2003 AMC 10A Problem 15

Start with all the even numbers, then remove the ones that are also multiples of $3$.

1. Even numbers from $1$ to $100$: $2, 4, 6, \ldots, 100$. There are $100 \div 2 = 50$ of them.
2. An even number that is also a multiple of $3$ is a multiple of $6$. Those are $6, 12, 18, \ldots$. Since $6 \times 16 = 96$ and $6 \times 17 = 102$ is too big, there are $16$ of them.
3. Even numbers that are not multiples of $3$: $50 - 16 = 34$.
4. The probability is
$$\frac{34}{100} = \frac{17}{50}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{17}{50}}$.

"Divisible by $2$ but not by $3$" means: take the multiples of $2$, then throw out the ones that are multiples of both, which are the multiples of $6$. The count of multiples of $6$ up to $100$ must be a whole number, $16$, so the answer is not exactly one third of the evens. Whenever the upper limit is not a multiple of the step, count carefully instead of using a fraction.

Multiples of $2$ up to $100$: $\lfloor 100/2 \rfloor = 50$.

Among these, the ones also divisible by $3$ are the multiples of $\operatorname{lcm}(2,3) = 6$: $\lfloor 100/6 \rfloor = 16$ (namely $6, 12, \dots, 96$).

Even numbers that are not multiples of $3$: $50 - 16 = 34$. The probability is $\dfrac{34}{100} = \dfrac{17}{50}$.

The answer is $\boxed{\textbf{(C)}\ \frac{17}{50}}$.