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:
Step-by-Step Solution
Key Concept: $i^{k}$ depends only on $k\bmod 4$. Each of $\{1,i,-1,-i\}$ occurs with probability $\tfrac{1}{4}.$ Sum is zero iff $\{i^{k_{1}},i^{k_{2}}\}=\{1,-1\}$ or $\{i,-i\}.$
$P(\text{sum}=0)$: ordered pairs $(1,-1),(-1,1),(i,-i),(-i,i)$ each with probability $\dfrac{1}{4}\cdot\dfrac{1}{4}=\dfrac{1}{16}.$
Total: $\dfrac{4}{16}=\dfrac{1}{4}.$
$P(\text{non-zero})=1-\dfrac{1}{4}=\dfrac{3}{4}.$
Correct Answer: 2