<Question>
A number is selected at random from 1 through 100, inclusive.  What is the probability that the number is a multiple of 3?
</Question>

<think>
<Plan_1>
Okay, let's see. The problem is asking for the probability that a number selected at random from 1 through 100 is a multiple of 3. Hmm, probability... right. So probability is usually the number of favorable outcomes divided by the total number of possible outcomes.
Let me make sure I remember that correctly. So in this case, the total number of possible outcomes is 100 because there are numbers from 1 to 100, inclusive. That part seems straightforward.

Now, the favorable outcomes here are the numbers that are multiples of 3. So I need to figure out how many numbers between 1 and 100 are divisible by 3. Once I have that count, I can divide it by 100 to get the probability.

But how do I find how many multiples of 3 there are up to 100? Let me think. A multiple of 3 can be written as 3 times some integer.
So the first multiple is 3×1=3, then 3×2=6, and so on. The last multiple of 3 that is less than or equal to 100 would be... let me calculate that. If I divide 100 by 3, that gives approximately 33.333. Since we can't have a fraction of a number, the integer part is 33. So the last multiple is 3×33=99.

So that means the multiples of 3 in this range are 3, 6, 9, ...., 99. To find the number of terms in this sequence, I can use the formula for the nth term of an arithmetic sequence.

<prompt_1.1> Use the formula for the nth term of an arithmetic sequence to find the number of multiples of 3 between 1 and 100. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
The nth term is given by a_n = a_1 + (n-1)d, where a_1 is the first term, d is the common difference, and n is the number of terms. Here, a_1 is 3, d is 3 (since each term is 3 more than the previous), and a_n is 99. Plugging these into the formula:

99 = 3 + (n - 1)×3

Subtract 3 from both sides: 96 = (n - 1)×3

Divide both sides by 3: 32 = n - 1

Add 1 to both sides: n = 33

So there are 33 terms. Therefore, there are 33 multiples of 3 between 1 and 99. Since 100 is not a multiple of 3, the total number of favorable outcomes is 33.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Wait a second, but let me verify that. If I list the multiples of 3, starting at 3, the nth term is 3n. So 3n ≤ 100. Solving for n, n ≤ 100/3 ≈ 33.333. So n can be 1 up to 33.
That gives 33 terms. So that checks out.

Therefore, the number of favorable outcomes is 33. Therefore, the probability is 33/100.

But hold on, is that correct? Let me think again. Let's take a smaller example to verify.

<prompt_2.1> Use smaller examples to verify the method of finding the number of multiples of 3. </prompt_2.1>

Therefore, in the case of 1 to 100, there are 33 multiples of 3. So the probability is 33/100.

But wait, another way to think about it: when you divide 100 by 3, you get 33 with a remainder of 1.

<prompt_2.2> Use division to verify the number of multiples of 3 between 1 and 100. </prompt_2.2>

So the probability is 33/100. But maybe we can simplify that fraction? Let's see.
33 and 100 have no common factors except 1, so 33/100 is already in simplest terms. Therefore, the probability is 33/100.

But wait, just to be thorough, let me manually count the multiples of 3 up to, say, 20, and see if the formula works.

<prompt_2.3> Manually count the multiples of 3 up to 20 to verify the formula. </prompt_2.3>
</Plan_2>

<Execution_2>
<execution_2.1>
Suppose we have numbers from 1 to 10. How many multiples of 3 are there? They are 3, 6, 9. That's 3 numbers. Using the formula: 10 divided by 3 is approximately 3.333, so the integer part is 3. That gives 3 numbers, which is correct. So that works.

Another example: 1 to 15. Multiples of 3 are 3,6,9,12,15. That's 5 numbers. 15 divided by 3 is exactly 5, so that works. So the formula seems to hold.
</execution_2.1>

<execution_2.2>
That means there are 33 multiples of 3 (since 3×33=99) and then the remaining numbers from 99 to 100 are 99, 100, which are 2 numbers, but since 99 is the last multiple, the next number 100 is not a multiple. So yes, 33 multiples.
</execution_2.2>

<execution_2.3>
Multiples of 3 between 1 and 20: 3,6,9,12,15,18. That's 6 numbers. Using the formula: 20 divided by 3 is approximately 6.666, so take the integer part, which is 6. Correct. So the formula works here. So similarly, 100 divided by 3 is approximately 33.333, so 33. Correct.

Therefore, the probability is 33/100.
</execution_2.3>
</Execution_2>

<Final_answer>
Therefore, the probability that a number selected at random from 1 through 100 is a multiple of 3 is $\boxed{\frac{33}{100}}$.
</Final_answer>
</think>