Verified Solution • 2025–26 Session moderate

QJEEM21Mafbc9: JEE Main 2021 Mathematics - Class 12 JEEM21Mafbc9 Solution: Let A be a set of all 4-digit natural numbers whose exactly one …

Let A be a set of all 4-digit natural numbers whose exactly one digit is 7. Then the probability that a randomly chosen element of A leaves remainder 2 when divided by 5 is :

Multiple Choice Options

  • A 29
  • B 15
  • C 122297
  • D 97297

Step-by-Step Verified Solution

n(s) = n(when 7 appears on thousands place)

+ n (7 does not appear on thousands place)

= 9 × 9 × 9 + 8 × 9 × 9 × 3

= 33 × 9 × 9

n(E) = n(last digit 7 & 7 appears once)

+n(last digit 2 when 7 appears once)

= 8 × 9 × 9 + (3 × 9 × 9 - 2 × 9)

∴ P(E)=8×9×9+9×2533×9×9=97297

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