Three defective oranges are accidentally mixed with seven good ones and on looking at them, it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If x denote the number of defective oranges, then the variance of x is :

### Related Formula Variance formula for discrete probability distributions: ### Core Logic Construct the discrete probability distribution matrix for drawing 2 items out of 10 total items (3 defective, 7 good): xi pi x=0 7C210C2 = (42/90) x=1 7C1 x 3C110C2 = (42/90) x=2 3C210C2 = (6/90) ### Step 1: Calculating the Mean ### Step 2: Evaluating the Variance Metric ### Pattern Recognition Discrete tables are best managed by computing component factor rows systematically before finalizing variance updates. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability

Three defective oranges are accidentally mixed with seven good ones and on looking at them, it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If x denote

### Related Formula Variance formula for discrete probability distributions: ### Core Logic Construct the discrete probability distribution matrix for drawing 2 items out of 10 total items (3 defective, 7 good): xi pi x=0 7C210C2 = (42/90) x=1 7C1 x 3C110C2 = (42/90) x=2 3C210C2 = (6/90) ### Step 1: Calculating the Mean ### Step 2: Evaluating the Variance Metric ### Pattern Recognition Discrete tables are best managed by computing component factor rows systematically before finalizing variance u

Three defective oranges are accidentally mixed with seven good ones and on looking at them, it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If x denote the number of defective oranges, then the variance of x is :

### Related Formula Variance formula for discrete probability distributions: ### Core Logic Construct the discrete probability distribution matrix for drawing 2 items out of 10 total items (3 defective, 7 good): xi pi x=0 7C210C2 = (42/90) x=1 7C1 x 3C110C2 = (42/90) x=2 3C210C2 = (6/90) ### Step 1: Calculating the Mean ### Step 2: Evaluating the Variance Metric ### Pattern Recognition Discrete tables are best managed by computing component factor rows systematically before finalizing variance updates. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability

Three defective oranges are accidentally mixed with seven good ones and on looking at them, it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If x denote

### Related Formula Variance formula for discrete probability distributions: ### Core Logic Construct the discrete probability distribution matrix for drawing 2 items out of 10 total items (3 defective, 7 good): xi pi x=0 7C210C2 = (42/90) x=1 7C1 x 3C110C2 = (42/90) x=2 3C210C2 = (6/90) ### Step 1: Calculating the Mean ### Step 2: Evaluating the Variance Metric ### Pattern Recognition Discrete tables are best managed by computing component factor rows systematically before finalizing variance u

Three defective oranges are accidentally mixed with seven good ones and on looking at them, it is not possible to…

Year	2026	2025	2024	Total
Questions	9	17	9	35

$xᵢ$ $x_i$	$pᵢ$ $p_i$
$x=0$ $x=0$	$⁷C₂¹⁰C₂ = (42)/(90)$
$x=1$ $x=1$	$⁷C₁ \times ³C₁¹⁰C₂ = (42)/(90)$
$x=2$ $x=2$	$³C₂¹⁰C₂ = (6)/(90)$

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

Q65 jee_main_2025_29_jan_evening Random Variables and Variance

Let

A = [aᵢⱼ]

be a

2 \times 2

matrix such that

aᵢⱼ in 0,1

for all i and j. Let the random variable

X

denote the possible values of the determinant of the matrix

A

. Then, the variance of

X

is:

A. $(1)/(4)$ $\frac{1}{4}$
B. $(3)/(8)$ $\frac{3}{8}$
C. $(5)/(8)$ $\frac{5}{8}$
D. $(3)/(4)$ $\frac{3}{4}$

Solution

Related Formula

Variance of a discrete random variable:

Var(X) = Σ Pᵢ Xᵢ² - (Σ Pᵢ Xᵢ)²

Core Logic

A $2 \times 2$ $2 \times 2$ matrix with binary elements $0,1$ ${0,1}$ has $2⁴ = 16$ $2^4 = 16$ total configurations. The determinant calculation yields outcomes matching values inside structural set $-1, 0, 1$ $\{-1, 0, 1\}$.

Probability distribution grid:

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

Step 1: Compute Variance

Plug components directly into statistical equations:

Var(X) = (3)/(8) - (0)² = (3)/(8)

Pattern Recognition

Symmetry in probability distributions centered across 0 means the expected mean $E[X]$ $\mathbb{E}[X]$ evaluates to zero immediately, saving half your calculation time during variance checks.

Chapter Mix

Class 12 Mathematics: Probability Class 12 Mathematics: Matrices and Determinants

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

Three defective oranges are accidentally mixed with seven good ones and on looking at them, it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If x denote the number of defective oranges, then the variance of x is :

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculating the Mean

Step 2: Evaluating the Variance Metric

Pattern Recognition

Chapter Mix

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: Compute Variance

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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