A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn at random is tossed and head…

Year	2026	2025	2024	Total
Questions	9	17	9	35

More Probability Previous-Year Questions — Page 5

Q56 jee_main_2025_07_april_evening Random Variables and Expectation

Let a random variable

X

take values 0, 1, 2, 3 with

P(X = 0) = P(X = 1) = p

P(X = 2) = P(X = 3)

and

E(X²) = 2E(X)

. Then the value of

8p - 1

is:

A. $0$ $0$
B. $2$ $2$
C. $1$ $1$
D. $3$ $3$

Solution

Related Formula

The sum of all probabilities in a probability distribution is strictly equal to $1$ $1$:

Σ P(Xᵢ) = 1

Core Logic

Let $P(X=2) = P(X=3) = q$ $P(X=2) = P(X=3) = q$. From the total probability rule:

P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1

p + p + q + q = 1 2p + 2q = 1 p + q = (1)/(2) (i)

Step 1: Computing Expectations

Compute $E(X)$ $E(X)$:

E(X) = 0\cdot p + 1\cdot p + 2\cdot q + 3\cdot q = p + 5q

Compute $E(X²)$ $E(X^2)$:

E(X²) = 0²\cdot p + 1²\cdot p + 2²\cdot q + 3²\cdot q = p + 13q

Step 2: Solve the Linear System

Given $E(X²) = 2E(X)$ $E(X^2) = 2E(X)$:

p + 13q = 2(p + 5q)

p + 13q = 2p + 10q p = 3q

Substitute $p = 3q$ $p = 3q$ into equation (i):

3q + q = (1)/(2) 4q = (1)/(2) q = (1)/(8)

Then, $p = 3((1)/(8)) = (3)/(8)$ .

Step 3: Calculate 8p - 1

Now we evaluate the required expression:

8p - 1 = 8((3)/(8)) - 1 = 3 - 1 = 2

Pattern Recognition

Always combine basic distribution axioms (sum of probabilities = 1) with structural definition equations ( $E(X) = Σ x P(x)$ $E(X) = \sum x P(x)$) to systematically eliminate unknown parameters.

Chapter Mix

Class 12 Mathematics: Probability

Q69 jee_main_2025_24_jan_evening Probability of Invertible Matrices

Let

A=[aᵢⱼ]

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:

A. $(5)/(8)$ $\frac{5}{8}$
B. $(3)/(16)$ $\frac{3}{16}$
C. $(1)/(8)$ $\frac{1}{8}$
D. $(3)/(8)$ $\frac{3}{8}$

Solution

Related Formula

A $2 \times 2$ $2 \times 2$ matrix $A = pmatrix a & b c & d pmatrix$ is invertible if and only if its determinant is non-zero:

(A) = ad - bc \neq 0

Step 1: Count Total Matrix Sample Space

Each of the 4 entry slots in the $2 \times 2$ $2 \times 2$ matrix has 2 binary choices (0 or 1) :

Total Matrices = 2⁴ = 16

Step 2: Count Favorable Non-Zero Determinant Matrices

Since elements are 0 or 1, the products $ad$ $ad$ and $bc$ $bc$ can only evaluate to 0 or 1. For $ad - bc \neq 0$ $ad - bc \neq 0$, we have two distinct cases:

Case I: $ad = 1$ $ad = 1$ and $bc = 0$ $bc = 0$ .

$ad = 1 \Rightarrow a = 1, d = 1$ $ad = 1 \Rightarrow a = 1, d = 1$ (1 configuration). $bc = 0 \Rightarrow (b, c) in (0,0), (0,1), (1,0)$ (3 configurations).

Ways = 1 \times 3 = 3 matrices

Case II: $ad = 0$ $ad = 0$ and $bc = 1$ $bc = 1$ .

$bc = 1 \Rightarrow b = 1, c = 1$ $bc = 1 \Rightarrow b = 1, c = 1$ (1 configuration). $ad = 0 \Rightarrow (a, d) in (0,0), (0,1), (1,0)$ (3 configurations).

Ways = 1 \times 3 = 3 matrices

Total Favorable Matrices = 3 + 3 = 6

Step 3: Calculate Probability

Divide the favorable count by the total sample size :

P(E) = (6)/(16) = (3)/(8)

Pattern Recognition

For low-order matrix configuration spaces with binary inputs, directly analyzing the product outcomes ( $1-0=1$ $1-0=1$ or $0-1=-1$ $0-1=-1$) prevents long manual lists of all 16 matrices.

Chapter Mix

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

Q61 jee_main_2025_24_jan_morning Infinite Geometric Series in Probability

A and B alternately throw a pair of dice. A wins if he throws a sum of

5

before B throws a sum of

8

, and B wins if he throws a sum of

8

before A throws a sum of

5

. The probability, that A wins if A makes the first throw, is :

A. $(9)/(17)$ $\frac{9}{17}$
B. $(9)/(19)$ $\frac{9}{19}$
C. $(8)/(17)$ $\frac{8}{17}$
D. $(8)/(19)$ $\frac{8}{19}$

Solution

Related Formula

For alternating multi-stage games continuing indefinitely, the total probability of winning is modeled as an infinite geometric series summation:

S\infty = (a)/(1 - r)

Core Logic

First, analyze the total sample outcomes for a pair of standard dice ( $n(S) = 36$ $n(S) = 36$):

Outcomes giving a sum of 5: $(1,4), (2,3), (3,2), (4,1) 4$ $(1,4), (2,3), (3,2), (4,1) \implies 4$ outcomes.

p(A) = (4)/(36) = (1)/(9) p(A') = 1 - (1)/(9) = (8)/(9)

Outcomes giving a sum of 8: $(2,6), (3,5), (4,4), (5,3), (6,2) 5$ outcomes.

p(B) = (5)/(36) p(B') = 1 - (5)/(36) = (31)/(36)

Step 1: Setup Infinite Game Series Path

For player $A$ $A$ to win on the first throw, third throw, fifth throw, etc., the probability sequence expands as follows:

P(A wins) = p(A) + p(A') \cdot p(B') \cdot p(A) + [p(A') \cdot p(B')]² \cdot p(A) + \infty

This is a geometric progression where the first term $a = p(A) = (1)/(9)$ $a = p(A) = \frac{1}{9}$ and the common ratio is:

r = p(A') \cdot p(B') = (8)/(9) \cdot (31)/(36) = (62)/(81)

Step 2: Calculate Invariant Sum

Apply the infinite GP sum formula directly:

P(A wins) = ((1)/(9))/(1 - (62)/(81)) = ((1)/(9))/((19)/(81)) = (1)/(9) \cdot (81)/(19) = (9)/(19)

Pattern Recognition

In infinite alternating games, player A's net probability can always be abbreviated shortcut-style to $(p₁)/(1 - (1-p₁)(1-p₂))$ $\frac{p_1}{1 - (1-p_1)(1-p_2)}$, where $p₁$ $p_1$ is A's success probability and $p₂$ $p_2$ is B's success probability.

Chapter Mix

Class 12 Mathematics: Probability

Q51 jee_main_2025_28_jan_evening Bayes Theorem

Bag

B₁

contains 6 white and 4 blue balls, Bag

B₂

contains 4 white and 6 blue balls, and Bag

B₃

contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability, that the ball is drawn from Bag

B₂

, is:

A. $(1)/(3)$ $\frac{1}{3}$
B. $(4)/(15)$ $\frac{4}{15}$
C. $(2)/(3)$ $\frac{2}{3}$
D. $(2)/(5)$ $\frac{2}{5}$

Solution

Related Formula

Bayes' Theorem formula:

P(E₂|A) = (P(E₂) \cdot P(A|E₂))/(P(E₁) \cdot P(A|E₁) + P(E₂) \cdot P(A|E₂) + P(E₃) \cdot P(A|E₃))

Core Logic

Let the events be defined as: $E₁$ $E_1$: Bag $B₁$ $B_1$ is selected $E₂$ $E_2$: Bag $B₂$ $B_2$ is selected $E₃$ $E_3$: Bag $B₃$ $B_3$ is selected A: The drawn ball is white

Since a bag is selected at random:

P(E₁) = P(E₂) = P(E₃) = (1)/(3)

Conditional probabilities of drawing a white ball from each bag:

P(A|E₁) = (6)/(10), P(A|E₂) = (4)/(10), P(A|E₃) = (5)/(10)

Step 1: Substitute and Calculate

Substituting the values into Bayes' theorem:

P(E₂|A) = ((1)/(3) \times (4)/(10))/((1)/(3) \times (6)/(10) + (1)/(3) \times (4)/(10) + (1)/(3) \times (5)/(10))

P(E₂|A) = (4)/(6 + 4 + 5) = (4)/(15)

Pattern Recognition

When bags have equal selection probability, the required conditional probability is simply the number of favorable white balls divided by the total number of white balls across all bags: $4 / (6 + 4 + 5) = 4/15$ $4 / (6 + 4 + 5) = 4/15$.

Chapter Mix

Class 12 Mathematics: Probability

Solution & Explanation

Related Formula

Core Logic

Step 1: Total Probability Calculation

Step 2: Apply Bayes Theorem

Step 3: Evaluate final expression

Pattern Recognition

Chapter Mix

Reference Study Guides

More Probability Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Computing Expectations

Step 2: Solve the Linear System

Step 3: Calculate 8p - 1

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Count Total Matrix Sample Space

Step 2: Count Favorable Non-Zero Determinant Matrices

Step 3: Calculate Probability

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Setup Infinite Game Series Path

Step 2: Calculate Invariant Sum

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Substitute and Calculate

Pattern Recognition

Chapter Mix

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