imc

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 $\frac{34}{100}=\frac{17}{50}\Rightarrow \boxed{\frac{17}{50}}$