JEE Mains · Maths · STD 11 - 14. probability
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
- A \(\frac{1}{2}\)
- B \(\frac{3}{4}\)
- C \(\frac{1}{4}\)
- D \(\frac{2}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{3}{4}\)
Step-by-step Solution
Detailed explanation
Let \(k_1=4 \lambda_1+r_1, r_1 \in\{0,1,2,3\}\) \(k_2=4 \lambda_2+r_2\) \(\begin{aligned} & (i)^{k_1}+(i)^{k_2}=(i)^{r_1}+(i)^{r_2} \\ & (i)^{r_1} \in\{1, i,-1,-i\} \end{aligned}\)…
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