1. Total number of 3-digit numbers

A 3-digit number ranges from $100$ to $999$. So, total number of 3-digit numbers is $999-100+1=900.$

2. Classify digits as odd/even

• Odd digits: $1,3,5,7,9$ $Rightarrow 5$ choices
• Even digits: $0,2,4,6,8$ $Rightarrow 5$ choices

But for the hundreds digit, $0$ is not allowed. So among digits $1$ to $9$:

• Odd hundreds digits: $1,3,5,7,9$ $Rightarrow 5$ choices
• Even hundreds digits: $2,4,6,8$ $Rightarrow 4$ choices

3. We need: at least two digits are odd

This means either:

• exactly $2$ digits are odd, or
• all $3$ digits are odd.

Case 1: Exactly 2 digits are odd

We count according to which digit is even.

(i) Hundreds digit even, tens and units odd

• Hundreds digit: $4$ choices ($2,4,6,8$)
• Tens digit: $5$ odd choices
• Units digit: $5$ odd choices

Number of such numbers: $4\times 5\times 5=100.$

(ii) Tens digit even, hundreds and units odd

• Hundreds digit: $5$ odd choices
• Tens digit: $5$ even choices ($0,2,4,6,8$)
• Units digit: $5$ odd choices

Number of such numbers: $5\times 5\times 5=125.$

(iii) Units digit even, hundreds and tens odd

• Hundreds digit: $5$ odd choices
• Tens digit: $5$ odd choices
• Units digit: $5$ even choices

Number of such numbers: $5\times 5\times 5=125.$

So, total with exactly 2 odd digits: $100+125+125=350.$

Case 2: All 3 digits are odd

• Hundreds digit: $5$ choices
• Tens digit: $5$ choices
• Units digit: $5$ choices

Number of such numbers: $5\times 5\times 5=125.$

4. Favourable numbers

Total favourable numbers: $350+125=475.$

5. Required probability

P=\frac{475}{900}=rac{19}{36}.

6. Compare with options

$\frac{19}{36}$ matches Option A.

7. Comparison with stored correct answer

Stored correct answer is A, which matches our result.