The probability, that in a randomly selected 3 -digit number at...

Total No of 3-digit Number $= 900$

Case -1 When there are $2$ odd digit and $1$ even digit

(i) Even odd odd $= 4 \times 5 \times 5 = 100$

{As $0$ cannot come at ${1}^{st}$ place}

(ii) Odd even odd $= 5 \times 5 \times 5 = 125$

(iii) Odd odd even $= 5 \times 5 \times 5 = 125$

Case-2

All $3$ odd digit $= 5 \times 5 \times 5 = 125$

So total number of selecting $3$-digit number having atleast $2$ odd digit will be addition of both cases which is$= 100 + 125 + 125 + 125 = 475$

So probability $= \dfrac{475}{900} = \dfrac{19}{36}$