All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial…

Year	2026	2025	2024	Total
Questions	9	17	9	35

Year

2026

2025

2024

Total

Questions

Solution & Explanation

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 $P(W₁) = x$ $P(W_1) = x$ [cite: 1431]. Since probabilities escalate geometrically [cite: 1433]: $Σi=1¹²⁰ P(Wᵢ) = x + 2x + 2²x + + 2¹¹⁹x = 1$ [cite: 1432, 1433] $x \cdot 2¹²⁰ - 12 - 1 = 1 x = 12¹²⁰ - 1$ [cite: 1434]

Step 1: Finding Dictionary Rank of CDBEA

Calculate alphabetical sequential position index values[cite: 1437]:

Words starting with $A$ $A$: $4! = 24$ $4! = 24$ [cite: 1438]
Words starting with $B$ $B$: $4! = 24$ $4! = 24$ [cite: 1438]
Words starting with $CA$ $CA$: $3! = 6$ $3! = 6$ [cite: 1438]
Words starting with $CB$ $CB$: $3! = 6$ $3! = 6$ [cite: 1439]
Words starting with $CDA$ $CDA$: $2! = 2$ $2! = 2$ [cite: 1439]
Next word alphabetically: $CDBAE$ $CDBAE$ (Rank 63) [cite: 1439]
Target word: $CDBEA$ $CDBEA$ (Rank 64) [cite: 1440]

Thus, target word index is exactly $n = 64$ $n = 64$[cite: 1442].

Step 2: Probabilistic coefficient calculation

Evaluate the specific targeted weighted entry value [cite: 1442]: $P(W₆₄) = 2⁶³ \cdot P(W₁) = 2⁶³2¹²⁰-1$ [cite: 1442, 1443]

Matching structural values with expression flags [cite: 1443]: $α = 63, β = 120$ $$\alpha = 63, \quad \beta = 120$$ [cite: 1443] $α + β = 63 + 120 = 183$ $$\alpha + \beta = 63 + 120 = 183$$ [cite: 1443]

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

More Probability Previous-Year Questions — Page 7

Q jee_main_2024_30_january_evening Bayes Theorem

Bag A contains

3

white,

7

red balls and bag B contains

3

white,

2

red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn is white, is :

A. $(1)/(4)$ $\frac{1}{4}$
B. $(1)/(9)$ $\frac{1}{9}$
C. $(1)/(3)$ $\frac{1}{3}$
D. $(3)/(10)$ $\frac{3}{10}$

Solution

Related Formula

Bayes' Theorem: P(E₁ | E) = (P(E₁)P(E | E₁))/(P(E₁)P(E | E₁) + P(E₂)P(E | E₂))

Core Logic

Let $E₁$ $E_1$ be the event that Bag A is selected, and $E₂$ $E_2$ be the event that Bag B is selected.

P(E₁) = P(E₂) = (1)/(2)

Let $E$ $E$ be the event that a white ball is drawn. From Bag A (3 white, 7 red, total 10): $P(E | E₁) = (3)/(10)$ $P(E | E_1) = \frac{3}{10}$ From Bag B (3 white, 2 red, total 5): $P(E | E₂) = (3)/(5)$ $P(E | E_2) = \frac{3}{5}$

Step 1: Calculating the Target Probability

We need to find the probability that the ball was drawn from Bag A given it is white, i.e., $P(E₁ | E)$ $P(E_1 | E)$.

P(E₁ | E) = ((1)/(2) \times (3)/(10))/((1)/(2) \times (3)/(10) + (1)/(2) \times (3)/(5))

Canceling out $(1)/(2)$ $\frac{1}{2}$ from the numerator and the denominator:

P(E₁ | E) = ((3)/(10))/((3)/(10) + (6)/(10)) = (3)/(3 + 6) = (3)/(9) = (1)/(3)

Pattern Recognition

Reverse probability with disjoint prior states directly signals Bayes' theorem. Canceling prior probability terms ( $P(E₁)=P(E₂)$ $P(E_1)=P(E_2)$) speeds up the calculation.

Chapter Mix

Class 12 Maths: Probability

Q11 jee_main_2024_30_jan_morning Classical Probability

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:

A. $(30)/(121)$ $\frac{30}{121}$
B. $(62)/(121)$ $\frac{62}{121}$
C. $(60)/(121)$ $\frac{60}{121}$
D. $(31)/(121)$ $\frac{31}{121}$

Solution

Related Formula

P(E) = Number of favorable outcomesTotal number of possible outcomes

Core Logic

Total possible selections for $(x, y)$ $(x, y)$ with replacement from $0, 1, , 10$ ${0, 1, \dots, 10}$ is $11 \times 11 = 121$ $11 \times 11 = 121$. We need pairs $(x, y)$ $(x, y)$ such that $|x - y| > 5$ $|x - y| > 5$, which means $x - y > 5$ $x - y > 5$ or $y - x > 5$ $y - x > 5$.

Step 1: Counting Favorable Cases

Assume $x < y$ $x < y$, so we need $y - x \geq 6$ $y - x \geq 6$. If $x = 0 \Rightarrow y in 6, 7, 8, 9, 10$ (5 ways) If $x = 1 \Rightarrow y in 7, 8, 9, 10$ (4 ways) If $x = 2 \Rightarrow y in 8, 9, 10$ $x = 2 \Rightarrow y \in \{8, 9, 10\}$ (3 ways) If $x = 3 \Rightarrow y in 9, 10$ $x = 3 \Rightarrow y \in \{9, 10\}$ (2 ways) If $x = 4 \Rightarrow y in 10$ $x = 4 \Rightarrow y \in \{10\}$ (1 way) If $x \geq 5$ $x \ge 5$, there are no possible values for $y$ $y$ strictly greater than $x$ $x$ satisfying the condition.

Step 2: Total Probability

The number of cases for $y > x$ $y > x$ is $5 + 4 + 3 + 2 + 1 = 15$ $5 + 4 + 3 + 2 + 1 = 15$. By symmetry, the number of cases for $x > y$ $x > y$ is also $15$ $15$. Total favorable cases $= 15 \times 2 = 30$ $= 15 \times 2 = 30$. Required probability $= (30)/(121)$ $= \frac{30}{121}$.

Pattern Recognition

Absolute difference conditions $|x - y| > k$ $|x - y| > k$ on discrete sets cleanly split into symmetric additive series $1+2+...+n$ $1+2+...+n$. Calculate one half and multiply by 2.

Chapter Mix

Class 12 Maths: Probability

Q18 jee_main_2024_31_jan_evening Binomial Distribution / Independent Events

A coin is based so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is-

A. $(2)/(9)$ $\frac{2}{9}$
B. $(1)/(9)$ $\frac{1}{9}$
C. $(2)/(27)$ $\frac{2}{27}$
D. $(1)/(27)$ $\frac{1}{27}$

Solution

Related Formula

P(X=k) = ⁿCk \cdot p^k \cdot qn-k

Core Logic

Given $P(H) = 2P(T)$ $P(H) = 2P(T)$. Since $P(H) + P(T) = 1$ $P(H) + P(T) = 1$, we get $2P(T) + P(T) = 1 3P(T) = 1 P(T) = (1)/(3)$ . Then, $P(H) = (2)/(3)$ $P(H) = \frac{2}{3}$.

The coin is tossed 3 times. We need the probability of getting exactly 2 tails and 1 head. Using binomial probability:

P(2T, 1H) = ³C₂ \times (P(T))² \times (P(H))¹

= 3 \times ((1)/(3))² \times ((2)/(3))

= 3 \times (1)/(9) \times (2)/(3) = (6)/(27) = (2)/(9)

Chapter Mix

Class 12 Maths: Probability

Q16 jee_main_2024_31_jan_morning Independent Events

Two marbles are drawn in succession from a box containing 10 red, 30 white, 20 blue and 15 orange marbles, with replacement being made after each drawing. Then the probability, that first drawn marble is red and second drawn marble is white, is

A. $(2)/(25)$ $$\frac{2}{25}$$
B. $(4)/(25)$ $$\frac{4}{25}$$
C. $(2)/(3)$ $$\frac{2}{3}$$
D. $(4)/(75)$ $$\frac{4}{75}$$

Solution

Core Logic

Total marbles $= 10 + 30 + 20 + 15 = 75$ $= 10 + 30 + 20 + 15 = 75$. Drawings are made with replacement, so the events are independent.

Step 1: Probability Calculation

Probability of first drawing a red marble: $P(R) = (10)/(75)$ $P(R) = \frac{10}{75}$. Probability of second drawing a white marble: $P(W) = (30)/(75)$ $P(W) = \frac{30}{75}$. Since they are independent: $P(R and W) = (10)/(75) \times (30)/(75) = (4)/(75)$ .

Chapter Mix

Class 12 Maths: Probability

Q19 jee_main_2024_31_jan_morning Variance of Random Variable

Three rotten apples are accidently mixed with fifteen good apples. Assuming the random variable

X

to be the number of rotten apples in a draw of two apples, the variance of

X

A. $(37)/(153)$ $$\frac{37}{153}$$
B. $(57)/(153)$ $$\frac{57}{153}$$
C. $(47)/(153)$ $$\frac{47}{153}$$
D. $(40)/(153)$ $$\frac{40}{153}$$

Solution

Core Logic

Total apples = 18 (3 rotten, 15 good). Random variable $X = 0, 1, 2$ $X = \{0, 1, 2\}$ representing the number of rotten apples.

Step 1: Probability Distribution

P(X = 0) = ¹⁵C₂¹⁸C₂ = (105)/(153)

P(X = 1) = ³C₁ \times ¹⁵C₁¹⁸C₂ = (45)/(153)

P(X = 2) = ³C₂¹⁸C₂ = (3)/(153)

Step 2: Expectation

E(X) = 0 \times (105)/(153) + 1 \times (45)/(153) + 2 \times (3)/(153) = (51)/(153) = (1)/(3)

Step 3: Variance

E(X²) = 0 \times (105)/(153) + 1 \times (45)/(153) + 4 \times (3)/(153) = (57)/(153)

Var(X) = E(X²) - (E(X))² = (57)/(153) - ((1)/(3))² = (57)/(153) - (17)/(153) = (40)/(153)

Chapter Mix

Class 12 Maths: Probability

Solution & Explanation

Related Formula

Core Logic

Step 1: Finding Dictionary Rank of CDBEA

Step 2: Probabilistic coefficient calculation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Probability Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 1: Calculating the Target Probability

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Counting Favorable Cases

Step 2: Total Probability

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Core Logic

Step 1: Probability Calculation

Chapter Mix

Solution

Core Logic

Step 1: Probability Distribution

Step 2: Expectation

Step 3: Variance

Chapter Mix

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