

   Year: Sep. 03, 2020 (II)

   Solution (Source): Total outcomes $= 9(10^4)$. Favourable outcomes $= {}^9C_2(2^5 - 2) + {}^9C_1(2^4 - 1) = 36(30) + 9(15)$. Probability $= \frac{36 \times 30 + 9 \times 15}{90000} = \frac{4 \times 30 + 15}{10^4} = \frac{135}{10^4}$.

   Step Solution:

    1.  Total Outcomes: The total number of 5-digit numbers is $9 \times 10^4 = 90,000$.

    2.  Case 1 (Two non-zero digits): Select 2 digits from $\{1, 2, \dots, 9\}$ in ${}^9C_2 = 36$ ways. Each of the 5 positions has 2 choices, but we subtract the 2 cases where all digits are the same: $36 \times (2^5 - 2) = 1080$.

    3.  Case 2 (One non-zero digit and zero): Select 1 non-zero digit in ${}^9C_1 = 9$ ways. The first digit must be the non-zero one, and the other 4 positions have 2 choices (0 or the digit), excluding the case where all are non-zero: $9 \times (2^4 - 1) = 135$.

    4.  Total Favorable: Sum both cases: $1080 + 135 = 1215$.

    5.  Calculate Probability: $P = \frac{1215}{90000} = \frac{135}{10000} = \frac{135}{10^4}$.

   Difficulty Level: Hard

   Concept Name: Combinatorial Probability / Case-based Counting

   Short cut solution: Use the formula $\frac{{}^9C_2(2^5-2) + {}^9C_1(2^4-1)}{90000}$ and simplify by dividing both numerator and denominator by 9 to get $\frac{135}{10^4}$.

Question 124

   Question: Out of 11 consecutive natural numbers if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference, is:

   Options: 

       A. 101/15

       B. 5/101

       C. 5/33

       D. 10/99

   Correct Answer: C

   Year: Sep. 06, 2020 (I)

   Solution (Source): For an A.P. $2b = a + c$ (even), so both a and c even numbers or odd numbers from given numbers and b number will be fixed automatically. Required probability $= \frac{{}^6C_2 + {}^5C_2}{{}^{11}C_3} = \frac{25}{165} = \frac{5}{33}$.

   Step Solution:

    1.  Total Outcomes: Selecting 3 distinct numbers from 11 is ${}^{11}C_3 = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165$.

    2.  AP Condition: For $a, b, c$ to be in AP, $a + c = 2b$, meaning $a + c$ must be even.

    3.  Identify Parity: In 11 consecutive numbers, there are 6 odd and 5 even numbers (or vice-versa).

    4.  Count Favorable Pairs: To make $a+c$ even, both must be odd (${}^6C_2 = 15$) or both even (${}^5C_2 = 10$). $15 + 10 = 25$ ways.

    5.  Final Probability: $P = \frac{25}{165} = \frac{5}{33}$.

   Difficulty Level: Medium

   Concept Name: Arithmetic Progression / Combinatorial Selection

   Short cut solution: In any set of $n$ consecutive integers, the number of ways to pick an AP of 3 is the number of ways to pick 2 numbers of the same parity.

Question 126

   Question: Let $S = \{1, 2, ....., 20\}$. A subset B of S is said to be “nice”, if the sum of the elements of B is 203. Than the probability that a randomly chosen subset of S is “nice” is :

   Options: 

       A. $7 / 2^{20}$

       B. $5 / 2^{20}$

       C. $4 / 2^{20}$

       D. $6 / 2^{20}$

   Correct Answer: B

   Year: Jan. 11, 2019 (II)

   Solution (Source): Since total number of subsets of the set $S = 2^{20}$. Now, the sum of all number from 1 to $20 = \frac{20 \times 21}{2} = 210$. Then, find the sets which has sum 7: (1) $\{7\}$, (2) $\{1, 6\}$, (3) $\{2, 5\}$, (4) $\{3, 4\}$, (5) $\{1, 2, 4\}$. There is only 5 sets which has sum 203. Hence required probability $= \frac{5}{2^{20}}$.

   Step Solution:

    1.  Total Sample Space: A set with 20 elements has $2^{20}$ possible subsets.

    2.  Calculate Total Sum: The sum of all elements in $S$ is $\sum_{i=1}^{20} i = \frac{20 \times 21}{2} = 210$.

    3.  Complementary Sum: A subset sums to 203 if and only if its complement subset sums to $210 - 203 = 7$.

    4.  Enumerate Sum-7 Subsets: The subsets summing to 7 are $\{7\}$, $\{1, 6\}$, $\{2, 5\}$, $\{3, 4\}$, and $\{1, 2, 4\}$ (5 subsets).

    5.  Final Probability: $P = \frac{\text{favorable subsets}}{\text{total subsets}} = \frac{5}{2^{20}}$.

   Difficulty Level: Medium

   Concept Name: Power Sets / Complementary Counting

   Short cut solution: Recognize that $P(\text{sum } X) = P(\text{sum } Total - X)$. Finding subsets that sum to 7 is much faster than finding those that sum to 203.

Question 131

   Question: An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noted. If the toss of the coin results in tail then a card from a well-shuffled pack of nine cards numbered $1, 2, 3, \ldots, 9$ is randomly picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is:

   Options: 

       A. 13/36 

       B. 15/72 

       C. 19/72 

       D. 19/36

   Correct Answer: C

   Year: Jan 10, 2019 (I)

   Solution (Source): $P(\text{Outcome is head}) = 1/2$. $P(\text{Outcome is tail}) = 1/2$. $P(\text{7 or 8 is the sum of two dice}) = 6/36 + 5/36 = 11/36$. $P(\text{7 or 8 is the number of card}) = 1/9 + 1/9 = 2/9$. Required probability $= 1/2 \times 11/36 + 1/2 \times 2/9 = \frac{1}{2} (\frac{11+8}{36}) = \frac{19}{72}$.

   Step Solution:

    1.  Define coin probabilities: The probability of Head $P(H)$ is $1/2$ and Tail $P(T)$ is $1/2$.

    2.  Calculate dice probability: For sum 7, there are 6 ways; for sum 8, there are 5 ways. $P(\text{Sum 7 or 8}) = (6+5)/36 = 11/36$.

    3.  Calculate card probability: Two specific numbers (7 and 8) out of 9 cards. $P(\text{Card 7 or 8}) = 2/9$.

    4.  Law of Total Probability: Total $P = P(H) \cdot P(\text{Dice}) + P(T) \cdot P(\text{Card})$.

    5.  Final Calculation: $P = (1/2)(11/36) + (1/2)(2/9) = 11/72 + 8/72 = 19/72$.

   Difficulty Level: Medium

   Concept Name: Law of Total Probability

   Short cut solution: Average the two conditional probabilities: $\frac{11/36 + 8/36}{2} = \frac{19/72}$.

Question 141

   Question: If three of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle formed with these chosen vertices is equilateral is:

   Options: 

       A. 1/10 

       B. 1/5 

       C. 3/10 

       D. 3/20

   Correct Answer: A

   Year: April 12, 2019 (I)

   Solution (Source): Total no. of triangles $= {}^6C_3$. Favorable no. of triangle i.e, equilateral triangles ($\Delta ACE$ and $\Delta BDF$) $= 2$. Hence, required probability $= \frac{2}{{}^6C_3} = \frac{1}{10}$.

   Step Solution:

    1.  Total Outcomes: Selecting 3 vertices out of 6 to form a triangle is $n(S) = {}^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.

    2.  Identify Equilateral Triangles: In a regular hexagon, equilateral triangles are formed by joining alternate vertices.

    3.  List Cases: Let vertices be $\{1, 2, 3, 4, 5, 6\}$. The triangles are $\{1, 3, 5\}$ and $\{2, 4, 6\}$.

    4.  Count Favorable: There are exactly 2 favorable cases.

    5.  Calculate Probability: $P = 2/20 = 1/10$.

   Difficulty Level: Easy

   Concept Name: Geometric Combinatorial Probability

   Short cut solution: In any regular $n$-gon where $n$ is a multiple of 3, the number of equilateral triangles is $n/3$. For $n=6$, $P = \frac{6/3}{{}^6C_3} = \frac{2}{20} = 0.1$.

Question 145

   Question: Two different families A and B are blessed with equal number of children. There are 3 tickets to be distributed amongst the children of these families so that no child gets more than one ticket. If the probability that all the tickets go to the children of family B is 1/12, then the number of children in each family is?

   Options: 

       A. 4 

       B. 6 

       C. 3 

       D. 5

   Correct Answer: D

   Year: Online April 16, 2018

   Solution (Source): Let number of children be $x$. Total children $= 2x$. $P = \frac{{}^xC_3}{{}^{2x}C_3} = 1/12$. $\frac{x(x-1)(x-2)}{2x(2x-1)(2x-2)} = 1/12 \Rightarrow \frac{x-2}{4(2x-1)} = 1/12 \Rightarrow x = 5$.

   Step Solution:

    1.  Define Variables: Let each family have $x$ children. Total population for tickets is $2x$.

    2.  Sample Space: Number of ways to give 3 tickets to $2x$ children is ${}^{2x}C_3$.

    3.  Favorable Event: Number of ways all 3 tickets go to the $x$ children of family B is ${}^xC_3$.

    4.  Set Equation: $\frac{{}^xC_3}{{}^{2x}C_3} = \frac{x(x-1)(x-2)}{2x(2x-1)(2x-2)} = \frac{x-2}{4(2x-1)} = \frac{1}{12}$.

    5.  Solve for $x$: $12(x-2) = 4(2x-1) \Rightarrow 3x - 6 = 2x - 1 \Rightarrow x = 5$.

   Difficulty Level: Hard

   Concept Name: Combinatorial Equations

   Short cut solution: Test the options. If $x=5$, $P = \frac{{}^5C_3}{{}^{10}C_3} = \frac{10}{120} = \frac{1}{12}$. This matches the given probability.

Question 151

   Question: From a group of 10 men and 5 women, four member committees are to be formed each of which must contain at least one woman. Then the probability for these committees to have more women than men, is :

   Options: 

       A. 21/220 

       B. 3/11 

       C. 1/11 

       D. 2/23

   Correct Answer: C

   Year: Online April 9, 2017

   Solution (Source): Probability of 4 member committee which contain atleast one woman includes cases (3M, 1W), (2M, 2W), (1M, 3W), and (0M, 4W). Total ways with at least one woman $= {}^{15}C_4 - {}^{10}C_4 = 1365 - 210 = 1155$. Favourable cases (more women than men) are (1M, 3W) and (0M, 4W). Probability $= \frac{{}^{10}C_1 \cdot {}^5C_3 + {}^{10}C_0 \cdot {}^5C_4}{1155} = \frac{100 + 5}{1155} = \frac{1}{11}$.

   Step Solution:

    1.  Total Sample Space: Total ways to select 4 people from 15 is ${}^{15}C_4 = 1365$.

    2.  Constraint Sample Space: Subtract committees with NO women (${}^{10}C_4 = 210$) to get committees with at least one woman: $n(S) = 1365 - 210 = 1155$.

    3.  Identify Favorable Cases: "More women than men" in a 4-person committee means 3 women and 1 man OR 4 women and 0 men.

    4.  Calculate Favorable Counts: (3W, 1M) $= {}^5C_3 \times {}^{10}C_1 = 10 \times 10 = 100$. (4W, 0M) $= {}^5C_4 \times {}^{10}C_0 = 5 \times 1 = 5$.

    5.  Final Probability: $P = \frac{100 + 5}{1155} = \frac{105}{1155} = \frac{1}{11}$.

   Difficulty Level: Hard

   Concept Name: Combinatorial Probability / Constrained Selection

   Short cut solution: Denominator is (Total - Men only). Numerator is (3W + 4W). $P = \frac{100+5}{1365-210} = 105/1155 = 1/11$.

Question 152

   Question: If two different numbers are taken from the set (0, 1, 2, 3, ......, 10), then the probability that their sum as well as absolute difference are both multiple of 4, is :

   Options: 

       A. 7/55 

       B. 6/55 

       C. 12/55 

       D. 14/55

   Correct Answer: B

   Year: 2017

   Solution (Source): Let $A \equiv \{0, 1, 2, \dots, 10\}$. $n(S) = {}^{11}C_2 = 55$. Let E be the given event. $E \equiv \{(0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10)\}$. $n(E) = 6$. $P(E) = 6/55$.

   Step Solution:

    1.  Total Outcomes: Selecting 2 distinct numbers from 11 is $n(S) = {}^{11}C_2 = \frac{11 \times 10}{2} = 55$.

    2.  Establish Conditions: Let the numbers be $x$ and $y$. We need $x+y = 4k$ and $|x-y| = 4m$.

    3.  Deduce Parity: Adding both equations gives $2x = 4(k+m) \Rightarrow x = 2(\text{integer})$. Thus $x$ and $y$ must both be even.

    4.  Test Even Pairs: Both numbers must be from $\{0, 2, 4, 6, 8, 10\}$. For $x+y$ to be a multiple of 4, $x/2 + y/2$ must be even. This means $x/2$ and $y/2$ must have the same parity.

    5.  List Pairs: 

           Both $x/2, y/2$ even: $\{0, 4, 8\} \rightarrow (0,4), (0,8), (4,8)$ (3 ways).

           Both $x/2, y/2$ odd: $\{2, 6, 10\} \rightarrow (2,6), (2,10), (6,10)$ (3 ways). Total $= 6$.

   Difficulty Level: Medium

   Concept Name: Number Theory / Classical Probability

   Short cut solution: For both sum and difference to be multiples of 4, both numbers must be even and separated by a multiple of 4 (e.g., 0 and 4, 2 and 6). Count these pairs directly from the even subset.

 Question 161

   Question: If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is :

   Options: 

       A. $220 (1/3)^{12}$ 

       B. $22 (1/3)^{11}$ 

       C. $\frac{55}{3} (2/3)^{11}$ 

       D. $55 (2/3)^{10}$

   Correct Answer: C

   Year: 2015

   Solution (Source): Note: The question should state ‘3 different’ boxes instead of ‘3 identical boxes’ and one particular box has 3 balls. Required probability $= \frac{{}^{12}C_3 \times 2^9}{3^{12}} = \frac{55}{3} (\frac{2}{3})^{11}$.

   Step Solution:

    1.  Interpret Constraints: Following the source's clarification, assume 3 distinct boxes and a specific box is chosen to have 3 balls.

    2.  Total Sample Space: Each of the 12 balls has 3 box options, so $n(S) = 3^{12}$.

    3.  Choose Balls for specific box: Number of ways to pick 3 balls from 12 is ${}^{12}C_3 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.

    4.  Distribute Remainder: The remaining 9 balls must go into the other 2 boxes, which gives $2^9$ ways.

    5.  Calculate and Simplify: $P = \frac{220 \times 2^9}{3^{12}} = \frac{55 \times 4 \times 2^9}{3 \times 3^{11}} = \frac{55 \times 2^{11}}{3 \times 3^{11}} = \frac{55}{3} \left(\frac{2}{3}\right)^{11}$.

   Difficulty Level: Hard

   Concept Name: Binomial Distribution / Distribution into Boxes

   Short cut solution: Use the Binomial Probability formula $P(X=r) = {}^{n}C_r p^r q^{n-r}$ where $n=12, r=3, p=1/3, q=2/3$. $P = {}^{12}C_3 (1/3)^3 (2/3)^9 = 220 \cdot \frac{2^9}{3^{12}} = \frac{55}{3} (2/3)^{11}$.

Question 162

   Question: Let $X$ be a set containing 10 elements and $P(X)$ be its power set. If $A$ and $B$ are picked up at random from $P(X)$ with replacement, then the probability that $A$ and $B$ have equal number elements, is:

   Options: 

       A. $(2^{10} - 1)$ 

       B. $\frac{{}^{20}C_{10}}{2^{10}}$ 

       C. $\frac{(2^{10} - 1)}{2^{20}}$ 

       D. $\frac{{}^{20}C_{10}}{2^{20}}$

   Correct Answer: D

   Year: Online April 10, 2015

   Solution (Source): Required probability is $\frac{({}^{10}C_0)^2 + ({}^{10}C_1)^2 + ({}^{10}C_2)^2 + \ldots \ldots + ({}^{10}C_{10})^2}{2^{10} \times 2^{10}} = \frac{{}^{20}C_{10}}{2^{20}}$.

   Step Solution:

    1.  Total Outcomes: A set with 10 elements has $2^{10}$ subsets. Choosing two subsets $A$ and $B$ with replacement gives $n(S) = 2^{10} \times 2^{10} = 2^{20}$.

    2.  Define Condition: Let the number of elements in both sets be $r$, where $r \in \{0, 1, \dots, 10\}$.

    3.  Count Ways for fixed $r$: The number of ways to choose subset $A$ with $r$ elements is ${}^{10}C_r$. For $B$ to have the same number, there are also ${}^{10}C_r$ ways. Total for fixed $r$ is $({}^{10}C_r)^2$.

    4.  Sum Favorable Cases: Total favorable cases $= \sum_{r=0}^{10} ({}^{10}C_r)^2$.

    5.  Apply Identity: Using the combinatorial identity $\sum_{r=0}^{n} ({}^{n}C_r)^2 = {}^{2n}C_n$, we get ${}^{20}C_{10}$. The probability is $\frac{{}^{20}C_{10}}{2^{20}}$.

   Difficulty Level: Hard

   Concept Name: Combinatorial Identities / Power Sets

   Short cut solution: In a set of size $n$, the number of pairs of subsets with equal size is always ${}^{2n}C_n$. Divide this by the total pairs of subsets $(2^n)^2$ to get the probability.

Question 164

   Question: A number $X$ is chosen at random from the set $\{1, 2, 3, 4, \dots \dots 100\}$. Define the event: $A =$ the chosen number $X$ satisfies $\frac{(x - 10)(x - 50)}{(x - 30)} \geq 0$. Then $P(A)$ is:

   Options: 

       A. 0.71 

       B. 0.70 

       C. 0.51 

       D. 0.20

   Correct Answer: A

   Year: Online April 12, 2014

   Solution (Source): Total value of $X$ between 10 to 30 is 20. Total values of $X$ between 50 to 100 including 50 and 100 is 51. Total values of $x = 51 + 20 = 71$. $P(A) = \frac{71}{100} = 0.71$.

   Step Solution:

    1.  Total Sample Space: There are 100 numbers in the set, so $n(S) = 100$.

    2.  Apply Wavy Curve Method: For the expression $\frac{(x-10)(x-50)}{(x-30)} \geq 0$, the critical points are 10, 30, and 50.

    3.  Determine Intervals: The expression is $\geq 0$ in the intervals $[10, 30)$ and $[50, \infty)$.

    4.  Count Integers in Interval 1: In $[10, 30)$, valid integers are $\{10, 11, \dots, 29\}$. Count $= 29 - 10 + 1 = 20$.

    5.  Count Integers in Interval 2: In $$, valid integers are $\{50, 51, \dots, 100\}$. Count $= 100 - 50 + 1 = 51$. Total favorable $= 20 + 51 = 71$. Probability $= 71/100 = 0.71$.

   Difficulty Level: Medium

   Concept Name: Wavy Curve Method / Algebraic Inequalities

   Short cut solution: Quickly sketch the sign of the fraction on a number line. Pick integers from the positive regions within the bounds.

 Question 165

   Question: A set $S$ contains 7 elements. A non-empty subset $A$ of $S$ and an element $x$ of $S$ are chosen at random. Then the probability that $x \in A$ is:

   Options: 

       A. $1/2$ 

       B. $64/127$ 

       C. $63/128$ 

       D. $31/128$

   Correct Answer: B

   Year: Online April 11, 2014

   Solution (Source): Total no. of non-empty subsets $= 128 - 1 = 127$. Subsets containing $x_i$ have 2 choices for each of the other 6 elements: $2^6 = 64$. Required prob $= \frac{64}{127}$.

   Step Solution:

    1.  Total Subset Choices: A set with 7 elements has $2^7 = 128$ subsets. Excluding the empty set gives 127 choices for $A$.

    2.  Total Element Choices: There are 7 choices for the element $x$ from set $S$. Total sample space pairs $(A, x) = 127 \times 7 = 889$.

    3.  Count Favorable for fixed $x$: For any specific element $x_i$, the number of subsets containing it is $2^{7-1} = 2^6 = 64$.

    4.  Total Favorable Pairs: Since there are 7 different elements that could be chosen as $x$, total favorable pairs $= 7 \times 64$.

    5.  Calculate Probability: $P = \frac{7 \times 64}{7 \times 127} = \frac{64}{127}$.

   Difficulty Level: Medium

   Concept Name: Power Sets / Inclusion Probability

   Short cut solution: The probability an element is in a non-empty subset is simply $\frac{\text{subsets containing } x}{\text{total non-empty subsets}} = \frac{2^{n-1}}{2^n - 1}$. For $n=7$, $P = 64/127$.

 Question 176

   Question: If six students, including two particular students A and B, stand in a row, then the probability that A and B are separated with one student in between them is.

   Options: 

       A. 8/15

       B. 4/15

       C. 2/15

       D. 1/15

   Correct Answer: B

   Year: Online May 19, 2012

   Solution (Source): Choice for a student between A and B is 4. A and B can interchange their places in 2 ways. Trio and remaining 3 students stand in 4! ways. Total ways $= 4 \times 2 \times 4!$. Total unrestricted ways $= 6!$. Probability $= \frac{4 \times 2 \times 4!}{6!} = \frac{4 \times 2}{6 \times 5} = \frac{4}{15}$.

   Step Solution:

    1.  Total Outcomes: The total number of ways to arrange 6 students in a row is $n(S) = 6! = 720$.

    2.  Select Intermediate Student: From the 4 students other than A and B, choose 1 to sit between them ($4$ ways).

    3.  Arrange the Trio: A and B can be on either side of the chosen student (A_B or B_A), providing $2$ ways.

    4.  Arrange Remaining Units: Treat the trio (A, intermediate, B) as a single block. Now arrange this block and the 3 remaining students ($4$ units total) in $4! = 24$ ways.

    5.  Calculate Probability: Total favorable $= 4 \times 2 \times 24 = 192$. $P = \frac{192}{720} = \frac{4}{15}$.

   Difficulty Level: Medium

   Concept Name: Permutations / String Method

   Short cut solution: In a row of $n$ people, the probability of having exactly $k$ people between two specific people is $\frac{2(n-k-1)}{n(n-1)}$. Here $n=6, k=1 \implies \frac{2(6-1-1)}{6 \times 5} = \frac{8}{30} = \frac{4}{15}$.

 Question 177

   Question: A number n is randomly selected from the set {1, 2, 3, ......, 1000}. The probability that $\frac{\sum_{i=1}^n i^2}{\sum_{i=1}^n i}$ is an integer is.

   Options: 

       A. 0.331

       B. 0.333

       C. 0.334

       D. 0.332

   Correct Answer: C

   Year: Online May 12, 2012

   Solution (Source): $\frac{\sum i^2}{\sum i} = \frac{n(n+1)(2n+1)/6}{n(n+1)/2} = \frac{2n+1}{3}$. For $n=1..1000$, values are $3/3, 5/3, 7/3, \dots, 2001/3$. Integral values are 1st term, 4th term, ..., 667th term. Total $= 333 + 1 = 334$. Required probability $= 334/1000 = 0.334$.

   Step Solution:

    1.  Simplify Expression: Use formulas for sum of squares and sum of first $n$ integers: $\frac{[n(n+1)(2n+1)]/6}{[n(n+1)]/2} = \frac{2n+1}{3}$.

    2.  Determine Condition: For the result to be an integer, $2n+1$ must be divisible by 3.

    3.  Apply Modular Arithmetic: $2n+1 \equiv 0 \pmod 3 \implies 2n \equiv -1 \equiv 2 \pmod 3$. This simplifies to $n \equiv 1 \pmod 3$.

    4.  Count Valid Integers: $n$ takes values $1, 4, 7, \dots, 1000$. This is an AP where $1000 = 1 + (k-1)3$.

    5.  Solve for k: $999 = 3(k-1) \implies 333 = k-1 \implies k = 334$. $P = 334/1000 = 0.334$.

   Difficulty Level: Easy

   Concept Name: Sum of Series / Modular Arithmetic

   Short cut solution: The expression $\frac{2n+1}{3}$ is an integer if $n = 3k+1$. In the range 1–1000, exactly one-third of the numbers satisfy such a linear congruence. Since $1000 = 3(333) + 1$, there are $333 + 1 = 334$ values.

 Question 183

   Question: Four numbers are chosen at random (without replacement) from the set {1, 2, 3, ...20}. Statement -1: The probability that the chosen numbers when arranged in some order will form an AP is 1/85. Statement -2: If the four chosen numbers form an AP, then the set of all possible values of common difference is {±1, ±2, ±3, ±4, ±5}.

   Options: 

       A. Statement -1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement -1

       B. Statement - 1 is true, Statment -2 is false

       C. Statement -1 is false, Statment -2 is true.

       D. Statement - 1 is true, Statement -2 is true ; Statement - 2 is a correct explanation for Statement -1.

   Correct Answer: B

   Year: 2010

   Solution (Source): $n(S) = {}^{20}C_4$. For AP: $d=1$ (17 ways), $d=2$ (14 ways), $d=3$ (11 ways), $d=4$ (8 ways), $d=5$ (5 ways), $d=6$ (2 ways). Total favorable $= 17+14+11+8+5+2 = 57$. Probability $= 57/4845 = 1/85$. Statement 2 is false because $d$ can be 6.

   Step Solution:

    1.  Total Sample Space: Choosing 4 numbers from 20 is $n(S) = {}^{20}C_4 = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} = 4845$.

    2.  Count APs with difference $d$: For an AP $\{a, a+d, a+2d, a+3d\}$, we need $a+3d \le 20$. The number of values for $a$ is $20 - 3d$.

    3.  Sum Favorable Cases: $d$ can range from 1 to 6 (since $3d < 20$). Sum $= (20-3) + (20-6) + (20-9) + (20-12) + (20-15) + (20-18) = 17+14+11+8+5+2 = 57$.

    4.  Verify Statement 1: $P = \frac{57}{4845}$. Dividing both by 57 gives $1/85$. Thus, Statement 1 is true.

    5.  Verify Statement 2: We found that $d$ can be 6 (e.g., $\{1, 7, 13, 19\}$). Statement 2 excludes 6, so it is false.

   Difficulty Level: Hard

   Concept Name: Arithmetic Progression / Combinatorial Counting

   Short cut solution: Use the formula for number of APs of length $k$ in $\{1, \dots, n\}$: $\sum_{d=1}^{\lfloor\frac{n-1}{k-1}\rfloor} (n - (k-1)d)$. For $n=20, k=4$: $\sum_{d=1}^{6} (20 - 3d)$. This sum is an arithmetic series: $17 + 14 + \dots + 2$, giving 57. $57/4845 = 1/85$.

Question 184

   Question: An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is.

   Options: 

       A. $2/7$

       B. $1/21$

       C. $2/23$

       D. $1/3$.

   Correct Answer: A.

   Year: 2010.

   Solution (Source): $n(S) = {}^9C_3$; $n(E) = {}^3C_1 \times {}^4C_1 \times {}^2C_1$. Probability $= \frac{3 \times 4 \times 2}{{}^9C_3} = \frac{24 \times 3!}{9!} \times 6! = \frac{24 \times 6}{9 \times 8 \times 7} = \frac{2}{7}$.

   Step Solution:

    1.  Total Outcomes: Calculate the number of ways to choose 3 balls from 9: $n(S) = {}^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.

    2.  Favorable Event: To have different colors, one must pick exactly 1 Red, 1 Blue, and 1 Green ball.

    3.  Count Favorable Ways: $n(E) = {}^3C_1 \times {}^4C_1 \times {}^2C_1 = 3 \times 4 \times 2 = 24$.

    4.  Calculate Probability: $P(E) = \frac{n(E)}{n(S)} = \frac{24}{84}$.

    5.  Simplify: $\frac{24}{84} = \frac{2}{7}$.

   Difficulty Level: Easy.

   Concept Name: Classical Probability / Combinations.

   Short cut solution: Use the formula $\frac{n_1 \cdot n_2 \cdot n_3}{{}^{(n_1+n_2+n_3)}C_3} = \frac{3 \times 4 \times 2}{84} = \frac{2}{7}$.

 Question 192

   Question: Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is.

   Options: 

       A. $2/9$

       B. $1/9$

       C. $8/9$

       D. $7/9$.

   Correct Answer: B.

   Year: 2005.

   Solution (Source): Probability of particular house being selected $= 1/3$. $P(\text{all the persons apply for the same house}) = (1/3 \times 1/3 \times 1/3) \times 3 = 1/9$.

   Step Solution:

    1.  Total Sample Space: Each of the 3 persons has 3 house choices. $n(S) = 3 \times 3 \times 3 = 27$.

    2.  Identify Condition: All three persons must apply for the exact same house.

    3.  Enumerate Favorable Cases: They all pick House 1, all pick House 2, or all pick House 3.

    4.  Count Favorable Ways: Total favorable ways $= 3$.

    5.  Final Probability: $P = \frac{3}{27} = \frac{1}{9}$.

   Difficulty Level: Easy.

   Concept Name: Classical Probability / Multiplication Principle.

   Short cut solution: The first person can pick any house ($P=1$). The second and third must pick that specific house ($1/3$ each). Probability $= 1 \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$.

Question 198

   Question: Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is.

   Options: 

       A. $5/2$ (Typo in source, calculated as $2/5$)

       B. $5/4$

       C. $5/3$

       D. $1/5$.

   Correct Answer: A (Note: The source lists Answer A but calculates the value as $2/5$).

   Year: 2003.

   Solution (Source): Total ways of selecting pair of horses $= {}^5C_2 = 10$. Any horse can win the race in 4 ways. Hence required probability $= \frac{4}{10} = \frac{2}{5}$.

   Step Solution:

    1.  Total Outcomes: Number of ways to select 2 horses from 5 is $n(S) = {}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$.

    2.  Define Success: Success occurs if one of the two chosen horses is the single winner.

    3.  Count Favorable Pairs: If Horse $H_1$ is the winner, the favorable pairs are $(H_1, H_2), (H_1, H_3), (H_1, H_4), (H_1, H_5)$.

    4.  Total Favorable: There are exactly 4 such pairs.

    5.  Calculate Probability: $P = \frac{4}{10} = \frac{2}{5}$.

   Difficulty Level: Easy.

   Concept Name: Combinatorial Probability / Selection.

   Short cut solution: In a set of $n$ items where one is special (the winner), the probability that the special item is in a random subset of size $k$ is simply $k/n$. Here, $2/5 = 0.4$.