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:

We need to find the set $$A$$ of all 4-digit natural numbers with exactly one digit equal to 7. A 4-digit number has digits $$d_1 d_2 d_3 d_4$$ where $$d_1 \in \{1, 2, \ldots, 9\}$$ and $$d_2, d_3, d_4 \in \{0, 1, \ldots, 9\}$$. Exactly one of these digits is 7.

We count $$|A|$$ by considering which position has the digit 7. If $$d_1 = 7$$: the remaining 3 digits are chosen from $$\{0,1,\ldots,9\} \setminus \{7\}$$, giving $$9^3 = 729$$ numbers. If $$d_2 = 7$$ (or $$d_3 = 7$$ or $$d_4 = 7$$): $$d_1$$ is from $$\{1,\ldots,9\}\setminus\{7\} = 8$$ choices, and the other two non-7 positions each have 9 choices. So each gives $$8 \times 9^2 = 648$$ numbers.

Thus $$|A| = 729 + 3 \times 648 = 729 + 1944 = 2673$$.

For the number to leave remainder 2 when divided by 5, the last digit $$d_4$$ must be 2 or 7 (since $$2 \equiv 2 \pmod{5}$$ and $$7 \equiv 2 \pmod{5}$$).

Case 1: $$d_4 = 7$$ (so the 7 is in the units place). Then $$d_1 \in \{1,\ldots,9\}\setminus\{7\}$$ (8 choices), $$d_2, d_3 \in \{0,\ldots,9\}\setminus\{7\}$$ (9 choices each). This gives $$8 \times 9 \times 9 = 648$$ numbers.

Case 2: $$d_4 = 2$$ (so the 7 is in one of the other three positions). Sub-case (a): $$d_1 = 7$$, then $$d_2, d_3 \in \{0,\ldots,9\}\setminus\{7\}$$ giving $$9 \times 9 = 81$$ numbers. Sub-case (b): $$d_2 = 7$$, then $$d_1 \in \{1,\ldots,9\}\setminus\{7\}$$ (8 choices), $$d_3 \in \{0,\ldots,9\}\setminus\{7\}$$ (9 choices), giving $$8 \times 9 = 72$$. Sub-case (c): $$d_3 = 7$$, similarly gives $$8 \times 9 = 72$$.

So Case 2 total = $$81 + 72 + 72 = 225$$.

Total favorable = $$648 + 225 = 873$$.

The probability = $$\frac{873}{2673} = \frac{97}{297}$$ (dividing numerator and denominator by 9).

Therefore, the required probability is $$\dfrac{97}{297}$$.