Two number k_1 and k_2 are randomly chosen from the set of natural numbers.

Question

Two number ${k}_{1}$ 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

Accepted Answer

$i^{k_1}+i^{k_2} \neq 0 \quad i^{k_1} \rightarrow 4$ option for $\mathrm{i},-1,-\mathrm{i}, 1$

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

Unfovourble cases $\Rightarrow \mathrm{i}^{\mathrm{k}_1}+\mathrm{i}^{\mathrm{k}_2}=0$

$$\left\{\[\begin{gathered} 1,-1 \ -1,1 \ i,-i \ -i, i \end{gathered}$$\right\}\]

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

Solutions · 1