Step 1: Count all 4-digit numbers.
Four-digit numbers run from $1000$ to $9999$, a total of $9000$ numbers. This is the sample space.
Step 2: Set up the favorable event.
We want exactly three odd digits and one even digit. Odd digits are $\{1,3,5,7,9\}$ (5 choices) and even digits are $\{0,2,4,6,8\}$ (5 choices).
Step 3: Choose the position of the even digit.
The single even digit can sit in any of the 4 positions, giving $\binom41=4$ placements.
Step 4: Fill the digits.
The three odd places give $5^3=125$ ways, and the even place gives 5 ways, so a naive count is $4\times125\times5=2500$.
Step 5: Correct for the leading digit.
The first digit may not be $0$. Subtracting the cases where the even digit is $0$ in the leading position removes $125$ arrangements, leaving the corrected favorable count $2375$.
Step 6: Compute the probability.
$P=\frac{2375}{9000}=\frac{19}{72}$.
\[ \boxed{\dfrac{19}{72}} \]