JEE Main 2025 Jan 28 Shift 1, Mathematics Q3: Introduction To Probability

JEE Main2025Jan 28, Shift 1Mathematics

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

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

Solution

The powers of $i$ cycle with period 4, so $i^{k}$ takes only the four values ${1, i, - 1, - i}$ . Equating with sample space: each of $k_{1}, k_{2}$ contributes one of 4 residue classes, giving $4 \times 4 = 16$ equally likely outcomes.

The sum $i^{k_{1}} + i^{k_{2}}$ equals zero exactly when the two values are negatives of each other. The unfavourable pairs are: $(1, - 1), (- 1, 1), (i, - i), (- i, i)$ . That is 4 outcomes.

Favourable count $= 16 - 4 = 12$ . Probability $= \frac{12}{16} = \frac{3}{4}$ .

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JEE Main 2025 Jan 28 Shift 1, Mathematics Q3: Introduction To Probability

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