Suppose a customer service rep gets on average exactly 2 calls per minute. What are the odds of getting 1 call in the next minute?
This problem calls (no pun intended) for the Possion distribution : $$ P\{X = k\} = e^{-2} \frac{2^k}{k!}$$
Here's where I'm stuck wondering. The probability of there being one call $$P\{X = 1\} = e^{-2} \frac{2^1}{1!} \approx .2707 $$ is the same as the probability of getting two calls: $$P\{X = 2\} = e^{-2} \frac{2^2}{2!} \approx .2707 $$
I understand that the calculations show they are the same (obviously), but conceptually, I don't see how they can be the same. And also, why should one have the same probability as the average, but not three (which has a probability of $.1804$). Unless, am I calculating it incorrectly?