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
Solution
<p>$i^{k_1}+i^{k_2} \neq 0 \quad i^{k_1} \rightarrow 4$ option for $\mathrm{i},-1,-\mathrm{i}, 1$</p>
<p>Total cases $\Rightarrow 4 \times 4=16$</p>
<p>Unfovourble cases $\Rightarrow \mathrm{i}^{\mathrm{k}_1}+\mathrm{i}^{\mathrm{k}_2}=0$</p>
<p>$$\left\{\begin{array}{c}
1,-1 \\
-1,1 \\
i,-i \\
-i, i
\end{array}\right\}$$</p>
<p>4 Cases $\Rightarrow$ Probability $=\frac{16-4}{16}=\frac{3}{4}$</p>
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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