Let A = [aij] be a 2 x 2 matrix such that aij 0,1 for all i and j. Let the random variable X denote the possible values…

Year	2026	2025	2024	Total
Questions	9	17	9	35

$Xᵢ$ $X_i$	$Pᵢ$ $P_i$	$Pᵢ Xᵢ$ $P_i X_i$	$Pᵢ Xᵢ²$ $P_i X_i^2$
$-1$ $-1$	$(3)/(16)$ $\frac{3}{16}$	$-(3)/(16)$ $-\frac{3}{16}$	$(3)/(16)$ $\frac{3}{16}$
$0$ $0$	$(10)/(16)$ $\frac{10}{16}$	$0$ $0$	$0$ $0$
$1$ $1$	$(3)/(16)$ $\frac{3}{16}$	$(3)/(16)$ $\frac{3}{16}$	$(3)/(16)$ $\frac{3}{16}$
Total	$1$ $1$	$Σ Pᵢ Xᵢ = 0$ $\sum P_i X_i = 0$	$Σ Pᵢ Xᵢ² = (3)/(8)$ $\sum P_i X_i^2 = \frac{3}{8}$

More Probability Previous-Year Questions — Page 3

Q74 jee_main_2025_02_april_morning Classical Definition of Probability

Three distinct numbers are selected randomly from the set

1, 2, 3, , 40

. If the probability that the selected numbers are in an increasing geometric progression is

(m)/(n)

where

gcd(m, n) = 1

, then

m + n

is equal to ________.

Numerical Answer. Answer: 4949 to 4949

Solution

Related Formula

Classical Probability equation:

P = Number of Favorable OutcomesTotal Outcomes in Sample Space

Core Logic

Calculate total outcomes via combinations $403$ $\binom{40}{3}$. Count the number of valid $3$ $3$-term geometric progressions $a, ar, ar² \leq 40$ $a, ar, ar^2 \le 40$ based on official integer common ratio assumptions.

Step 1: Count Total Sample Space Outcomes

Total Outcomes = 403 = (40 \times 39 \times 38)/(3 \times 2 \times 1) = 9880

Step 2: Count Favorable GP Sets (Integer Ratios)

Let the elements be $a, ar, ar² \leq 40$ $a, ar, ar^2 \le 40$.

If $r = 2 4a \leq 40 a in 1, 2, , 10 arrow 10 progressions.$
If $r = 3 9a \leq 40 a in 1, 2, 3, 4 arrow 4 progressions.$
If $r = 4 16a \leq 40 a in 1, 2 arrow 2 progressions.$
If $r = 5 25a \leq 40 a = 1 arrow 1 progression.$
If $r = 6 36a \leq 40 a = 1 arrow 1 progression.$

Sum of integer ratio progressions = $10 + 4 + 2 + 1 + 1 = 18$ $10 + 4 + 2 + 1 + 1 = 18$.

Step 3: Final Fraction Evaluation (NTA Answer Keys)

Following the official NTA answer calculation criteria based exclusively on integer ratios:

P = (18)/(9880) = (9)/(4940) = (m)/(n)

Since $gcd(9, 4940) = 1$ $\text{gcd}(9, 4940) = 1$:

m + n = 9 + 4940 = 4949

Pattern Recognition

The question assumes integer common ratios ( $r in N$ $r \in \mathbb{N}$) according to the primary NTA verification engine, drastically narrowing down the manual search space for valid bounding values.

Chapter Mix

Class 12 Mathematics: Probability Class 11 Mathematics: Sequences and Series

Q54 jee_main_2025_08_april_evening Conditional Probability

If A and B are two events such that

P(A) = 0.7

P(B) = 0.4

and

P(A B) = 0.5

, where

B

denotes the complement of B, then

P(B | (A B))

is equal to:

A. $(1)/(4)$ $\frac{1}{4}$
B. $(1)/(2)$ $\frac{1}{2}$
C. $(1)/(6)$ $\frac{1}{6}$
D. $(1)/(3)$ $\frac{1}{3}$

Solution

Related Formula

P(X|Y) = (P(X Y))/(P(Y))

P(A B) = P(A) - P(A B)

Core Logic

Utilize set probability laws to derive component values like the intersection $P(A B)$ $P(A \cap B)$ and basic union forms to simplify conditional constraints.

Step 1: Evaluate Component Intersections

Given $P(A B) = 0.5$ $P(A \cap \overline{B}) = 0.5$ and $P(A) = 0.7$ $P(A) = 0.7$:

P(A B) = P(A) - P(A B) 0.5 = 0.7 - P(A B)

P(A B) = 0.2

Step 2: Calculate Set Union

Compute the total area of the conditional domain set:

P(A B) = P(A) + P( B) - P(A B)

P(A B) = 0.7 + (1 - 0.4) - 0.5 = 0.7 + 0.6 - 0.5 = 0.8

Step 3: Resolve Final Conditional Probability

Using distribution laws on intersection fields:

P(B (A B)) = P((B A) (B B)) = P(A B) + 0 = 0.2

P(B | (A B)) = P(A B)P(A B) = (0.2)/(0.8) = (1)/(4)

Pattern Recognition

In conditional sets containing expressions like $X (Y X)$ $X \cap (Y \cup \overline{X})$, the disjoint nature of $X X$ $X \cap \overline{X}$ means it collapses quickly to standard overlap intersections $X Y$ $X \cap Y$.

Chapter Mix

Class 12 Mathematics: Probability

Q66 jee_main_2025_29_jan_evening Total Probability Theorem

Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability, that the ball drawn is white, is 29/45, then n is equal to:

A. $3$ $3$
B. $4$ $4$
C. $5$ $5$
D. $6$ $6$

Solution

Related Formula

Total Probability Law:

P(W) = P(W|B₁)P(B₁) + P(W|B₂)P(B₂)

Core Logic

Bag 1 contents: $4W, 5B$ $\{4W, 5B\}$ (Total 9 balls). Bag 2 contents initially: $nW, 3B$ $\{nW, 3B\}$ (Total $n+3$ $n+3$ balls).

Case 1: Transferred ball is white ( $P = (4)/(9)$ $P = \frac{4}{9}$): Bag 2 now has $(n+1)W$ $(n+1)W$ and $3B$ $3B$ (Total $n+4$ $n+4$). Probability of drawing white $= (n+1)/(n+4)$ $= \frac{n+1}{n+4}$.

Case 2: Transferred ball is black ( $P = (5)/(9)$ $P = \frac{5}{9}$): Bag 2 now has $nW$ $nW$ and $4B$ $4B$ (Total $n+4$ $n+4$). Probability of drawing white $= (n)/(n+4)$ $= \frac{n}{n+4}$.

Step 1: Set up Equation and Solve

Aggregate components via Total Probability Formula:

((4)/(9) \times (n+1)/(n+4)) + ((5)/(9) \times (n)/(n+4)) = (29)/(45)

(4(n+1) + 5n)/(9(n+4)) = (29)/(45)

(9n + 4)/(n+4) = (29)/(5)

5(9n + 4) = 29(n + 4)

45n + 20 = 29n + 116

16n = 96 n = 6

Pattern Recognition

Notice how the denominators inside conditional stages match up identically ( $n+4$ $n+4$). Clear constants before running fraction line conversions to speed up single-variable systems.

Chapter Mix

Class 12 Mathematics: Probability

Q jee_main_2025_28_jan_morning Classical Probability and Complex Powers

Two number

k₁

and

k₂

are randomly chosen from the set of natural numbers. Then, the probability that the value of

ik₁ + ik₂

(i = \sqrt(-1))

is non-zero, equals

A. $(1)/(2)$ $\frac{1}{2}$
B. $(1)/(4)$ $\frac{1}{4}$
C. $(3)/(4)$ $\frac{3}{4}$
D. $(2)/(3)$ $\frac{2}{3}$

Solution

Related Formula

Probability of non-zero outcome:

P(E) = 1 - P(E^ )

where $E^$ $E^\prime$ represents the condition $ik₁ + ik₂ = 0$ $i^{k_1} + i^{k_2} = 0$.

Core Logic

Powers of $i$ $i$ repeat periodically every 4 cycles: $i, -1, -i, 1$ $\{i, -1, -i, 1\}$. Total possible outcomes for the pair $(ik₁, ik₂)$ $(i^{k_1}, i^{k_2})$ are $4 \times 4 = 16$ $4 \times 4 = 16$ cases.

Step 1: Finding Unfavorable Cases

The sum is zero ( $ik₁ + ik₂ = 0$ $i^{k_1} + i^{k_2} = 0$) when the values are additive inverses:

$(1, -1)$ $(1, -1)$
$(-1, 1)$ $(-1, 1)$
$(i, -i)$ $(i, -i)$
$(-i, i)$ $(-i, i)$

This yields exactly 4 unfavorable cases.

Step 2: Computing Probability

Probability = (16 - 4)/(16) = (12)/(16) = (3)/(4)

Pattern Recognition

Whenever modular cycles exist (like exponents of $i$ $i$ mod 4), compress infinite natural choices safely into a single cycle map.

Chapter Mix

Class 11 Maths: Complex Numbers Class 12 Maths: Probability

Let A = [aᵢⱼ]$\mathrm{A} = [\mathrm{a}_{\mathrm{ij}}]$ be a 2 × 2$2 \times 2$ matrix such that aᵢⱼ in 0,1$\mathrm{a}_{\mathrm{ij}} \in \{0,1\}$ for all i and j. Let the random variable X$X$ denote the possible values of the determinant of the matrix A$A$. Then, the variance of X$X$ is:

Solution & Explanation

Related Formula

Core Logic

Step 1: Compute Variance

Pattern Recognition

Chapter Mix

Reference Study Guides

More Probability Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Count Total Sample Space Outcomes

Step 2: Count Favorable GP Sets (Integer Ratios)

Step 3: Final Fraction Evaluation (NTA Answer Keys)

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluate Component Intersections

Step 2: Calculate Set Union

Step 3: Resolve Final Conditional Probability

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Set up Equation and Solve

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding Unfavorable Cases

Step 2: Computing Probability

Pattern Recognition

Chapter Mix

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Let $A = [aᵢⱼ]$ be a $2 \times 2$ $2 \times 2$ matrix such that $aᵢⱼ in 0,1$ $\mathrm{a}_{\mathrm{ij}} \in \{0,1\}$ for all i and j. Let the random variable $X$ $X$ denote the possible values of the determinant of the matrix $A$ $A$. Then, the variance of $X$ $X$ is: