Mathematics · Probability

JEE Main 2025 — 28 January, Morning Shift — Question 3

Two number $\mathrm{k}_{1}$ and $\mathrm{k}_{2}$ are randomly chosen from the set of natural numbers. Then, the probability that the value of $\mathrm{i}^{\mathrm{k}_{1}}+\mathrm{i}^{\mathrm{k}_{2}},(\mathrm{i}=\sqrt{-1})$ is non-zero, equals

Option A:
$\frac{1}{2}$
Option B:
$\frac{1}{4}$
Option C:
$\frac{3}{4}$
Correct
Option D:
$\frac{2}{3}$

Answer: C

Step-by-step solution

$i^{k_{1}}+i^{k_{2}} \neq 0 \quad i^{k_{1}} \rightarrow 4$ option for $\mathrm{i},-1,-i, 1$

Total cases $\Rightarrow 4 \times 4=16$

Unfovourble cases $\Rightarrow \mathrm{i}^{\mathrm{k}_{1}}+\mathrm{i}^{\mathrm{k}_{2}}=0$

$\left\{\begin{array}{l}1,-1 -1,1 i,-i -i, i\end{array}\right\}$

4 Cases $\Rightarrow$ Probability $=\frac{16-4}{16}=\frac{3}{4}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 28 January, Morning Shift
Subject: Mathematics
Chapter: Probability
Topic: Problems based on P & C

Three defective oranges are accidently mixed with seven good ones and on looking at them, it is not possible to differentiate between them.…

Answer: C

Step-by-step solution

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