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

The number of four digit numbers having 7 as one digit in that four

If the number begins with 7, then the number of four-digit numbers is $1\times 9\times 9\times 9$

The number of four-digit numbers, in which 7 is not in the first place and 7 must be one of the digits in the remaining three digits, is$8\times 3_{C_1}\times 9\times 9$

Hence the total number four-digit numbers having 7 as one of the digits is$n\left(s\right)=9\times 9\times 9+24\times 9\times 9=81\times \left(33\right)=2673$

Among all the above numbers, find the number of numbers whose unit digit is either 2 or 7

The number of four-digit numbers having 2 as unit digit and 7 as first digit is 81, the number of four digit numbers having 7 in unit digit is $8\times 9\times 9=648$

And the number of four-digit numbers having 2 as unit digit and 7 at any place other than first digit place is $8\times 2C_1\times 9=144$

The number of four-digit numbers whose unit digit is either 2 or 7 is$n\left(A\right)=81+648+144=873$

Therefore, the required probability is $\frac{n\left(A\right)}{n\left(S\right)}=\frac{873}{2673}=\boxed{\frac{97}{297}}$