JEE Main 2021 — Probability Question with Solution
From: JEE Main 2021 (Online) 25th February Evening Shift
Question
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 :
Choose an option
Show full solutionCorrect option: D
Correct answer
D
Step-by-step explanation
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)
+ 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)
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This is a previous-year question from JEE Main 2021, covering the Probability chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.