The probability, that in a randomly selected 3-digit number at least two digits are odd, is

A 3-digit number ranges from 100 to 999. Total count = 900.

We need P(at least 2 digits are odd). It is easier to compute:

$$P(\text{at least 2 odd}) = 1 - P(\text{0 odd}) - P(\text{exactly 1 odd})$$

Case 1: No odd digits (all even)

Even digits: {0, 2, 4, 6, 8}. The first digit cannot be 0, so first digit has 4 choices (2, 4, 6, 8). Second and third digits each have 5 choices.

Count = $$4 \times 5 \times 5 = 100$$

Case 2: Exactly 1 odd digit

Odd digits: {1, 3, 5, 7, 9} (5 choices). Even digits: {0, 2, 4, 6, 8}.

Sub-case 2a: First digit is odd, rest even

First digit: 5 choices. Second: 5 choices. Third: 5 choices.

Count = $$5 \times 5 \times 5 = 125$$

Sub-case 2b: Second digit is odd, rest even

First digit: 4 choices (even, non-zero). Second: 5 choices. Third: 5 choices.

Count = $$4 \times 5 \times 5 = 100$$

Sub-case 2c: Third digit is odd, rest even

First digit: 4 choices. Second: 5 choices. Third: 5 choices.

Count = $$4 \times 5 \times 5 = 100$$

Total for exactly 1 odd = $$125 + 100 + 100 = 325$$

At least 2 odd digits:

$$900 - 100 - 325 = 475$$

$$P = \frac{475}{900} = \frac{19}{36}$$

Hence the correct answer is Option A: $$\dfrac{19}{36}$$.