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…

Year	2026	2025	2024	Total
Questions	9	17	9	35

Q55 jee_main_2025_07_april_evening Bayes Theorem

A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn at random is tossed and head turns up. If the probability that the drawn coin was unbiased, is

mathrmmmathrmn, (mathrmm,n) = 1

, then

n² -m²

is equal to :

A. $80$ $80$
B. $60$ $60$
C. $72$ $72$
D. $64$ $64$

Solution

Related Formula

Bayes' Theorem for conditional probability is formulated as:

P(E₁|H) = (P(E₁) \cdot P(H|E₁))/(P(E₁) \cdot P(H|E₁) + P(E₂) \cdot P(H|E₂))

Core Logic

Let the events be: $E₁$ $E_1$: Selection of an unbiased coin. $E₂$ $E_2$: Selection of the two-headed (biased) coin. $H$ $H$: Head turns up on the toss.

Syllabus values:

P(E₁) = (19)/(20), P(E₂) = (1)/(20)

P(H|E₁) = (1)/(2), P(H|E₂) = 1

Step 1: Total Probability Calculation

The overall probability of obtaining a head is:

P(H) = P(E₁)P(H|E₁) + P(E₂)P(H|E₂)

P(H) = (19)/(20) \cdot (1)/(2) + (1)/(20) \cdot 1 = (19)/(40) + (2)/(40) = (21)/(40)

Step 2: Apply Bayes Theorem

We need the probability that the coin is unbiased given a head showed up:

P(E₁|H) = ((19)/(40))/((21)/(40)) = (19)/(21)

Thus, $(m)/(n) = (19)/(21) m = 19, n = 21$ since $(19, 21) = 1$ $\gcd(19, 21) = 1$.

Step 3: Evaluate final expression

Calculate $n² - m²$ $n^2 - m^2$:

n² - m² = 21² - 19² = 441 - 361 = 80

Pattern Recognition

Bayes' Theorem split problems are easily handled by constructing paths: $unbiased path = 19 \times 1 = 19$ , $biased path = 1 \times 2 = 2$ $\text{biased path} = 1 \times 2 = 2$. Probability = $(19)/(19+2) = (19)/(21)$ $\frac{19}{19+2} = \frac{19}{21}$.

Chapter Mix

Class 12 Mathematics: Probability

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

If A and B are two events such that P(A) = 0.7$P(A) = 0.7$, P(B) = 0.4$P(B) = 0.4$ and P(A B) = 0.5$P(A \cap \overline{B}) = 0.5$, where B$\overline{B}$ denotes the complement of B, then P(B | (A B))$P(B | (A \cup \overline{B}))$ is equal to:

Solution & Explanation

Related Formula

Core Logic

Step 1: Evaluate Component Intersections

Step 2: Calculate Set Union

Step 3: Resolve Final Conditional Probability

Pattern Recognition

Chapter Mix

Reference Study Guides

More Probability Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Total Probability Calculation

Step 2: Apply Bayes Theorem

Step 3: Evaluate final expression

Pattern Recognition

Chapter Mix

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

More Probability Questions — jee_main_2025_08_april_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

If A and B are two events such that $P(A) = 0.7$ $P(A) = 0.7$, $P(B) = 0.4$ $P(B) = 0.4$ and $P(A B) = 0.5$ $P(A \cap \overline{B}) = 0.5$, where $B$ $\overline{B}$ denotes the complement of B, then $P(B | (A B))$ $P(B | (A \cup \overline{B}))$ is equal to: