<Question>
What is the probability that a randomly selected integer in the set $$\{1,2,3,\ldots,100\}$$  is divisible by 2 and not divisible by 3? Express your answer as a common fraction.
</Question>

<think>
<Plan_1>
Okay, so I need to find the probability that a randomly selected integer from 1 to 100 is divisible by 2 but not by 3. Let me break this down step by step.

First, probability is generally the number of favorable outcomes divided by the total number of possible outcomes. Here, the total number of possible outcomes is 100 because there are 100 numbers in the set. So, the denominator of my probability fraction will be 100. Now, I need to figure out the numerator, which is the count of numbers between 1 and 100 that are divisible by 2 but not by 3.

Let me start by finding how many numbers between 1 and 100 are divisible by 2. Divisible by 2 means the number is even. Every second number is even, so from 1 to 100, there should be 100 divided by 2, which is 50 numbers.
So, 50 numbers are divisible by 2.

But now, some of these numbers are also divisible by 3, and we need to exclude those. So, the next step is to find how many numbers between 1 and 100 are divisible by both 2 and 3. Because if a number is divisible by both, we need to subtract those from our initial count of 50.

<prompt_1.1> Calculate how many numbers between 1 and 100 are divisible by both 2 and 3. </prompt_1.1>
-----

Now, from these 50 even numbers, we need to exclude the 16 that are also divisible by 3. So, 50 - 16 = 34. Therefore, there are 34 numbers between 1 and 100 that are divisible by 2 but not by 3.

Wait, but let me check with an example. Let's take a smaller range, say 1 to 10, to see if the method works.

<prompt_1.2> Verify the method using numbers from 1 to 10. </prompt_1.2>
-----

Alternatively, maybe I can approach this using the principle of inclusion-exclusion. The number of numbers divisible by 2 or 3 is equal to the number divisible by 2 plus the number divisible by 3 minus the number divisible by both. But wait, we don't need that. Wait, actually, we want numbers divisible by 2 but not by 3. So, that's equivalent to (numbers divisible by 2) minus (numbers divisible by both 2 and 3). Which is exactly what we did before. So, 50 - 16 = 34.

Another way to think about it: the numbers we want are even numbers not divisible by 3. So, even numbers are 2,4,6,....,100. There are 50 of them. From these, remove the ones divisible by 3. Which are the numbers divisible by 6, as we established.
So, numbers divisible by 6 up to 100: 16 numbers. So, 50 -16=34.

But let me just verify with actual numbers.
Take the first few even numbers: 2,4,6,8,10,12,14,16,18,20,.... 

Now, numbers divisible by 3 in this list are 6,12,18,24,... So every third even number is divisible by 3. So, starting from 6, every 6th number. So, how many are there?

<prompt_1.3> Verify the count of numbers using the observation that every third even number is divisible by 3. </prompt_1.3>
-----

Alternatively, maybe using probability rules. The probability that a number is divisible by 2 is 1/2. Given that, the probability that it's not divisible by 3 is 1 - 1/3 = 2/3. But since divisibility by 2 and 3 are independent events (since 2 and 3 are coprime), the combined probability is 1/2 * 2/3 = 1/3. Wait, but this gives 1/3, which is approximately 33.333...., but our actual count is 34/100, which is 0.34.

<prompt_1.4> Compare the probability method with the actual count and resolve any discrepancies. </prompt_1.4>
</Plan_1>

<Execution_1>
<execution_1.1>
Numbers divisible by both 2 and 3 are divisible by their least common multiple (LCM). The LCM of 2 and 3 is 6. So, I need to find how many numbers between 1 and 100 are divisible by 6.

To find the count of numbers divisible by 6 in this range, I can divide 100 by 6. Let's see, 6*16=96, and 6*17=102, which is over 100. So, there are 16 numbers divisible by 6 up to 96. Therefore, there are 16 numbers between 1 and 100 that are divisible by both 2 and 3.
</execution_1.1>

<execution_1.2>
In that case, numbers divisible by 2: 2,4,6,8,10. That's 5 numbers. Numbers divisible by both 2 and 3: 6. So, one number. So, numbers divisible by 2 but not 3: 5 -1 =4. Let's list them: 2,4,8,10. Yes, that's four numbers. So, the method works here. 5 -1 =4.
</execution_1.2>

<execution_1.3>
So, from 1 to 100, how many multiples of 6 are there? As before, 6*1, 6*2,....,6*16=96, so 16 numbers. So, removing 16 from 50 gives 34.
</execution_1.3>

<execution_1.4>
Hmm, there's a discrepancy here. Wait, why is there a difference?

Wait, maybe because the two events are not independent when considering the finite set. In reality, the events "divisible by 2" and "divisible by 3" are not independent in a finite set like 1 to 100. In an infinite set, the density might approach 1/3, but in a finite set, the exact count can differ. So, the discrepancy is because 100 is not a multiple of 6, perhaps? Wait, 100 divided by 6 is 16 with a remainder of 4. So, the number of multiples of 6 up to 96 (which is 16*6) is 16, and then 100 is left, but 100 is not a multiple of 6. So, that's why it's 16.
</execution_1.4>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Therefore, going back to our original problem, the numbers divisible by 2 but not by 3 would be the numbers divisible by 2 minus those divisible by both.
So, 50 - 16 = 34. Therefore, there are 34 numbers between 1 and 100 that are divisible by 2 but not by 3.

Hence, the probability would be 34/100. But this fraction can be simplified.
Dividing numerator and denominator by 2, we get 17/50.

Wait, let me double-check my calculations to make sure I didn't make a mistake.

<prompt_2.1> Double-check the calculations for the number of even numbers and those divisible by 6. </prompt_2.1>
-----

So, 34/100 reduces to 17/50. Therefore, the probability is 17/50.
-----

But wait, let me check with actual numbers. Let's list some numbers between 1 and 20 and see if the ratio holds. Maybe that will help confirm.

<prompt_2.2> Verify the probability using numbers from 1 to 20. </prompt_2.2>
-----

None
</Plan_2>

<Execution_2>
<execution_2.1>
First, numbers divisible by 2: 2,4,6,....,100. The count is 100/2 = 50. That's correct. Then numbers divisible by 6: 6,12,18,....,96. Let's check 6*16=96, so 16 numbers. So, 50 -16=34. Yes, that seems right. So, 34 numbers. Therefore, 34/100 simplifies to 17/50.
</execution_2.1>

<execution_2.2>
In 1-20, numbers divisible by 2: 10 numbers (2,4,6,8,10,12,14,16,18,20). Numbers divisible by both 2 and 3 (i.e., 6): 6,12,18. So, 3 numbers. Therefore, numbers divisible by 2 but not 3: 10 -3 =7. So, 7/20. If we apply the same method: numbers divisible by 6 up to 20: 3 (since 6*3=18, next is 24 which is over 20). So, 10 -3=7. Which is correct. The probability here is 7/20. If we use the formula: (numbers divisible by 2) - (numbers divisible by 6) =10 -3=7. So, 7/20. Which is correct.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_1.1, execution_2.1:
Therefore, the answer should be 17/50.
-----

Therefore, 50 -16=34. Hence, 34 numbers. So, probability is 34/100, which reduces to 17/50.

But let me verify once again with another approach. Let's consider the inclusion-exclusion principle.

<prompt_3.1> Verify the result using the inclusion-exclusion principle. </prompt_3.1>
-----

Alternatively, we can think of it as the intersection of two sets: even numbers and numbers not divisible by 3. The total numbers in the intersection would be the numbers that are even and not multiples of 3.

<prompt_3.2> Consider alternative methods to confirm the solution. </prompt_3.2>
-----

Based on execution_2.2:
So, applying the same logic to 1-100: numbers divisible by 2 (50) minus numbers divisible by 6 (16) gives 34. 34/100=17/50.

<prompt_3.3> Apply the verified logic to the original problem and conclude. </prompt_3.3>
</Plan_3>

<Execution_3>
<execution_3.1>
The number of numbers divisible by 2 or 3 is equal to the number divisible by 2 plus the number divisible by 3 minus the number divisible by both. But wait, we don't need that. Wait, actually, we need numbers divisible by 2 but not by 3. So, that's equivalent to (numbers divisible by 2) minus (numbers divisible by both 2 and 3). Which is exactly what we did before.

Numbers divisible by 2: 50. Numbers divisible by both 2 and 3:16. So, 50 -16=34. So, 34 numbers. Therefore, 34/100 reduces to 17/50.
</execution_3.1>

<execution_3.2>
But how does that help? Maybe not directly. Alternatively, perhaps using complementary counting. The total numbers divisible by 2 is 50. From these, subtract those divisible by 3. Which is exactly what we did. So, yes, 50 -16=34.

Alternatively, think of the numbers as being in arithmetic progressions. The even numbers form an arithmetic progression starting at 2, with common difference 2. The numbers divisible by 6 form an arithmetic progression starting at 6, with common difference 6. The number of terms in the first progression is 50, the second is 16.
</execution_3.2>

<execution_3.3>
Therefore, the answer is 17/50.

But just to be thorough, let me actually list some numbers and count them to ensure that 34 is the correct count.

Let's list the even numbers from 1 to 100 and cross out those divisible by 3. The even numbers are 2,4,6,8,10,12,14,16,18,20,....,100. Now, the ones divisible by 3 are 6,12,18,24,....,96. As we saw earlier, there are 16 such numbers. So, removing these 16 from the 50 even numbers, we get 34 numbers that are even but not divisible by 3.

Alternatively, perhaps I can compute it using the formula for arithmetic sequences. The numbers divisible by 2 but not by 3 form an arithmetic sequence starting at 2, with a common difference of 2, but excluding terms that are multiples of 6. Wait, but how does that help? Maybe not directly. Alternatively, the numbers we want are even numbers not divisible by 3. So, in other words, numbers congruent to 2 or 4 modulo 6. Because if a number is even and not divisible by 3, then when divided by 6, it leaves a remainder of 2 or 4. For example, 2,4,8,10,14,16, etc. Let's check:

2 divided by 6 is 0 with remainder 2.

4 divided by 6 is 0 with remainder 4.

8 divided by 6 is 1 with remainder 2.

10 divided by 6 is 1 with remainder 4.

14 divided by 6 is 2 with remainder 2.

16 divided by 6 is 2 with remainder 4.

And so on. So, the pattern is that every two numbers, starting at 2 and 4, then adding 6 each time. So, each cycle of 6 numbers contributes 2 numbers that are even and not divisible by 3. Therefore, in 100 numbers, how many such cycles are there?

Each cycle is 6 numbers. 100 divided by 6 is approximately 16.666.... So, 16 full cycles, which account for 16*6=96 numbers. Each cycle gives 2 numbers, so 16 cycles give 16*2=32 numbers. Then, there are 4 remaining numbers: 97,98,99,100. Checking these:

97: odd, so not even.

98: even, check if divisible by 3. 9+8=17, 17 is not divisible by 3, so 98 is even and not divisible by 3. So, count 1.

99: divisible by 3 (9+9=18), so not counted.

100: even, check divisibility by 3. 1+0+0=1, not divisible by 3. So, 100 is even and not divisible by 3. Count 1.

So, in the remaining 4 numbers, we have 98 and 100, which are 2 numbers. Therefore, total numbers are 32 +2=34. Exactly matches our previous result. Therefore, this confirms that there are 34 numbers between 1 and 100 that are divisible by 2 but not by 3.

Therefore, the probability is 34/100, which simplifies by dividing numerator and denominator by 2: 17/50. Since 17 is a prime number, and 50 is 2*5^2, there are no common factors, so 17/50 is the simplified common fraction.

Just to recap:

1. Total numbers: 100

2. Divisible by 2: 50 numbers

3. Divisible by both 2 and 3 (i.e., 6): 16 numbers

4. Subtract to get numbers divisible by 2 but not 3: 50 -16=34

5. Probability: 34/100=17/50

All steps check out with multiple methods and verifications. Therefore, I can confidently conclude that the answer is 17/50.
</execution_3.3>
</Execution_3>

<Plan_4>
Based on execution_3.3:
**Final Answer**
The probability is \boxed{\dfrac{17}{50}}.
</Plan_4>

<Execution_4>
No prompt to execute.
</Execution_4>

<Final_answer>
Therefore, the probability that a randomly selected integer from the set $\{1,2,3,\ldots,100\}$ is divisible by 2 and not divisible by 3 is $\boxed{\frac{17}{50}}$.
</Final_answer>
</think>