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JEE Main 202529 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Probability

JEE Main 2025 · 29 January · Evening Shift · Mathematics · Probability

Let A=[a_ i j ] be a 2 × 2 matrix such that a_ i j ∈\ 0,1\ for all...

Let A=[aij]bea\mathrm{A}=\left[\mathrm{a}_{i j}\right] be a2 \times 2matrixsuchthatmatrix such that\mathrm{a}_{i j} \in\{0,1\}forallfor alli and jj. Let the random variable X denote the possible values of the determinant of the matrix AA. Then, the variance of XX is :

Options

  1. A34\frac{3}{4}
  2. B58\frac{5}{8}
  3. C38\frac{3}{8}
  4. D14\frac{1}{4}

Correct answer

C. 38\frac{3}{8}

Step-by-step solution

x01−1P(x)1016316316\begin{array}{|c|c|c|c|} \hline x & 0 & 1 & -1 \\ \hline P(x) & \frac{10}{16} & \frac{3}{16} & \frac{3}{16} \\ \hline \end{array} Var⁡(x)=E(x2)−[E(x)]2=∑i=13xi2P(xi)−(μ)2=1×316+1×316[μ=0]=616=38\begin{aligned} & \operatorname{Var}(x)=E \left(x^2\right)-[E(x)]^2 \\ & =\sum_{i=1}^3 x_i^2 P \left(x_i\right)-(\mu)^2 \\ & =1 \times \frac{3}{16}+1 \times \frac{3}{16} \quad[\mu=0] \\ & =\frac{6}{16}=\frac{3}{8}\end{aligned}$
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