The probability distribution of a random variable X is given below : X 4k (30/7)k (32/7)k (34/7)k (36/7)k (38/7)k (40/7)k 6k P(X) (2/15) (1/15) (2/15) (1/5) (1/15) (2/15) (1/5) (1/15) If E(X) = (263/15), then P(X < 20) is equal to:

Calculate the expected value: Factor out k and sum the products to solve for k using the given condition E(X) = (263/15). Summing the terms: Given E(X) = (263/15): Substitute k to find the actual values of X: 4k = 14, (30/7)k = 15, (32/7)k = 16, (34/7)k = 17, (36/7)k = 18, (38/7)k = 19, (40/7)k = 20, 6k = 21. We need P(X < 20): When E(X) is provided with a parameterized state variable k, summing the discrete expected value directly unlocks the exact sample space scaling. Class 12 Maths: Probability

The probability distribution of a random variable X is given below : X 4k (30/7)k (32/7)k (34/7)k (36/7)k (38/7)k (40/7)k 6k P(X) (2/15) (1/15) (2/15) (1/5) (1/15) (2/15) (1/5) (1/15) If E(X) = (263/15), then P(X < 20) is equal to:

Calculate the expected value: Factor out k and sum the products to solve for k using the given condition E(X) = (263/15). Summing the terms: Given E(X) = (263/15): Substitute k to find the actual values of X: 4k = 14, (30/7)k = 15, (32/7)k = 16, (34/7)k = 17, (36/7)k = 18, (38/7)k = 19, (40/7)k = 20, 6k = 21. We need P(X < 20): When E(X) is provided with a parameterized state variable k, summing the discrete expected value directly unlocks the exact sample space scaling. Class 12 Maths: Probability

The probability distribution of a random variable X is given below : X 4k (30/7)k (32/7)k (34/7)k (36/7)k (38/7)k…

Q: The probability distribution of a random variable X is given below : X 4k (30/7)k (32/7)k (34/7)k (36/7)k (38/7)k (40/7)k 6k P(X) (2/15) (1/15) (2/15) (1/5) (1/15) (2/15) (1/5) (1/15) If E(X) = (263/1

Calculate the expected value: Factor out k and sum the products to solve for k using the given condition E(X) = (263/15). Summing the terms: Given E(X) = (263/15): Substitute k to find the actual values of X: 4k = 14, (30/7)k = 15, (32/7)k = 16, (34/7)k = 17, (36/7)k = 18, (38/7)k = 19, (40/7)k = 20, 6k = 21. We need P(X < 20): When E(X) is provided with a parameterized state variable k, summing the discrete expected value directly unlocks the exact sample space scaling. Class 12 Maths: Probabil

Year	2026	2025	2024	Total
Questions	9	17	9	35

$X$ $X$	$4k$ $4k$	$(30)/(7)k$ $\frac{30}{7}k$	$(32)/(7)k$ $\frac{32}{7}k$	$(34)/(7)k$ $\frac{34}{7}k$	$(36)/(7)k$ $\frac{36}{7}k$	$(38)/(7)k$ $\frac{38}{7}k$	$(40)/(7)k$ $\frac{40}{7}k$	$6k$ $6k$
$P(X)$ $P(X)$	$(2)/(15)$ $\frac{2}{15}$	$(1)/(15)$ $\frac{1}{15}$	$(2)/(15)$ $\frac{2}{15}$	$(1)/(5)$ $\frac{1}{5}$	$(1)/(15)$ $\frac{1}{15}$	$(2)/(15)$ $\frac{2}{15}$	$(1)/(5)$ $\frac{1}{5}$	$(1)/(15)$ $\frac{1}{15}$

Q jee_main_2025_28_jan_morning Probability Distribution and Variance

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 :

A. 28/75
B. 14/25
C. 26/75
D. 18/25

Solution

Related Formula

Variance formula for discrete probability distributions:

σ² = Σ pᵢ xᵢ² - μ²

Core Logic

Construct the discrete probability distribution matrix for drawing 2 items out of 10 total items (3 defective, 7 good):

Probability Distribution and Variance diagram for Q69 - JEE Main 2025 Morning

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

Step 1: Calculating the Mean

μ = Σ xᵢ pᵢ = 0((42)/(90)) + 1((42)/(90)) + 2((6)/(90)) = (54)/(90) = (3)/(5)

Step 2: Evaluating the Variance Metric

σ² = Σ pᵢ xᵢ² - μ² = [0 + 1²((42)/(90)) + 2²((6)/(90))] - ((3)/(5))²

σ² = (66)/(90) - (9)/(25) = (11)/(15) - (9)/(25) = (28)/(75)

Pattern Recognition

Discrete tables are best managed by computing component factor rows systematically before finalizing variance updates.

Chapter Mix

Class 12 Maths: Probability

Q71 jee_main_2025_03_april_morning Dictionary Rank with Geometrically Weighted Probabilities

All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number n be denoted by

Wₙ

. Let the probability

P(Wₙ)

of choosing the word

Wₙ

satisfy

P(Wₙ)=2P(Wₙ₋₁)

,

n>1

. If

P(CDBEA)= 2α2β-1

, where

α, β in N

, then

α+β

is equal to:

Numerical Answer. Answer: 183 to 183

Solution

Related Formula

Sum of geometric sequence series configuration:

SN = (a(r^N - 1))/(r - 1)

Total counts of permutations of 5 unique letter combinations $= 5! = 120$ $= 5! = 120$.

Core Logic

Set initial base state configuration parameter $P(W₁) = x$ $P(W_1) = x$. Since probabilities escalate geometrically:

Σi=1¹²⁰ P(Wᵢ) = x + 2x + 2²x + + 2¹¹⁹x = 1

x \cdot 2¹²⁰ - 12 - 1 = 1 x = 12¹²⁰ - 1

Step 1: Finding Dictionary Rank of CDBEA

Calculate alphabetical sequential position index values:

Words starting with $A$ $A$: $4! = 24$ $4! = 24$
Words starting with $B$ $B$: $4! = 24$ $4! = 24$
Words starting with $CA$ $CA$: $3! = 6$ $3! = 6$
Words starting with $CB$ $CB$: $3! = 6$ $3! = 6$
Words starting with $CDA$ $CDA$: $2! = 2$ $2! = 2$
Next word alphabetically: $CDBAE$ $CDBAE$ (Rank 63)
Target word: $CDBEA$ $CDBEA$ (Rank 64)

Thus, target word index is exactly $n = 64$ $n = 64$.

Step 2: Probabilistic coefficient calculation

Evaluate the specific targeted weighted entry value:

P(W₆₄) = 2⁶³ \cdot P(W₁) = 2⁶³2¹²⁰-1

Matching structural values with expression flags:

α = 63, β = 120

α + β = 63 + 120 = 183

Pattern Recognition

When probability states map through clean binary power scale jumps, their total summation creates standard Mersenne number forms.

Chapter Mix

Class 11 Mathematics: Permutations and Combinations Class 12 Mathematics: Probability

Q73 jee_main_2025_04_april_evening Bayes Theorem

A card from a pack of 52 cards is lost. From the remaining 51 cards, n cards are drawn and are found to be spades. If the probability of the lost card to be a spade is

(11)/(50)

, then n is equal to

Numerical Answer. Answer: 2 to 2

Solution

Core Logic

Let $E₁$ $E_1$ be the event that the lost card is a spade, and $E₂$ $E_2$ be the event that the lost card is not a spade.

P(E₁) = (13)/(52) = (1)/(4) and P(E₂) = (39)/(52) = (3)/(4)

Let $A$ $A$ be the event that $n$ $n$ cards drawn from the remaining $51$ $51$ cards are all spades.

If the lost card was a spade ( $E₁$ $E_1$), there are $12$ $12$ spades left out of $51$ $51$:

P(A|E₁) = 12n 51n

If the lost card was not a spade ( $E₂$ $E_2$), there are $13$ $13$ spades left out of $51$ $51$:

P(A|E₂) = 13n 51n

Step 1: Applying Bayes Theorem

We are given the posterior probability that the lost card is a spade, $P(E₁|A) = (11)/(50)$ $P(E_1|A) = \frac{11}{50}$:

P(E₁|A) = (P(E₁)P(A|E₁))/(P(E₁)P(A|E₁) + P(E₂)P(A|E₂)) = (11)/(50)

(1)/(4) \cdot 12n 51n(1)/(4) \cdot 12n 51n + (3)/(4) \cdot 13n 51n = (11)/(50)

Canceling out the shared fractions $(1)/(4)$ $\frac{1}{4}$ and $51n$ $\binom{51}{n}$:

12n 12n + 3 13n = (11)/(50)

Step 2: Simplifying Binomial Coefficients

Express $13n$ $\binom{13}{n}$ in terms of $12n$ $\binom{12}{n}$ using the identity $13n = (13)/(13-n) 12n$ :

12n 12n + 3 \cdot [ (13)/(13-n) 12n ] = (11)/(50)

Canceling $12n$ $\binom{12}{n}$ from numerator and denominator:

(1)/(1 + (39)/(13-n)) = (11)/(50) (13-n)/(13-n + 39) = (11)/(50)

(13-n)/(52-n) = (11)/(50)

Cross-multiplying:

50(13 - n) = 11(52 - n)

650 - 50n = 572 - 11n 39n = 78 n = 2

Pattern Recognition

When expanding binomial dynamic ratios like $An A+1n$ $\frac{\binom{A}{n}}{\binom{A+1}{n}}$, always reduce the larger term fractionally using the absorption property to cancel out the factorial variables quickly.

Chapter Mix

Class 12 Mathematics: Probability

Q55 jee_main_2025_04_april_morning Classical Definition of Probability

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:

A. $(129)/(182)$ $\frac{129}{182}$
B. $(103)/(182)$ $\frac{103}{182}$
C. $(17)/(26)$ $\frac{17}{26}$
D. $(19)/(26)$ $\frac{19}{26}$

Solution

Related Formula

Classical Probability:

P(E) = Favorable CasesTotal Cases = (n(E))/(n(S))

Core Logic

Total pool size $= 4 + 2 + 10 = 16$ $= 4 + 2 + 10 = 16$ people. We choose 12. Total cases $n(S) = ¹⁶C₁₂ = 1820$ $n(S) = {}^{16}\mathrm{C}_{12} = 1820$.

List distinct combinations satisfying 'at least 3 engineers' and 'at least 1 doctor':

3 Engineers, 1 Doctor, 8 Professors: $⁴C₃ \times ²C₁ \times ¹⁰C₈ = 4 \times 2 \times 45 = 360$
3 Engineers, 2 Doctors, 7 Professors: $⁴C₃ \times ²C₂ \times ¹⁰C₇ = 4 \times 1 \times 120 = 480$
4 Engineers, 1 Doctor, 7 Professors: $⁴C₄ \times ²C₁ \times ¹⁰C₇ = 1 \times 2 \times 120 = 240$
4 Engineers, 2 Doctors, 6 Professors: $⁴C₄ \times ²C₂ \times ¹⁰C₆ = 1 \times 1 \times 210 = 210$

Step 1: Calculate Final Probability

Summing favorable cases:

n(E) = 360 + 480 + 240 + 210 = 1290

P(E) = (1290)/(1820) = (129)/(182)

Pattern Recognition

When constraints involve 'at least' for multiple groups simultaneously with a large committee size, creating structured exhaustive case distributions is safer and less error-prone than standard bijection/subtraction methods.

Chapter Mix

Class 11 Mathematics: Permutations and Combinations Class 12 Mathematics: Probability

Q69 jee_main_2025_04_april_morning Random Variables and Probability Distributions

A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let X denote the number of defective pens. Then the variance of X is

A. $(11)/(15)$ $\frac{11}{15}$
B. $(28)/(75)$ $\frac{28}{75}$
C. $(2)/(15)$ $\frac{2}{15}$
D. $(3)/(5)$ $\frac{3}{5}$

Solution

Related Formula

Variance formula for discrete random variables:

Var(X) = Σ xᵢ² P(xᵢ) - μ², μ = Σ xᵢ P(xᵢ)

Core Logic

Total selection size is $¹⁰C₂ = 45$ ${}^{10}\mathrm{C}_2 = 45$. $X$ $X$ can take values 0, 1, 2. Set up probability distribution table:

$xᵢ$ $x_i$	$x = 0$ $x = 0$	$x = 1$ $x = 1$	$x = 2$ $x = 2$
$P(xᵢ)$ $P(x_i)$	$⁷C₂¹⁰C₂ = (21)/(45) = (7)/(15)$	$⁷C₁ \times ³C₁¹⁰C₂ = (21)/(45) = (7)/(15)$	$³C₂¹⁰C₂ = (3)/(45) = (1)/(15)$

Step 1: Calculate Mean

μ = 0((7)/(15)) + 1((7)/(15)) + 2((1)/(15)) = (9)/(15) = (3)/(5)

Step 2: Calculate Variance

Σ xᵢ² P(xᵢ) = 0²((7)/(15)) + 1²((7)/(15)) + 2²((1)/(15)) = (7 + 4)/(15) = (11)/(15)

Var(X) = (11)/(15) - ((3)/(5))² = (11)/(15) - (9)/(25) = (55 - 27)/(75) = (28)/(75)

Pattern Recognition

This setup maps identically to a Hypergeometric Distribution. Checking fractions against a common denominator ( $15$ $15$ or $45$ $45$) helps avoid simple fractional reduction errors.

Chapter Mix

Class 12 Mathematics: Probability

Solution & Explanation

Related Formula

Core Logic

Execution

Pattern Recognition

Chapter Mix

Reference Study Guides

More Probability Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Calculating the Mean

Step 2: Evaluating the Variance Metric

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding Dictionary Rank of CDBEA

Step 2: Probabilistic coefficient calculation

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Applying Bayes Theorem

Step 2: Simplifying Binomial Coefficients

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Final Probability

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Mean

Step 2: Calculate Variance

Pattern Recognition

Chapter Mix

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