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Probability question

2021 · 27 Jul · Shift 1 · Q37
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  5. /2021 · 27 Jul · Shift 1 · Q37

Probability question

2021 · 27 Jul · Shift 1 · Q37

JEE MainMathematicsProbabilityMCQ+4 / −1
The probability that a randomly selected 2-digit number belongs to the set {n ∈\in∈ N : (2n −-− 2) is a multiple of 3} is equal to :
  1. A
    16{1 \over 6}61​
  2. B
    23{2 \over 3}32​
  3. C
    12{1 \over 2}21​
  4. D
    13{1 \over 3}31​
View written solutionFree

Correct answer: D

  1. Total number of 2-digit numbers

A 2-digit number can be any integer from 101010 to 999999.

So, total count is 99−10+1=90.99-10+1=90.99−10+1=90.

  1. Condition for the number nnn

We need (2n−2)(2n-2)(2n−2) to be a multiple of 333.

That means 2n−2≡0(mod3).2n-2 \equiv 0 \pmod{3}.2n−2≡0(mod3).

Simplify: 2n≡2(mod3).2n \equiv 2 \pmod{3}.2n≡2(mod3).

Since 2−1≡2(mod3)2^{-1} \equiv 2 \pmod{3}2−1≡2(mod3) (because 2⋅2=4≡1(mod3)2\cdot 2=4\equiv 1 \pmod{3}2⋅2=4≡1(mod3)), multiply both sides by 222: n≡1(mod3).n \equiv 1 \pmod{3}.n≡1(mod3).

So we need 2-digit numbers congruent to 111 modulo 333.

  1. Count 2-digit numbers of the form 3k+13k+13k+1

The 2-digit numbers satisfying n≡1(mod3)n\equiv 1\pmod{3}n≡1(mod3) are: 10,13,16,…,97.10,13,16,\dots,97.10,13,16,…,97.

This is an arithmetic progression with:

  • first term a=10a=10a=10
  • last term l=97l=97l=97
  • common difference d=3d=3d=3

Number of terms: 97−103+1=873+1=29+1=30.\frac{97-10}{3}+1=\frac{87}{3}+1=29+1=30.397−10​+1=387​+1=29+1=30.

  1. Compute the probability

P=favourable outcomestotal outcomes=3090=13.P=\frac{\text{favourable outcomes}}{\text{total outcomes}}=\frac{30}{90}=\frac{1}{3}.P=total outcomesfavourable outcomes​=9030​=31​.

  1. Compare with options

Thus the correct option is D: 13.\boxed{\text{D: }\frac{1}{3}}.D: 31​​.

  1. Comparison with stored answer

Stored correct answer is C: 12\frac{1}{2}21​, but our derived answer is D: 13\frac{1}{3}31​.

Hence, the stored answer appears to be incorrect.

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