Let A be a set of all 4 -digit natural numbers whose exactly one digit is 7 . Then the probability that a…

$n\left(s\right)=n$( when $7$ appears on thousands place ) $+n$($7$ does not appear on thousands place)

$=9\times 9\times 9+8\times 9\times 9\times 3$

$=33\times 9\times 9$

$n\left(E\right)=n$( last digit $7 \& 7$ appears once)$+n$( last digit $2$ when $7$ appears once)

$=8\times 9\times 9+\left(9\times 9+8\times 9\times 2\right)$

$\therefore P\left(E\right)=\frac{8\times 9\times 9+9\times 25}{33\times 9\times 9}=\frac{97}{297}$