I am just confused about all the formulas about exponential distribution and poisson distribution. Here's the problem: Average number of hits each minute is 0.5, what is the probability that the waiting time for the next hit is more than 3 minute?
did has given the answer with little explanation.
If you have a Poisson distribution with mean $\lambda$ then the probability of no hits in the time interval is $\exp(-\lambda)$ and so the probability of no hits in $t$ time intervals is $(\exp(-\lambda))^t = \exp(- \lambda t)$.
That explanation was based on $t$ being an integer. An alternative approach would be to say that if the mean is $\lambda$ in a particular time interval then the mean in a time interval $t$ times as long (with $t$ being a non-negative real number) is $\lambda t$ and the probability of no hits in the new time interval is $\exp(- \lambda t)$.
A third approach would be to use the exponential distribution with rate parameter $\lambda$. The cumulative distribution function is $1-\exp(-\lambda t)$ for the probability of a hit by time $t$. You want the complement of this $ \exp(-\lambda t)$.
These all give the same answer to your question, namely $$\exp(- 0.5 \times 3) = \exp(-1.5) \approx 0.223.$$