Let $A=[a_ij]$ $A=[a_{ij}]$ 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: [cite: 3338, 3339]

A. $frac\58$ $\frac{\5}{8}$
B. $frac\316$ $\frac{\3}{16}$
C. $frac\18$ $\frac{\1}{8}$
D. $frac\38$ $\frac{\3}{8}$

Solution & Explanation

### Related Formula A

2 times 2

matrix

A = beginpmatrix a & b \\ c & d endpmatrix

is invertible if and only if its determinant is non-zero:

det(A) = ad - bc neq 0

### Step 1: Count Total Matrix Sample Space Each of the 4 entry slots in the

2 times 2

matrix has 2 binary choices (0 or 1) :

textTotal Matrices = 2^4 = 16

### Step 2: Count Favorable Non-Zero Determinant Matrices Since elements are 0 or 1, the products

ad

and

bc

can only evaluate to 0 or 1. For

ad - bc neq 0

, we have two distinct cases [cite: 4051, 4054]: - **Case I:**

ad = 1

and

bc = 0

ad = 1 Rightarrow a = 1, d = 1

(1 configuration).

bc = 0 Rightarrow (b, c) in \(0,0), (0,1), (1,0)\

(3 configurations).

textWays = 1 times 3 = 3 text matrices

- **Case II:**

ad = 0

and

bc = 1

bc = 1 Rightarrow b = 1, c = 1

(1 configuration).

ad = 0 Rightarrow (a, d) in \(0,0), (0,1), (1,0)\

(3 configurations).

textWays = 1 times 3 = 3 text matrices

textTotal Favorable Matrices = 3 + 3 = 6

### Step 3: Calculate Probability Divide the favorable count by the total sample size :

P(E) = frac616 = frac38

### Pattern Recognition For low-order matrix configuration spaces with binary inputs, directly analyzing the product outcomes (

1-0=1

0-1=-1

) prevents long manual lists of all 16 matrices. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Probability Class 12 Mathematics: Matrices and Determinants

Reference Study Guides

Probability

More Probability Previous-Year Questions

Q60 2025 Bayes' Theorem

Given three identical bags each containing 10 balls, whose colours are as follows:

beginarray|l|l|l|l| hline & textbfRed & textbfBlue & textbfGreen \\ hline textbfBag I & 3 & 2 & 5 \\ hline textbfBag II & 4 & 3 & 3 \\ hline textbfBag III & 5 & 1 & 4 \\ hline endarray

A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from bag I is

p

and if the ball is Green, the probability that it is from bag III is

q

, then the value of

left(frac1p + frac1qright)

is:

A. 6
B. 9
C. 7
D. 8

Solution

### Related Formula

textBayes' Theorem: P(E_1|A) = fracP(E_1) P(A|E_1)sum_i=1^n P(E_i) P(A|E_i)

### Core Logic This is a conditional probability problem. We apply Bayes' Theorem twice: first for the Red ball, then for the Green ball. ### Step 1: Solve for p (Red ball) Let

E_1, E_2, E_3

be the events of choosing Bag I, Bag II, and Bag III respectively. Since bags are identical,

P(E_1) = P(E_2) = P(E_3) = frac13

. The probabilities of drawing a Red ball from each bag are: -

P(R|E_1) = frac310

P(R|E_2) = frac410

P(R|E_3) = frac510

Applying Bayes' Theorem:

p = P(E_1|R) = fracP(E_1) P(R|E_1)P(E_1)P(R|E_1) + P(E_2)P(R|E_2) + P(E_3)P(R|E_3)

p = fracfrac310frac310 + frac410 + frac510 = frac312 = frac14

Thus,

frac1p = 4

. ### Step 2: Solve for q (Green ball) The probabilities of drawing a Green ball from each bag are: -

P(G|E_1) = frac510

P(G|E_2) = frac310

P(G|E_3) = frac410

Applying Bayes' Theorem:

q = P(E_3|G) = fracP(E_3) P(G|E_3)P(E_1)P(G|E_1) + P(E_2)P(G|E_2) + P(E_3)P(G|E_3)

q = fracfrac410frac510 + frac310 + frac410 = frac412 = frac13

Thus,

frac1q = 3

. ### Step 3: Calculate the requested value Sum the inverse values:

frac1p + frac1q = 4 + 3 = 7

### Pattern Recognition Simplification of Bayes' denominator: Since all prior events have identical probability

P(E_i) = 1/k

, they cancel out of the Bayes' fraction entirely, allowing you to work directly with the raw ball counts. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Probability

Q74 2025 Classical Definition of Probability

Three distinct numbers are selected randomly from the set

\1, 2, 3, dots, 40\

. If the probability that the selected numbers are in an increasing geometric progression is

fracmn

where

textgcd(m, n) = 1

, then

m + n

is equal to ________.

Numerical Answer. Answer: 4949 to 4949

Solution

### Related Formula Classical Probability equation:

P = fractextNumber of Favorable OutcomestextTotal Outcomes in Sample Space

### Core Logic Calculate total outcomes via combinations

binom403

. Count the number of valid

3

-term geometric progressions

a, ar, ar^2 le 40

based on official integer common ratio assumptions. ### Step 1: Count Total Sample Space Outcomes

textTotal Outcomes = binom403 = frac40 times 39 times 383 times 2 times 1 = 9880

### Step 2: Count Favorable GP Sets (Integer Ratios) Let the elements be

a, ar, ar^2 le 40

. * If

r = 2 implies 4a le 40 implies a in \1, 2, dots, 10\ rightarrow 10 text progressions.

* If

r = 3 implies 9a le 40 implies a in \1, 2, 3, 4\ rightarrow 4 text progressions.

* If

r = 4 implies 16a le 40 implies a in \1, 2\ rightarrow 2 text progressions.

* If

r = 5 implies 25a le 40 implies a = 1 rightarrow 1 text progression.

* If

r = 6 implies 36a le 40 implies a = 1 rightarrow 1 text progression.

Sum of integer ratio progressions =

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 = frac189880 = frac94940 = fracmn

Since

textgcd(9, 4940) = 1

m + n = 9 + 4940 = 4949

### Pattern Recognition The question assumes integer common ratios (

r in mathbbN

) according to the primary NTA verification engine, drastically narrowing down the manual search space for valid bounding values. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Probability Class 11 Mathematics: Sequences and Series

Q54 2025 Conditional Probability

If A and B are two events such that

P(A) = 0.7

P(B) = 0.4

and

P(A cap overlineB) = 0.5

, where

overlineB

denotes the complement of B, then

P(B | (A cup overlineB))

is equal to:

A. $frac14$ $\frac{1}{4}$
B. $frac12$ $\frac{1}{2}$
C. $frac16$ $\frac{1}{6}$
D. $frac13$ $\frac{1}{3}$

Solution

### Related Formula

P(X|Y) = fracP(X cap Y)P(Y)

P(A cap overlineB) = P(A) - P(A cap B)

### Core Logic Utilize set probability laws to derive component values like the intersection

P(A cap B)

and basic union forms to simplify conditional constraints. ### Step 1: Evaluate Component Intersections Given

P(A cap overlineB) = 0.5

and

P(A) = 0.7

P(A cap overlineB) = P(A) - P(A cap B) implies 0.5 = 0.7 - P(A cap B)

P(A cap B) = 0.2

### Step 2: Calculate Set Union Compute the total area of the conditional domain set:

P(A cup overlineB) = P(A) + P(overlineB) - P(A cap overlineB)

P(A cup overlineB) = 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 cap (A cup overlineB)) = P((B cap A) cup (B cap overlineB)) = P(A cap B) + 0 = 0.2

P(B | (A cup overlineB)) = fracP(A cap B)P(A cup overlineB) = frac0.20.8 = frac14

### Pattern Recognition In conditional sets containing expressions like

X cap (Y cup overlineX)

, the disjoint nature of

X cap overlineX

means it collapses quickly to standard overlap intersections

X cap Y

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Probability

Q65 2025 Random Variables and Variance

Let

mathrmA = [mathrma_mathrmij]

be a

2 times 2

matrix such that

mathrma_mathrmij 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. $frac14$ $\frac{1}{4}$
B. $frac38$ $\frac{3}{8}$
C. $frac58$ $\frac{5}{8}$
D. $frac34$ $\frac{3}{4}$

Solution

### Related Formula Variance of a discrete random variable:

textVar(X) = sum P_i X_i^2 - left(sum P_i X_iright)^2

### Core Logic A

2 times 2

matrix with binary elements

0,1

has

2^4 = 16

total configurations. The determinant calculation yields outcomes matching values inside structural set

\-1, 0, 1\

. Probability distribution grid:

$X_i$ $X_i$	$P_i$ $P_i$	$P_i X_i$ $P_i X_i$	$P_i X_i^2$ $P_i X_i^2$
$-1$ $-1$	$frac316$ $\frac{3}{16}$	$-frac316$ $-\frac{3}{16}$	$frac316$ $\frac{3}{16}$
$0$ $0$	$frac1016$ $\frac{10}{16}$	$0$ $0$	$0$ $0$
$1$ $1$	$frac316$ $\frac{3}{16}$	$frac316$ $\frac{3}{16}$	$frac316$ $\frac{3}{16}$
Total	$1$ $1$	$sum P_i X_i = 0$ $\sum P_i X_i = 0$	$sum P_i X_i^2 = frac38$ $\sum P_i X_i^2 = \frac{3}{8}$

### Step 1: Compute Variance Plug components directly into statistical equations:

textVar(X) = frac38 - (0)^2 = frac38

### Pattern Recognition Symmetry in probability distributions centered across 0 means the expected mean

mathbbE[X]

evaluates to zero immediately, saving half your calculation time during variance checks. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Probability Class 12 Mathematics: Matrices and Determinants

Q66 2025 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|B_1)P(B_1) + P(W|B_2)P(B_2)

### Core Logic Bag 1 contents:

\4W, 5B\

(Total 9 balls). Bag 2 contents initially:

\nW, 3B\

(Total

n+3

balls). Case 1: Transferred ball is white (

P = frac49

): Bag 2 now has

(n+1)W

and

3B

(Total

n+4

). Probability of drawing white

= fracn+1n+4

. Case 2: Transferred ball is black (

P = frac59

): Bag 2 now has

nW

and

4B

(Total

n+4

). Probability of drawing white

= fracnn+4

. ### Step 1: Set up Equation and Solve Aggregate components via Total Probability Formula:

left(frac49 times fracn+1n+4right) + left(frac59 times fracnn+4right) = frac2945

frac4(n+1) + 5n9(n+4) = frac2945

frac9n + 4n+4 = frac295

5(9n + 4) = 29(n + 4)

45n + 20 = 29n + 116

16n = 96 implies n = 6

### Pattern Recognition Notice how the denominators inside conditional stages match up identically (

n+4

). Clear constants before running fraction line conversions to speed up single-variable systems. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Probability

We Map Every Repeating Question in Competitive Exams.

Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.

Select Your Target Exam

Choose an exam track below to find formulas per chapter and patterns.

Syncing Exam Intelligence

Mapping formulas and patterns across all tracks…

PATH A — FULL LENGTH PRACTICE

Full Mock Test Hub

Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.

Under Development

PATH B — TARGETED PRACTICE

Topic-wise Practice Hub

Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.

Loading Questions... Browse Topics

Latest from the Blog

View all →

Loading articles...

Review & Tools

Mistakes To Revise

No conceptual mistakes detected.

🔥 Practice Session Live

Showing All Questions

Exam Questions

Test your concepts live. Choose options or enter numerical values, then verify your answer to reveal the double-box solution matrix.

Let $A=[a_ij]$ $A=[a_{ij}]$ 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: [cite: 3338, 3339]

Solution & Explanation

Reference Study Guides

More Probability Previous-Year Questions

Solution

Solution

Solution

Solution

Solution

More Probability Questions — jee_main_2025_24_jan_evening

We Map Every Repeating Question in Competitive Exams.

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Let A=[a_ij]$A=[a_{ij}]$ 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: [cite: 3338, 3339]

Solution & Explanation

Reference Study Guides

More Probability Previous-Year Questions

Solution

Solution

Solution

Solution

Solution

More Probability Questions — jee_main_2025_24_jan_evening

We Map Every Repeating Question in Competitive Exams.

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Let $A=[a_ij]$ $A=[a_{ij}]$ 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: [cite: 3338, 3339]