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:

Step-by-step 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 + 9 × 9 + 8 × 9 × 2 ∴ P E = 8 × 9 × 9 + 9 × 25 33 × 9 × 9 = 97 297