Explanation:
Okay, so I need to find the probability that a randomly selected integer from the set {1, 2, 3, ..., 100} is divisible by 2 and not divisible by 3. Hmm, let me think about how to approach this.

First, probability is generally the number of favorable outcomes divided by the total number of possible outcomes. In this case, 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, the numerator is the number of integers between 1 and 100 that are divisible by 2 but not divisible by 3. Let me break this down. I need to count how many numbers are divisible by 2, then subtract those numbers that are divisible by both 2 and 3 because those would be the ones that are divisible by 2 but also by 3, which we don't want.

Wait, actually, another way to think about it is that numbers divisible by 2 and not divisible by 3 are the numbers divisible by 2 minus the numbers divisible by both 2 and 3. Yes, that seems correct. So, maybe I can calculate the number of numbers divisible by 2 and then subtract the number of numbers divisible by both 2 and 3. That should give me the count of numbers divisible by 2 but not 3.

Let me recall that the number of integers divisible by a number 'n' in the range from 1 to N is given by the floor of N divided by n. But since N is 100, which is a multiple of 2 and 3, maybe I can compute it without floor function?

For example, numbers divisible by 2: from 1 to 100, the count is 100 divided by 2, which is 50. So, there are 50 numbers divisible by 2.

Now, numbers divisible by both 2 and 3. Well, numbers divisible by both 2 and 3 are numbers divisible by the least common multiple (LCM) of 2 and 3. The LCM of 2 and 3 is 6. Therefore, numbers divisible by 6. So, the count of numbers divisible by 6 between 1 and 100 is 100 divided by 6. Wait, 100 divided by 6 is 16.666..., so since we can't have a fraction of a number, we take the floor of that. But since 6*16=96 and 6*17=102, which is beyond 100. So, there are 16 numbers divisible by 6 in the set {1, ..., 100}.

Therefore, the number of numbers divisible by 2 but not by 3 is 50 - 16 = 34. Hence, the probability would be 34/100. But let me check if I did that correctly.

Wait, is 100 divided by 6 exactly 16.666...? Yes, so floor(100/6) is 16. But let me verify by actually counting how many multiples of 6 are there up to 100. The first multiple is 6, then 12, 18,..., up to the largest one less than or equal to 100. The largest multiple of 6 less than or equal to 100 is 96, since 6*16=96 and 6*17=102. Therefore, there are 16 multiples. So that seems correct.

Therefore, 50 numbers divisible by 2, 16 of which are also divisible by 3. Therefore, subtracting gives 50 - 16 = 34 numbers that are divisible by 2 but not by 3. Therefore, the probability is 34/100. Simplifying that fraction, divide numerator and denominator by 2: 17/50. So 17/50 is the simplified form. Let me check if that is correct.

Alternatively, maybe another approach. Let's think of numbers divisible by 2 but not by 3. So another way to think about it is numbers that are divisible by 2 and not divisible by 3. So, for numbers divisible by 2, which are even numbers, we need to exclude those that are divisible by 3. So, in the even numbers, how many are divisible by 3?

Alternatively, since even numbers are numbers congruent to 0 mod 2. Divisible by 3 means congruent to 0 mod 3. So numbers that are divisible by both 2 and 3 are divisible by 6, as before. Therefore, same result. Therefore, the count is 50 - 16 = 34. So 34/100 reduces to 17/50. Let me check if that is the simplest form.

Yes, 17 is a prime number, and 50 is 2*5^2, and 17 does not divide 50, so 17/50 is the simplified fraction.

Wait, but just to make sure, let me verify by another method. Let me think about the count of numbers divisible by 2 but not by 3. So, perhaps using inclusion-exclusion?

Alternatively, think of numbers not divisible by 3. How many numbers are not divisible by 3? Total numbers 100, numbers divisible by 3 are floor(100/3) = 33.333..., so 33 numbers. Therefore, numbers not divisible by 3 are 100 - 33 = 67. But how many of these are divisible by 2?

Wait, but that approach might complicate. Wait, maybe not. If I want numbers divisible by 2 and not divisible by 3, is there a way to compute that? Maybe using the principle of inclusion-exclusion. Let me think.

Alternatively, think of the set of numbers divisible by 2 as 50 numbers. Now, how many of these 50 numbers are not divisible by 3? That's exactly the count we need. So, if I can compute how many of the 50 numbers are not divisible by 3, that's the answer.

Alternatively, since the numbers divisible by 2 are 2, 4, 6, 8, ..., 100. Now, among these numbers, some are divisible by 3. So, how many numbers in this list are divisible by 3? Which would be numbers divisible by 6, as we saw, which is 16 numbers. Therefore, the numbers divisible by 2 but not by 3 is 50 - 16 = 34. So same result.

Alternatively, think of the numbers divisible by 2 but not 3 as numbers congruent to 0 mod 2 and not congruent to 0 mod 3. So, perhaps we can think of residues.

But perhaps another way is to think of the numbers from 1 to 100, and for each number, check if it's even and not divisible by 3. But that would be tedious.

Alternatively, maybe using the principle of inclusion-exclusion. The number of numbers divisible by 2 or 3 is equal to numbers divisible by 2 + numbers divisible by 3 - numbers divisible by both 2 and 3. But that's not directly helpful here. Wait, but maybe if I want numbers divisible by 2 but not by 3, it's equal to numbers divisible by 2 minus numbers divisible by both 2 and 3, which is exactly what I did earlier.

Alternatively, think of the probability. The probability that a number is divisible by 2 is 1/2. The probability that it is divisible by both 2 and 3 is probability divisible by 6, which is 1/6. Therefore, the probability that it is divisible by 2 and not by 3 is 1/2 - 1/6 = 1/3. Wait, but 1/3 is approximately 0.333..., but 34/100 is 0.34. Wait, there's a discrepancy here. Which one is correct?

Wait, hold on. Let me check. If I use probability, the probability that a number is divisible by 2 is 50/100 = 1/2. The probability that a number is divisible by 6 is 16/100 = 4/25. Therefore, the probability of being divisible by 2 and not divisible by 3 would be 1/2 - 4/25 = (25/50 - 16/50) = 9/50? Wait, hold on, wait:

Wait 1/2 is 50/100. 16/100 is numbers divisible by 6. Therefore, 50/100 - 16/100 = 34/100 = 17/50 ≈ 0.34. But earlier, if I thought of probability as 1/2 - 1/6, that would be 1/2 - 1/6 = 1/3 ≈ 0.333..., but that's different from 17/50. Which is correct?

Wait, why is there a discrepancy? Because when I said that probability of divisible by 6 is 1/6, but actually, in reality, numbers divisible by 6 between 1-100 is 16, not 16.666... So, 16/100 is not exactly 1/6. 1/6 is approximately 0.1666..., but 16/100 is 0.16, which is close but not exact. Therefore, the exact probability is 16/100, so the exact value is 50/100 - 16/100 = 34/100 = 17/50. Therefore, 17/50 is correct.

But why does the other approach give me 1/2 - 1/6? Because in reality, the probability that a number is divisible by 6 is not exactly 1/6, because 100 isn't a multiple of 6. If we were considering numbers from 1 to 6, the probability would be 1/6, but since we are going up to 100, which isn't a multiple, it's slightly different. So, in general, if N is a multiple of 6, then the number of multiples of 6 would be exactly N/6. But when N isn't a multiple of 6, it's floor(N/6). Therefore, for N=100, floor(100/6)=16, so the probability is 16/100, not exactly 1/6. Therefore, the exact answer is 17/50. So, the mistake was in assuming that the probability is exactly 1/6, but actually, since we are dealing with finite numbers up to 100, which isn't a multiple of 6, the exact count is needed. Therefore, 17/50 is the correct probability.

Let me check with another approach. Let me list out the numbers divisible by 2 and then check how many are not divisible by 3.

Numbers divisible by 2 from 1 to 100: 2, 4, 6, 8, 10, ..., 100. There are 50 numbers. Now, among these, the ones divisible by 3 are those numbers that are multiples of 6, as mentioned. So, starting from 6, 12, 18,..., 96.

Let me count them again. 6*1=6, 6*2=12,..., 6*16=96. So, 16 numbers. Therefore, the numbers divisible by 2 but not 3 are 50-16=34. So, 34 numbers. Therefore, probability is 34/100=17/50. So, that's confirming again.

But maybe let me do another sanity check. Let's take a smaller set where I can manually count. Let's say numbers from 1 to 6. How many are divisible by 2 and not by 3?

Numbers: 1,2,3,4,5,6.

Divisible by 2: 2,4,6.

From these, divisible by 3: 6. So numbers divisible by 2 but not 3: 2,4. So 2 numbers. Therefore, probability is 2/6=1/3.

But according to the formula: total numbers divisible by 2 is 3. Numbers divisible by both 2 and 3: floor(6/6)=1. Therefore, 3 - 1 = 2. So probability is 2/6=1/3. Which matches.

If I use the same formula for N=6, then 1/2 - 1/6=1/3. Which matches. So in this case, even though N is a multiple of 6, it works.

But when N is not a multiple of 6, like N=7. Let's check.

Numbers from 1-7. Divisible by 2: 2,4,6. Divisible by 3: 3,6. Divisible by both: 6. So numbers divisible by 2 but not 3: 2,4. So again 2 numbers. Probability is 2/7. According to formula: numbers divisible by 2: floor(7/2)=3. Numbers divisible by 6: floor(7/6)=1. So 3-1=2. 2/7. If I use probability formula: 1/2 - floor(7/6)/7 = 1/2 - 1/7 = 5/14 ≈ 0.357... But actual probability is 2/7 ≈ 0.2857. Wait, so here the formula 1/2 - 1/6 is not correct. But that's because in reality, the numbers divisible by 6 up to N is floor(N/6). Therefore, in the general case, the probability is (floor(N/2) - floor(N/6))/N. So, for N=7, floor(7/2)=3, floor(7/6)=1, so (3-1)/7=2/7.

But if I try to use fractions, like 1/2 - 1/6=1/3≈0.333, which would be different. So, in reality, when N is not a multiple of 6, the exact count requires floor functions. Therefore, in our problem with N=100, since 100 is not a multiple of 6, we need to calculate floor(100/2) - floor(100/6)=50 - 16=34. Therefore, 34/100=17/50.

Therefore, that seems correct.

Alternatively, if N was a multiple of 6, say N=6, then floor(6/2)=3, floor(6/6)=1, so 3-1=2. Which is correct. For N=12, floor(12/2)=6, floor(12/6)=2, so 6-2=4. Let's check numbers 1-12. Numbers divisible by 2: 6 numbers (2,4,6,8,10,12). Numbers divisible by both 2 and 3: 6, 12. So 2 numbers. Therefore, numbers divisible by 2 but not by 3: 6-2=4. Which are 2,4,8,10. Correct. Probability is 4/12=1/3. Which is same as 1/2 - 1/6=1/3. So in this case, since N is multiple of 6, floor(N/6)=N/6, so the formula works. So in general, when N is a multiple of 6, the formula 1/2 - 1/6=1/3 gives the exact probability, but when N isn't, we need to use floor functions.

Therefore, in our problem, since N=100 is not a multiple of 6, we need to compute exactly numbers divisible by 2, subtract numbers divisible by 6, and get 34/100=17/50.

Alternatively, perhaps there's a way to compute it by considering the residues. Let me think. For numbers from 1 to 100, each number can be categorized based on their divisibility by 2 and 3.

Since we are dealing with divisibility by 2 and 3, we can consider modulo 6, since 6 is the LCM of 2 and 3.

The residues mod 6 are 0,1,2,3,4,5.

Numbers divisible by 2 are those congruent to 0 or 2 or 4 mod 6. So, residues 0,2,4.

Numbers divisible by 3 are those congruent to 0 or 3 mod 6.

Therefore, numbers divisible by 2 and not by 3 are residues 2 and 4 mod 6.

Therefore, in each block of 6 numbers, there are 2 numbers that are divisible by 2 but not by 3. For example, in numbers 1-6, the numbers 2 and 4 are divisible by 2 but not by 3. Similarly, in numbers 7-12, numbers 8 and 10, etc. So, in each group of 6 numbers, there are 2 such numbers.

Now, total numbers from 1 to 100. How many complete blocks of 6 are there? 100 divided by 6 is 16 with a remainder of 4. Because 16*6=96, so 100-96=4. Therefore, there are 16 complete blocks and then 4 extra numbers (97, 98, 99, 100).

In each complete block of 6 numbers, there are 2 numbers that are divisible by 2 but not by 3. Therefore, 16 blocks contribute 16*2=32 numbers.

Now, we need to check the remaining 4 numbers: 97, 98, 99, 100. Let's check which of these are divisible by 2 and not by 3.

- 97: odd, so not divisible by 2.
- 98: divisible by 2 (since even). Check if divisible by 3: 9+8=17, 17 isn't divisible by 3. Therefore, 98 is divisible by 2 and not by 3.
- 99: odd, so not divisible by 2.
- 100: divisible by 2. Check divisibility by 3: 1+0+0=1, not divisible by 3. So 100 is divisible by 2 and not by 3.

Therefore, in the remaining 4 numbers, there are 2 numbers (98 and 100) that are divisible by 2 and not by 3.

Therefore, total numbers divisible by 2 and not by 3 is 32 + 2 = 34. Therefore, same as before. So, 34 numbers. Therefore, probability is 34/100=17/50. Therefore, this method also confirms the same answer.

Therefore, regardless of the method I use—counting directly, using inclusion-exclusion, or breaking into residues mod 6—the answer comes out to 34/100, which simplifies to 17/50.

I think that's thorough. Let me check once more for any possible mistake. For example, maybe I miscalculated the number of numbers divisible by 6?

Wait, numbers divisible by 6 up to 100: 6*1=6, 6*2=12,...,6*16=96, and 6*17=102 which is over 100. So, yes, 16 numbers. Therefore, 50-16=34. Correct. Also, in the mod 6 approach, 16 blocks of 6 numbers, each contributing 2 numbers, gives 32, plus 2 more in the last partial block, total 34. Correct. So, seems solid.

Another way: Let me compute the total numbers divisible by 2, not by 3. So, for any number, if it's even (divisible by 2), then it's not divisible by 3 if it's not divisible by 3. So, for the even numbers, the probability that they are not divisible by 3 is equal to 1 minus the probability that an even number is divisible by 3. Since numbers divisible by 3 among even numbers are those divisible by 6. So, in the set of even numbers, what fraction is divisible by 3?

There are 50 even numbers. How many are divisible by 3? As we saw, 16. So, the fraction is 16/50. Therefore, the fraction of even numbers not divisible by 3 is 1 - 16/50 = 34/50 = 17/25. Wait, but the probability over the entire set is 34/100, which is same as 17/50. Wait, why is there a discrepancy here?

Wait, hold on. If I consider the set of even numbers, the probability that a randomly selected even number is not divisible by 3 is 1 - (number divisible by 3)/50 = 1 - 16/50 = 34/50 = 17/25. However, since the entire set is 100 numbers, the probability that a number is even and not divisible by 3 is (number of such numbers)/100 = 34/100 = 17/50. Therefore, there's no discrepancy. 17/25 is the conditional probability given that the number is even, whereas 17/50 is the probability over the entire set. Therefore, correct.

So, that's consistent. So, no issue there. Therefore, all the different methods lead to the same answer.

Hence, I can be confident that the probability is 17/50.

**Final Answer**
The probability is \boxed{\dfrac{17}{50}}.

To determine the probability that a randomly selected integer from the set $\{1, 2, 3, \ldots, 100\}$ is **divisible by 2 and not divisible by 3**, we follow a structured approach using concepts from number theory:

---

### Step 1: Total Number of Integers

The set has:
$$
100 \text{ integers}
$$
So, the **total number of possible outcomes** is $100$.

---

### Step 2: Count of Numbers Divisible by 2

Numbers divisible by 2 in the range from $1$ to $100$ are the even numbers:
$$
2, 4, 6, \ldots, 100
$$
This is an arithmetic sequence with first term $2$, common difference $2$, and last term $100$.

Using the formula for the number of terms in such a sequence:
$$
\text{Number of terms} = \frac{\text{Last term} - \text{First term}}{\text{Common difference}} + 1 = \frac{100 - 2}{2} + 1 = 50
$$
So, there are **50 integers divisible by 2**.

---

### Step 3: Count of Numbers Divisible by 2 **and** 3

Numbers divisible by both 2 and 3 are divisible by the **least common multiple (LCM)** of 2 and 3, which is:
$$
\text{LCM}(2,3) = 6
$$
So, we now count how many numbers from $1$ to $100$ are divisible by $6$.

These are:
$$
6, 12, 18, \ldots, 96
$$
This is also an arithmetic sequence with first term $6$, common difference $6$, and last term $96$.

Number of terms:
$$
\frac{96 - 6}{6} + 1 = \frac{90}{6} + 1 = 15 + 1 = 16
$$
So, there are **16 integers divisible by both 2 and 3**.

---

### Step 4: Count of Numbers Divisible by 2 and **Not** by 3

To find how many numbers are divisible by 2 **but not** by 3, subtract the count of those divisible by both from the count of those divisible by 2:
$$
\text{Numbers divisible by 2 but not by 3} = 50 - 16 = 34
$$

---

### Step 5: Compute the Probability

The probability is given by:
$$
\text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{34}{100}
$$

Simplifying:
$$
\frac{34}{100} = \frac{17}{50}
$$

---

### Final Answer

$$
\boxed{\dfrac{17}{50}}
$$

[Assistant]
Explanation:
Okay, so I need to find the probability that a randomly selected integer from the set {1, 2, 3, ..., 100} is divisible by 2 and not divisible by 3. Hmm, let me think about how to approach this.

First, probability is generally the number of favorable outcomes divided by the total number of possible outcomes. In this case, 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, the numerator is the number of integers between 1 and 100 that are divisible by 2 but not divisible by 3. Let me break this down. I need to count how many numbers are divisible by 2, then subtract those numbers that are divisible by both 2 and 3 because those would be the ones that are divisible by 2 but also by 3, which we don't want.

Wait, actually, another way to think about it is that numbers divisible by 2 and not divisible by 3 are the numbers divisible by 2 minus the numbers divisible by both 2 and 3. Yes, that seems correct. So, maybe I can calculate the number of numbers divisible by 2 and then subtract the number of numbers divisible by both 2 and 3. That should give me the count of numbers divisible by 2 but not 3.

Let me recall that the number of integers divisible by a number 'n' in the range from 1 to N is given by the floor of N divided by n. But since N is 100, which is a multiple of 2 and 3, maybe I can compute it without floor function?

For example, numbers divisible by 2: from 1 to 100, the count is 100 divided by 2, which is 50. So, there are 50 numbers divisible by 2.

Now, numbers divisible by both 2 and 3. Well, numbers divisible by both 2 and 3 are numbers divisible by the least common multiple (LCM) of 2 and 3. The LCM of 2 and 3 is 6. Therefore, numbers divisible by 6. So, the count of numbers divisible by 6 between 1 and 100 is 100 divided by 6. Wait, 100 divided by 6 is 16.666..., so since we can't have a fraction of a number, we take the floor of that. But since 6*16=96 and 6*17=102, which is beyond 100. So, there are 16 numbers divisible by 6 in the set {1, ..., 100}.

Therefore, the number of numbers divisible by 2 but not by 3 is 50 - 16 = 34. Hence, the probability would be 34/100. But let me check if I did that correctly.

Wait, is 100 divided by 6 exactly 16.666...? Yes, so floor(100/6) is 16. But let me verify by actually counting how many multiples of 6 are there up to 100. The first multiple is 6, then 12, 18,..., up to the largest one less than or equal to 100. The largest multiple of 6 less than or equal to 100 is 96, since 6*16=96 and 6*17=102. Therefore, there are 16 multiples. So that seems correct.

Therefore, 50 numbers divisible by 2, 16 of which are also divisible by 3. Therefore, subtracting gives 50 - 16 = 34 numbers that are divisible by 2 but not by 3. Therefore, the probability is 34/100. Simplifying that fraction, divide numerator and denominator by 2: 17/50. So 17/50 is the simplified form. Let me check if that is correct.

Alternatively, maybe another approach. Let's think of numbers divisible by 2 but not by 3. So another way to think about it is numbers that are divisible by 2 and not divisible by 3. So, for numbers divisible by 2, which are even numbers, we need to exclude those that are divisible by 3. So, in the even numbers, how many are divisible by 3?

Alternatively, since even numbers are numbers congruent to 0 mod 2. Divisible by 3 means congruent to 0 mod 3. So numbers that are divisible by both 2 and 3 are divisible by 6, as before. Therefore, same result. Therefore, the count is 50 - 16 = 34. So 34/100 reduces to 17/50. Let me check if that is the simplest form.

Yes, 17 is a prime number, and 50 is 2*5^2, and 17 does not divide 50, so 17/50 is the simplified fraction.

Wait, but just to make sure, let me verify by another method. Let me think about the count of numbers divisible by 2 but not by 3. So, perhaps using inclusion-exclusion?

Alternatively, think of numbers not divisible by 3. How many numbers are not divisible by 3? Total numbers 100, numbers divisible by 3 are floor(100/3) = 33.333..., so 33 numbers. Therefore, numbers not divisible by 3 are 100 - 33 = 67. But how many of these are divisible by 2?

Wait, but that approach might complicate. Wait, maybe not. If I want numbers divisible by 2 and not divisible by 3, is there a way to compute that? Maybe using the principle of inclusion-exclusion. Let me think.

Alternatively, think of the set of numbers divisible by 2 as 50 numbers. Now, how many of these 50 numbers are not divisible by 3? That's exactly the count we need. So, if I can compute how many of the 50 numbers are not divisible by 3, that's the answer.

Alternatively, since the numbers divisible by 2 are 2, 4, 6, 8, ..., 100. Now, among these numbers, some are divisible by 3. So, how many numbers in this list are divisible by 3? Which would be numbers divisible by 6, as we saw, which is 16 numbers. Therefore, the numbers divisible by 2 but not by 3 is 50 - 16 = 34. So same result.

Alternatively, think of the numbers divisible by 2 but not 3 as numbers congruent to 0 mod 2 and not congruent to 0 mod 3. So, perhaps we can think of residues.

But perhaps another way is to think of the numbers from 1 to 100, and for each number, check if it's even and not divisible by 3. But that would be tedious.

Alternatively, maybe using the principle of inclusion-exclusion. The number of numbers divisible by 2 or 3 is equal to numbers divisible by 2 + numbers divisible by 3 - numbers divisible by both 2 and 3. But that's not directly helpful here. Wait, but maybe if I want numbers divisible by 2 but not by 3, it's equal to numbers divisible by 2 minus numbers divisible by both 2 and 3, which is exactly what I did earlier.

Alternatively, think of the probability. The probability that a number is divisible by 2 is 1/2. The probability that it is divisible by both 2 and 3 is probability divisible by 6, which is 1/6. Therefore, the probability that it is divisible by 2 and not by 3 is 1/2 - 1/6 = 1/3. Wait, but 1/3 is approximately 0.333..., but 34/100 is 0.34. Wait, there's a discrepancy here. Which one is correct?

Wait, hold on. Let me check. If I use probability, the probability that a number is divisible by 2 is 50/100 = 1/2. The probability that a number is divisible by 6 is 16/100 = 4/25. Therefore, the probability of being divisible by 2 and not divisible by 3 would be 1/2 - 4/25 = (25/50 - 16/50) = 9/50? Wait, hold on, wait:

Wait 1/2 is 50/100. 16/100 is numbers divisible by 6. Therefore, 50/100 - 16/100 = 34/100 = 17/50 ≈ 0.34. But earlier, if I thought of probability as 1/2 - 1/6, that would be 1/2 - 1/6 = 1/3 ≈ 0.333..., but that's different from 17/50. Which is correct?

Wait, why is there a discrepancy? Because when I said that probability of divisible by 6 is 1/6, but actually, in reality, numbers divisible by 6 between 1-100 is 16, not 16.666... So, 16/100 is not exactly 1/6. 1/6 is approximately 0.1666..., but 16/100 is 0.16, which is close but not exact. Therefore, the exact probability is 16/100, so the exact value is 50/100 - 16/100 = 34/100 = 17/50. Therefore, 17/50 is correct.

But why does the other approach give me 1/2 - 1/6? Because in reality, the probability that a number is divisible by 6 is not exactly 1/6, because 100 isn't a multiple of 6. If we were considering numbers from 1 to 6, the probability would be 1/6, but since we are going up to 100, which isn't a multiple, it's slightly different. So, in general, if N is a multiple of 6, then the number of multiples of 6 would be exactly N/6. But when N isn't a multiple of 6, it's floor(N/6). Therefore, for N=100, floor(100/6)=16, so the probability is 16/100, not exactly 1/6. Therefore, the exact answer is 17/50. So, the mistake was in assuming that the probability is exactly 1/6, but actually, since we are dealing with finite numbers up to 100, which isn't a multiple of 6, the exact count is needed. Therefore, 17/50 is the correct probability.

Let me check with another approach. Let me list out the numbers divisible by 2 and then check how many are not divisible by 3.

Numbers divisible by 2 from 1 to 100: 2, 4, 6, 8, 10, ..., 100. There are 50 numbers. Now, among these, the ones divisible by 3 are those numbers that are multiples of 6, as mentioned. So, starting from 6, 12, 18,..., 96.

Let me count them again. 6*1=6, 6*2=12,..., 6*16=96. So, 16 numbers. Therefore, the numbers divisible by 2 but not 3 are 50-16=34. So, 34 numbers. Therefore, probability is 34/100=17/50. So, that's confirming again.

But maybe let me do another sanity check. Let's take a smaller set where I can manually count. Let's say numbers from 1 to 6. How many are divisible by 2 and not by 3?

Numbers: 1,2,3,4,5,6.

Divisible by 2: 2,4,6.

From these, divisible by 3: 6. So numbers divisible by 2 but not 3: 2,4. So 2 numbers. Therefore, probability is 2/6=1/3.

But according to the formula: total numbers divisible by 2 is 3. Numbers divisible by both 2 and 3: floor(6/6)=1. Therefore, 3 - 1 = 2. So probability is 2/6=1/3. Which matches.

If I use the same formula for N=6, then 1/2 - 1/6=1/3. Which matches. So in this case, even though N is a multiple of 6, it works.

But when N is not a multiple of 6, like N=7. Let's check.

Numbers from 1-7. Divisible by 2: 2,4,6. Divisible by 3: 3,6. Divisible by both: 6. So numbers divisible by 2 but not 3: 2,4. So again 2 numbers. Probability is 2/7. According to formula: numbers divisible by 2: floor(7/2)=3. Numbers divisible by 6: floor(7/6)=1. So 3-1=2. 2/7. If I use probability formula: 1/2 - floor(7/6)/7 = 1/2 - 1/7 = 5/14 ≈ 0.357... But actual probability is 2/7 ≈ 0.2857. Wait, so here the formula 1/2 - 1/6 is not correct. But that's because in reality, the numbers divisible by 6 up to N is floor(N/6). Therefore, in the general case, the probability is (floor(N/2) - floor(N/6))/N. So, for N=7, floor(7/2)=3, floor(7/6)=1, so (3-1)/7=2/7.

But if I try to use fractions, like 1/2 - 1/6=1/3≈0.333, which would be different. So, in reality, when N is not a multiple of 6, the exact count requires floor functions. Therefore, in our problem with N=100, since 100 is not a multiple of 6, we need to calculate floor(100/2) - floor(100/6)=50 - 16=34. Therefore, 34/100=17/50.

Therefore, that seems correct.

Alternatively, if N was a multiple of 6, say N=6, then floor(6/2)=3, floor(6/6)=1, so 3-1=2. Which is correct. For N=12, floor(12/2)=6, floor(12/6)=2, so 6-2=4. Let's check numbers 1-12. Numbers divisible by 2: 6 numbers (2,4,6,8,10,12). Numbers divisible by both 2 and 3: 6, 12. So 2 numbers. Therefore, numbers divisible by 2 but not by 3: 6-2=4. Which are 2,4,8,10. Correct. Probability is 4/12=1/3. Which is same as 1/2 - 1/6=1/3. So in this case, since N is multiple of 6, floor(N/6)=N/6, so the formula works. So in general, when N is a multiple of 6, the formula 1/2 - 1/6=1/3 gives the exact probability, but when N isn't, we need to use floor functions.

Therefore, in our problem, since N=100 is not a multiple of 6, we need to compute exactly numbers divisible by 2, subtract numbers divisible by 6, and get 34/100=17/50.

Alternatively, perhaps there's a way to compute it by considering the residues. Let me think. For numbers from 1 to 100, each number can be categorized based on their divisibility by 2 and 3.

Since we are dealing with divisibility by 2 and 3, we can consider modulo 6, since 6 is the LCM of 2 and 3.

The residues mod 6 are 0,1,2,3,4,5.

Numbers divisible by 2 are those congruent to 0 or 2 or 4 mod 6. So, residues 0,2,4.

Numbers divisible by 3 are those congruent to 0 or 3 mod 6.

Therefore, numbers divisible by 2 and not by 3 are residues 2 and 4 mod 6.

Therefore, in each block of 6 numbers, there are 2 numbers that are divisible by 2 but not by 3. For example, in numbers 1-6, the numbers 2 and 4 are divisible by 2 but not by 3. Similarly, in numbers 7-12, numbers 8 and 10, etc. So, in each group of 6 numbers, there are 2 such numbers.

Now, total numbers from 1 to 100. How many complete blocks of 6 are there? 100 divided by 6 is 16 with a remainder of 4. Because 16*6=96, so 100-96=4. Therefore, there are 16 complete blocks and then 4 extra numbers (97, 98, 99, 100).

In each complete block of 6 numbers, there are 2 numbers that are divisible by 2 but not by 3. Therefore, 16 blocks contribute 16*2=32 numbers.

Now, we need to check the remaining 4 numbers: 97, 98, 99, 100. Let's check which of these are divisible by 2 and not by 3.

- 97: odd, so not divisible by 2.
- 98: divisible by 2 (since even). Check if divisible by 3: 9+8=17, 17 isn't divisible by 3. Therefore, 98 is divisible by 2 and not by 3.
- 99: odd, so not divisible by 2.
- 100: divisible by 2. Check divisibility by 3: 1+0+0=1, not divisible by 3. So 100 is divisible by 2 and not by 3.

Therefore, in the remaining 4 numbers, there are 2 numbers (98 and 100) that are divisible by 2 and not by 3.

Therefore, total numbers divisible by 2 and not by 3 is 32 + 2 = 34. Therefore, same as before. So, 34 numbers. Therefore, probability is 34/100=17/50. Therefore, this method also confirms the same answer.

Therefore, regardless of the method I use—counting directly, using inclusion-exclusion, or breaking into residues mod 6—the answer comes out to 34/100, which simplifies to 17/50.

I think that's thorough. Let me check once more for any possible mistake. For example, maybe I miscalculated the number of numbers divisible by 6?

Wait, numbers divisible by 6 up to 100: 6*1=6, 6*2=12,...,6*16=96, and 6*17=102 which is over 100. So, yes, 16 numbers. Therefore, 50-16=34. Correct. Also, in the mod 6 approach, 16 blocks of 6 numbers, each contributing 2 numbers, gives 32, plus 2 more in the last partial block, total 34. Correct. So, seems solid.

Another way: Let me compute the total numbers divisible by 2, not by 3. So, for any number, if it's even (divisible by 2), then it's not divisible by 3 if it's not divisible by 3. So, for the even numbers, the probability that they are not divisible by 3 is equal to 1 minus the probability that an even number is divisible by 3. Since numbers divisible by 3 among even numbers are those divisible by 6. So, in the set of even numbers, what fraction is divisible by 3?

There are 50 even numbers. How many are divisible by 3? As we saw, 16. So, the fraction is 16/50. Therefore, the fraction of even numbers not divisible by 3 is 1 - 16/50 = 34/50 = 17/25. Wait, but the probability over the entire set is 34/100, which is same as 17/50. Wait, why is there a discrepancy here?

Wait, hold on. If I consider the set of even numbers, the probability that a randomly selected even number is not divisible by 3 is 1 - (number divisible by 3)/50 = 1 - 16/50 = 34/50 = 17/25. However, since the entire set is 100 numbers, the probability that a number is even and not divisible by 3 is (number of such numbers)/100 = 34/100 = 17/50. Therefore, there's no discrepancy. 17/25 is the conditional probability given that the number is even, whereas 17/50 is the probability over the entire set. Therefore, correct.

So, that's consistent. So, no issue there. Therefore, all the different methods lead to the same answer.

Hence, I can be confident that the probability is 17/50.

**Final Answer**
The probability is \boxed{\dfrac{17}{50}}.

To determine the probability that a randomly selected integer from the set $\{1, 2, 3, \ldots, 100\}$ is **divisible by 2 and not divisible by 3**, we follow a structured approach using concepts from number theory:

---

### Step 1: Total Number of Integers

The set has:
$$
100 \text{ integers}
$$
So, the **total number of possible outcomes** is $100$.

---

### Step 2: Count of Numbers Divisible by 2

Numbers divisible by 2 in the range from $1$ to $100$ are the even numbers:
$$
2, 4, 6, \ldots, 100
$$
This is an arithmetic sequence with first term $2$, common difference $2$, and last term $100$.

Using the formula for the number of terms in such a sequence:
$$
\text{Number of terms} = \frac{\text{Last term} - \text{First term}}{\text{Common difference}} + 1 = \frac{100 - 2}{2} + 1 = 50
$$
So, there are **50 integers divisible by 2**.

---

### Step 3: Count of Numbers Divisible by 2 **and** 3

Numbers divisible by both 2 and 3 are divisible by the **least common multiple (LCM)** of 2 and 3, which is:
$$
\text{LCM}(2,3) = 6
$$
So, we now count how many numbers from $1$ to $100$ are divisible by $6$.

These are:
$$
6, 12, 18, \ldots, 96
$$
This is also an arithmetic sequence with first term $6$, common difference $6$, and last term $96$.

Number of terms:
$$
\frac{96 - 6}{6} + 1 = \frac{90}{6} + 1 = 15 + 1 = 16
$$
So, there are **16 integers divisible by both 2 and 3**.

---

### Step 4: Count of Numbers Divisible by 2 and **Not** by 3

To find how many numbers are divisible by 2 **but not** by 3, subtract the count of those divisible by both from the count of those divisible by 2:
$$
\text{Numbers divisible by 2 but not by 3} = 50 - 16 = 34
$$

---

### Step 5: Compute the Probability

The probability is given by:
$$
\text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{34}{100}
$$

Simplifying:
$$
\frac{34}{100} = \frac{17}{50}
$$

---

### Final Answer

$$
\boxed{\dfrac{17}{50}}
$$

Solution: Since $100 = 50\cdot 2$, there are 50 integers in the set that are divisible by 2.  The numbers among these that are also divisible by 3 are the multiples of 6 in the set.  Dividing 100 by 6 gives $16\frac23$, so there are 16 multiples of 6 in the set, which leaves $50-16 = 34$ multiples of 2 that are not also multiples of 3.  There are 100 numbers in the set so the desired probability is $\dfrac{34}{100} = \boxed{\dfrac{17}{50}}$.