Question about probability distributions I've recently came across this question:
You are trying to hitch-hike. Cars pass at random intervals, at an average rate of 1 
per minute. The probability of a car giving you a lift is 1%. What is the probability
that you will still be waiting after one hour?

My first thought was to use the Binomial distribution. After one hour an average of 60 cars will pass, so the probability of all 60 cars not giving a lift is $\mathcal{P}_{binom}(60;0.999,60)=0.999^{60}=0.547$.
To check this I then retried the calculation using the Poisson distribution, If cars pass at a rate of 1 per minute then $\lambda=0.01\times60=0.6$, so the probability of not getting a lift is $\mathcal{P}_{pois}(0,\lambda=0.6)=e^{-\lambda}=0.549$. 
Both answers are very close to each other, but why aren't they exactly equal?

Because they model two different things.
A Binomial Distribution, $\mathcal{Bin}(n, p)$, is that the count of successes in of $n$ Bernoulli trials with each trial having independent and identical probability of success $p$.   The count of such successes can range from $0$ to $n$.
A Poisson Distribution, $\mathcal{Pois}(\lambda, \Delta t)$ is that of the count of events occurring within a fixed interval, $\Delta t$, if the subsequent events occur independently of the span since their prior event at a constant rate $\lambda$ per unit interval.   The count of such events can range from $0$ to indefinite; no upper bound.
Only for large $n$ do the two intervals approach the same 'shape'.
