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

$\begin{array}{rcl} & & \boxed{ 7 \overline{) \overline{) \overline{) }}}} 9\times 9\times 9\\ & & \boxed{ 7 \overline{) \overline{) \overline{) }}}} 8\times 9\times 9\times 3\\ \mathrm{n}\left(\mathrm{S}\right)& =& 9^2\times (9+24)=33\times 9^2\end{array}$

E=number divided by 5 leaves remainder 2

$\begin{array}{c}\boxed{ 7 \overline{) \overline{) \overline{) 2 }}}} 9\times 9\\ \boxed{ \overline{) 7 \overline{) \overline{) 2 }}}} 8\times 9\times 9\end{array} \overline{) \begin{array}{c}\boxed{ \overline{) \overline{) \overline{) 7 }}}} 8\times 9\times 9\\ \end{array} }$

$\begin{array}{rcl}\mathrm{n}\left(\mathrm{E}\right)& =& 9^2+916+9\times 72\\ & =& 9\times (9+16+72)\\ \mathrm{P}\left(\mathrm{E}\right)& =& \frac{97}{33\times 9}=\frac{97}{297}\end{array}$