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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?

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The problem is not clear. Is the waiting time between hits iid exponential? –  Gautam Shenoy Dec 16 '12 at 6:36
I think it's just the first hit. –  Andy Dec 16 '12 at 7:00
e^(-1.5). $ $ $ $ –  Did Dec 16 '12 at 9:16
how do you get that? where does 1.5 come from? Do we need integration? –  Andy Dec 16 '12 at 18:08
Andy: When are you going to stop putting questions on the site without any explanation about where you are stuck, what you tried, what you know? –  Did Dec 16 '12 at 18:55
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1 Answer

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 $$\exp(- 0.5 \times 3) = \exp(-1.5) \approx 0.223.$$

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