JEE Main 2025 Jan 29 Shift 2, Mathematics Q15: Introduction To Probability

JEE Main2025Jan 29, Shift 2Mathematics

Let $A = [a_{ij}]$ be a $2 \times 2$ matrix such that $a_{ij} \in {0, 1}$ 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
$\frac{1}{4}$
B
$\frac{3}{8}$
C
$\frac{5}{8}$
D
$\frac{3}{4}$

Solution

There are $2^{4} = 16$ equally likely matrices. The determinant is $a_{11} a_{22} - a_{12} a_{21}$ , which can only take values $- 1, 0, 1$ .

$X = 1$ : requires $a_{11} = a_{22} = 1$ and $a_{12} a_{21} = 0$ . With both diagonal entries fixed, the pair $(a_{12}, a_{21})$ has $3$ admissible values $(0, 0), (0, 1), (1, 0)$ . Count $= 3$ .

$X = - 1$ : by symmetry $a_{12} = a_{21} = 1$ and $a_{11} a_{22} = 0$ . Count $= 3$ .

$X = 0$ : remaining $16 - 6 = 10$ matrices.

$P (X = 1) = P (X = - 1) = \frac{3}{16}$ , $P (X = 0) = \frac{10}{16}$ .

$E [X] = \frac{1 \cdot 3 + ( - 1 ) \cdot 3}{16} = 0$ , $E [X^{2}] = \frac{1 \cdot 3 + 1 \cdot 3}{16} = \frac{6}{16} = \frac{3}{8}$ .

$∴ Var (X) = E [X^{2}] - (E [X])^{2} = \frac{3}{8}$ .

Concept behind this question

Introduction To ProbabilityNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Jan 29 Shift 2, Mathematics Q15: Introduction To Probability

Concept behind this question

More previous year questions on Introduction To Probability

Practice more Mathematics