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)=33×9×98×9×9+9×25=29797
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:
Held on 25 Feb 2021 · Verified 6 Jul 2026.
51
297122
29797
92
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