What is the probability that an integer in the set $\{1,2,3,...,100\}$ is divisible by $2$ and not divisible by $3$? 
$\mathrm{(A) \ } \frac{1}{6}\qquad \mathrm{(B) \ }  \frac{33}{100}\qquad \mathrm{(C) \ }  \frac{17}{50}\qquad \mathrm{(D) \ }  \frac{1}{2}\qquad \mathrm{(E) \ }  \frac{18}{25}$
There are $100$ integers in the set. 
Since every $2^{\text{nd}}$ integer is divisible by $2$, there are $\lfloor\frac{100}{2}\rfloor=50$ integers divisible by $2$ in the set. 
To be divisible by both $2$ and $3$, a number must be divisible by $(2,3)=6$.
Since every $6^{\text{th}}$ integer is divisible by $6$, there are $\lfloor\frac{100}{6}\rfloor=16$ integers divisible by both $2$ and $3$ in the set. 
So there are $50-16=34$ integers in this set that are divisible by $2$ and not divisible by $3$. 
Therefore, the desired probability is $\frac{34}{100}=\frac{17}{50}\Rightarrow\boxed{\mathrm{(C)}\ \frac{17}{50}}$