Question ID: #430
Two numbers $k_1$ and $k_2$ are randomly chosen from the set of natural numbers. Then, the probability that the value of $i^{k_1} + i^{k_2}$, $(i = \sqrt{-1})$ is non-zero, equals
- (1) $\frac{1}{2}$
- (2) $\frac{1}{4}$
- (3) $\frac{3}{4}$
- (4) $\frac{2}{3}$
Solution:
Let $S$ be the sample space for values of $i^k$.
The value of $i^k$ repeats in a cycle of 4: $\{1, i, -1, -i\}$.
Total possible outcomes for pair $(k_1, k_2)$ in terms of remainders:
$$ n(S) = 4 \times 4 = 16 $$
For the sum $i^{k_1} + i^{k_2}$ to be zero, the terms must cancel out.
The favorable pairs $(i^{k_1}, i^{k_2})$ for sum $= 0$ are:
$$ (1, -1), (-1, 1), (i, -i), (-i, i) $$
$$ \text{Number of cases where sum is zero} = 4 $$
Probability that sum is zero:
$$ P(\text{Sum} = 0) = \frac{4}{16} = \frac{1}{4} $$
Probability that sum is non-zero:
$$ P(\text{Sum} \neq 0) = 1 – P(\text{Sum} = 0) $$
$$ = 1 – \frac{1}{4} = \frac{3}{4} $$
Ans. (3)
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