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Probability question

2025 · 28 Jan · Shift 1 · Q35
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  5. /2025 · 28 Jan · Shift 1 · Q35

Probability question

2025 · 28 Jan · Shift 1 · Q35

JEE MainMathematicsProbabilityMCQ+4 / −1
Two number k1\mathrm{k}_1k1​ and k2\mathrm{k}_2k2​ 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})ik1​+ik2​,(i=−1​) is non-zero, equals
  1. A
    34\frac{3}{4}43​
  2. B
    12\frac{1}{2}21​
  3. C
    14\frac{1}{4}41​
  4. D
    23\frac{2}{3}32​
View written solutionFree

Correct answer: A

  1. Understand the cycle of powers of iii

    The powers of iii repeat every 444:

    i1=i,i2=−1,i3=−i,i4=1,i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1,i1=i,i2=−1,i3=−i,i4=1,

    and then the pattern repeats.

    So the value of iki^kik depends only on k(mod4)k \pmod 4k(mod4).

  2. When is ik1+ik2=0i^{k_1}+i^{k_2}=0ik1​+ik2​=0?

    For the sum to be zero, we need the two terms to be additive inverses:

    ik2=−ik1.i^{k_2}=-i^{k_1}.ik2​=−ik1​.

    From the cycle {1,i,−1,−i}\{1,i,-1,-i\}{1,i,−1,−i}, each value is negated by shifting exponent by 222 modulo 444:

    ik+2=−ik.i^{k+2}=-i^k.ik+2=−ik.

    Hence,

    ik1+ik2=0  ⟺  k1−k2≡2(mod4)i^{k_1}+i^{k_2}=0 \iff k_1-k_2 \equiv 2 \pmod 4ik1​+ik2​=0⟺k1​−k2​≡2(mod4)

    (equivalently, k1k_1k1​ and k2k_2k2​ differ by 222 modulo 444).

  3. Find the probability of this happening

    A randomly chosen natural number is equally likely to be in any residue class modulo 444:

    1,2,3,0(mod4).1,2,3,0 \pmod 4.1,2,3,0(mod4).

    So for each of k1k_1k1​ and k2k_2k2​, there are 444 equally likely residue classes.

    Total possible residue pairs:

    4×4=16.4\times 4 =16.4×4=16.

    Favorable pairs for zero sum satisfy difference 222 modulo 444:

    (1,3),(2,0),(3,1),(0,2).(1,3),(2,0),(3,1),(0,2).(1,3),(2,0),(3,1),(0,2).

    So there are 444 such pairs.

    Therefore,

    P(ik1+ik2=0)=416=14.P\big(i^{k_1}+i^{k_2}=0\big)=\frac{4}{16}=\frac14.P(ik1​+ik2​=0)=164​=41​.
  4. Required probability: non-zero sum

    P(ik1+ik2≠0)=1−14=34.P\big(i^{k_1}+i^{k_2}\neq 0\big)=1-\frac14=\frac34.P(ik1​+ik2​=0)=1−41​=43​.
  5. Check options

    • A: 34\frac3443​ ✅
    • B: 12\frac1221​
    • C: 14\frac1441​
    • D: 23\frac2332​

    Therefore, the correct option is A.

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