Kind of hard!

Because we are working with numbers in the triple digits, our numbers with at least one 0 will have that 0 in either the units digit or the tens digit (or both, though they will only be counted once).

We know that our numbers are inclusive, so our first number will be 100, and will include every number from 100 though 109. That gives us 10 numbers so far.

From here, we can see that the first 10 numbers of 200, 300, 400, 500, 600, 700, 800, and 900 will be included as well, giving us a total of:

10*9

90 so far.

Now we also must include every number that ends in 0. For the first 100 (NOT including 100, which we have already counted!), we would have:

110, 120, 130, 140, 150, 160, 170, 180, 190

This gives us 9 more numbers, which we can also expand to include 9 more in the 200’s, 300’s, 400’s, 500’s, 600’s, 700’s, 800’s, and 900’s. This gives us a total of:

9*9

81

Now, let us add our totals (all the numbers with a units digit of 0 and all the numbers with a tens digit of 0) together:

90+81

171

There are a total of 900 numbers between 100 and 999, inclusive, so our final probability will be:

$\frac{171}{900}$

Our final answer is D, $\large \color{#EC7300} \frac{171}{900}$

There are $999-99=900$ numbers in the range from $100$ to $999$ inclusive. The number of three-digit numbers using only digits $1$ through $9$ inclusive is $9 \times 9 \times 9=729.$ Thus the probability that a random three-digit number has no zeroes in its decimal representation is

$\frac{9\times 9 \times9}{900} = \frac{9 \times 9}{100} = 81\%$

and so the probability that it has at least one zero is

$100\%-81\%=19\%$

which is the same as

$\boxed{\dfrac{171}{900}} = \frac{19}{100} = 19\%.$

Let $X$ be a random variable denoting the number of zeroes in a $3$ digit number.

We are asked to find $P(X\geq 1)$ ,

$\begin{aligned} P(X\geq1) &=1-P(X=0)\\ P(X=0)&=\dfrac{9^3}{900}\hspace{4mm}\color{#3D99F6}\text{There are } 9\text{ ways to choose a digit such that it is}\neq0\\ \implies P(X\geq1)&=1-\dfrac{9^3}{900}\\ &=\color{#EC7300}\boxed{\color{#333333}\dfrac{171}{900}}\end{aligned}$