1. Count all 4-digit numbers with exactly one digit equal to 7

A 4-digit number has positions: thousands, hundreds, tens, units.

We need exactly one digit to be $7$.

• Choose the position of the digit $7$: $4$ ways.
• Fill the other three positions so that none of them is $7$.
• Also, the thousands digit cannot be $0$.

We count casewise.

Case 1: Thousands digit is $7$

Then the other three digits can be anything except $7$.

• Hundreds: $9$ choices ($0$ to $9$, except $7$)
• Tens: $9$ choices
• Units: $9$ choices

So count = $9^3 = 729$

Case 2: Hundreds digit is $7$

Then:

• Thousands digit: can be $1$ to $9$, except $7$ $\\Rightarrow 8$ choices
• Tens digit: $9$ choices (anything except $7$)
• Units digit: $9$ choices

So count = $8 \cdot 9 \cdot 9 = 648$

Case 3: Tens digit is $7$

Similarly, $8 \cdot 9 \cdot 9 = 648$

Case 4: Units digit is $7$

Similarly, $8 \cdot 9 \cdot 9 = 648$

Thus total number of elements in $A$ is $729+648+648+648 = 2673$

2. Now count numbers leaving remainder $2$ when divided by $5$

A number leaves remainder $2$ on division by $5$ iff its units digit is either $2$.

So we need numbers in $A$ with:

• exactly one digit equal to $7$
• units digit $=2$

Since units digit is $2$, the digit $7$ must occur in one of the first three positions.

Subcase 1: Thousands digit is $7$

Number form: $7abc$ with unit digit fixed as $2$.

• Hundreds digit: any digit except $7$ $\Rightarrow 9$ choices
• Tens digit: any digit except $7$ $\Rightarrow 9$ choices
• Units digit: fixed as $2$

Count = $9 \cdot 9 = 81$

Subcase 2: Hundreds digit is $7$

Number form: $a7b2$

• Thousands digit: from $1$ to $9$, except $7$ $\Rightarrow 8$ choices
• Tens digit: any digit except $7$ $\Rightarrow 9$ choices

Count = $8 \cdot 9 = 72$

Subcase 3: Tens digit is $7$

Number form: $ab72$

• Thousands digit: $8$ choices
• Hundreds digit: $9$ choices

Count = $8 \cdot 9 = 72$

Therefore favorable numbers = $81+72+72 = 225$

3. Compute the probability

$P=\frac{225}{2673}$

Simplify: $\frac{225}{2673} = \frac{25}{297}$

since $2673 = 9 \cdot 297, \quad 225 = 9 \cdot 25.$

Hence, $P=\frac{25}{297}$

4. Check against options

Given options are:

• A: $\frac{2}{9}$
• B: $\frac{1}{5}$
• C: $\frac{122}{297}$
• D: $\frac{97}{297}$

But our computed probability is $\frac{25}{297}$ which is not among the options.

5. Compare with stored correct answer

Stored correct answer: D = $\frac{97}{297}$

This does not match the correct counting. Therefore the stored answer appears to be incorrect.