Probability

Complementary counting. The number of numbers that don't have zero anywhere is $9^3 = 729$ (9 possibilities for each individual digit), and there are $9 \cdot 10^2 = 900$ numbers overall (9 possibilities for hundreds, 10 for the tens and units), so there are $900 - 729 = 171$ numbers with at least a zero and thus $\frac{171}{900} = \frac{19}{100}$ probability.

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:

171/900

Our final answer is D, 171/900

Let $A$ be a set of number between 100 and 999 whose unit digit is zero, and Let $B$ be a set of number (same range) whose tenth digit is zero.

Since we can select only nine numbers in hundredth digit (zero is excluded), and at the digit with zero, there can only be one way to do so, set A will have a member of $9\times10\times1 = 90$ members. Similarly, set B will have the member of $9\times1\times10 = 90$ members. Combining gives $90+90 = 180$ members.

The solution above counts the case where the number ends with $00$ twice. This can happen in $9$ cases. Hence, $90+90-9 = 171$ numbers will contain zero at some digit. The possibility is then $\displaystyle\frac{171}{900} = 0.19$

Not sure why the 2 examples on this page are including separate math problems aside from the probability portion. Are these types of counting problems related to probability?

contains 0 scenarios:

(A) XX0 => first digit can be 1 to 9 and second digit can be 0 to 9 => 9 * 10 => 90

(B) X0X => first digit can be 1 to 9 and third digit can be 0 to 9 => 9 * 10 => 90

Both the second and third digits are zeroes are double counted (appeared in both (A) and (B)), so we have to reduce it back once.

X00 => first digit can be 1 to 9 => 9

total = 90 + 90 - 9 = 171

all possible = 999 - 100 + 1 = 900

probability = 171 / 900