What is the probability of not choosing a number divisible by both 4 and 3?
If a number is divisible by both 4 and 3, then, since 3 and 4 are relatively prime, it must be divisible by 12.
Technically, since natural numbers and multiples of 12 are both countably infinite sets, the probability of choosing a multiple of 12 is undefined. However, if there is a high upper limit on the numbers we can choose, the probability would become approximately $\displaystyle \frac{1}{12}$.