Okay, let's see. The problem is asking for the probability that a randomly selected number from 1 to 100 is a multiple of 3. Hmm, probability is usually the number of successful outcomes divided by the total number of possible outcomes. So first, I need to figure out how many numbers between 1 and 100 are multiples of 3. Then, divide that by 100 to get the probability.

Wait, but how do I find the number of multiples of 3 in that range? Maybe I can start by listing some multiples of 3 and see if there's a pattern. Let's see, 3, 6, 9, 12... and so on. The nth multiple of 3 is 3n. So I need to find the largest n such that 3n is less than or equal to 100. That would give me the total number of multiples.

Let me calculate that. If 3n ≤ 100, then n ≤ 100/3. Dividing 100 by 3 gives approximately 33.333. But since n has to be an integer, the largest possible n is 33. That means there are 33 multiples of 3 between 1 and 99. But wait, 3*33 is 99, which is less than 100. So does that mean 99 is included? Yes, and 100 isn't a multiple of 3. So the multiples are 3, 6, ..., 99. That's 33 numbers.

But hold on, let me check if 3*33 is indeed 99. 3*30 is 90, plus 3*3 is 9, so 90+9=99. Yep, that's correct. So there are 33 numbers. Therefore, the number of successful outcomes is 33, and the total possible outcomes are 100. So the probability is 33/100. But wait, 33 divided by 100 is 0.33, which is 33%. But let me verify this again.

Alternatively, maybe there's a formula for the number of multiples of a number within a range. The formula is floor division of the upper bound by the number. So floor(100/3) would give the number of multiples. Let me compute 100 divided by 3. 3*33=99, so 100/3 is 33.333..., so the floor of that is 33. So yes, 33 multiples. Therefore, the probability is 33/100. 

But wait, another way to think about it: the probability that a number is a multiple of 3 is 1/3, right? Because every third number is a multiple of 3. But 100 isn't a multiple of 3, so the exact number might be slightly less than 1/3 of 100. Since 100 divided by 3 is 33.333..., so 33 numbers. So 33/100 is 0.33, which is exactly 1/3. Wait, but 33/100 is 0.33, which is approximately 1/3, but not exactly. Hmm, but 1/3 would be approximately 33.333... So since we can't have a fraction of a number, we take the floor. So 33 is correct. Therefore, 33/100 is the exact probability.

But let me confirm with an example. Let's take numbers from 1 to 6. Multiples of 3 are 3 and 6. That's 2 numbers, which is 2/6 = 1/3. So in that case, it's exactly 1/3. But in the case of 1 to 100, since 100 isn't a multiple of 3, it's 33/100, which is approximately 0.33. So that's slightly less than 1/3. So depending on how the question is phrased, maybe the answer is 33/100 or rounded to the nearest hundredth. But the question just asks for the probability, so 33/100 is exact. 

Alternatively, maybe I can use modular arithmetic to check. For numbers from 1 to 100, each number can be expressed as 3k, 3k+1, or 3k+2. The number of numbers congruent to 0 mod 3 is floor(100/3) = 33. Numbers congruent to 1 mod 3 would be floor((100-1)/3) +1 = floor(99/3)+1=33+1=34. Similarly, numbers congruent to 2 mod 3 would be 33. Wait, that adds up to 34+33+33=100. So that's correct. Therefore, numbers divisible by 3 are 33. So the probability is 33/100. 

So, in conclusion, the probability is 33/100. But let me just check once more. If I list the multiples: 3,6,9,...99. How many terms are there? The nth term is 3n. So 3n=99 => n=33. So yes, 33 terms. Therefore, 33/100. 

I think that's solid. So the answer should be 33/100.
Therefore, the probability that a randomly selected number from 1 through 100 is a multiple of 3 is $\boxed{\frac{33}{100}}$.
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