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JEE Main 202528 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Probability

JEE Main 2025 · 28 January · Morning Shift · Mathematics · Probability

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

Two number k1\mathrm{k}_1 and k2\mathrm{k}_2 are randomly chosen from the set of natural numbers. Then, the probability that the value of ik1+ik2,(i=−1)\mathrm{i}^{\mathrm{k}_1}+\mathrm{i}^{\mathrm{k}_2},(\mathrm{i}=\sqrt{-1}) is non-zero, equals

Options

  1. A12\frac{1}{2}
  2. B34\frac{3}{4}
  3. C14\frac{1}{4}
  4. D23\frac{2}{3}

Correct answer

B. 34\frac{3}{4}

Step-by-step solution

Let k1=4λ1+r1,r1∈{0,1,2,3}k2=4λ2+r2k_1=4 \lambda_1+r_1, r_1 \in\{0,1,2,3\} k_2=4 \lambda_2+r_2(i)k1+(i)k2=(i)r1+(i)r2(i)r1∈{1,i,−1,−i}\begin{aligned} & (i)^{k_1}+(i)^{k_2}=(i)^{r_1}+(i)^{r_2} \\ & (i)^{r_1} \in\{1, i,-1,-i\} \end{aligned}\text{ Zero } \Rightarrow 1,(-1) \text{ pair } \quad \Rightarrow\left\{1,−1i,−i−i,+i−1,1\begin{array}{cc} 1, & -1 \\ i, & -i \\ -i,+i \\ -1, & 1 \end{array}\right\} i,(-i)pairpair Zero probability =44C1⋅4C1=14 Probability (non-zero) =1−14=34\begin{aligned} &\text{ Zero probability }=\frac{4}{{ }^4 C_1 \cdot{ }^4 C_1}=\frac{1}{4} \\ &\text{ Probability (non-zero) }=1-\frac{1}{4}=\frac{3}{4}\end{aligned}$
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