How Many Whole Numbers Are Between 100 and 999?

The integers from 100 through 999 form a set of 900 possible outcomes. If the selection is uniform, each one is equally likely. The range alone does not specify an event—such as choosing an even number—so it does not determine one particular probability. Once you know the event, count how many numbers satisfy it and divide by 900.

Count the possible outcomes

For an inclusive range of integers from $a$ to $b$, the number of integers is $b-a+1$. Therefore,

Equivalently, these are exactly the three-digit whole numbers. Their digits have different roles: the hundreds digit can be 1 through 9, while each of the tens and units digits can be 0 through 9. Thus there are $9\cdot10\cdot10=900$ such numbers.

The phrase “selected at random” is usually interpreted to mean that every number in the range has the same chance of being selected. Under that assumption, the sample space has 900 equally likely outcomes, and each individual number has probability $1/900$.

Calculate a probability once the event is known

For any event $A$ consisting of numbers in this range,

The main task is to count the favorable outcomes correctly. Here are two examples.

Example 1: The number is even

An integer is even when its units digit is one of $0,2,4,6,8$. There are 9 choices for the hundreds digit, 10 choices for the tens digit, and 5 choices for the units digit. Thus there are

even numbers, and

Example 2: All three digits are different

Choose the hundreds digit first: there are 9 choices, from 1 to 9. The tens digit can be any of the 10 digits except the hundreds digit, giving 9 choices. The units digit must differ from both previous digits, leaving 8 choices. The count is

so

The digit choices are counted in order because the hundreds, tens, and units positions are distinct. For example, 123 and 321 are different numbers. This is why multiplying the choices is appropriate here.

Common mistakes

• Leaving out an endpoint: “Between 100 and 999 inclusive” includes both 100 and 999. The count is $999-100+1$, not $999-100$.
• Treating the range as 1,000 outcomes: There are 900 three-digit numbers, not 1,000. The hundreds digit cannot be zero.
• Assuming an unstated event: Knowing the sample space does not tell you the probability of being prime, divisible by 7, or having a certain digit pattern. The event must be given.
• Assuming uniformity without checking: The calculation uses equal likelihood for all 900 integers. If the selection mechanism favors some numbers, you need its probability model instead.

If a problem supplies a specific condition, translate it into a count of qualifying numbers, then divide that count by 900. For a complement event, it can be quicker to count the numbers that do not satisfy the condition and subtract that count from 900.