A bike has probability of breaking down $p$, on any given day.

In this case, to determine the number of times that a bike breaks down in a year, I have been told that it would be best modelled with a Poisson distribution, with $\lambda = 365\,p$.

I am wondering why it would be incorrect to use a binomial distribution, with $n=365$. After all, isn't Poisson really an approximation of a sum of Bernoulli random variables?


  • $\begingroup$ As I understand, if bernoulli we assume if the bike breaks in a given day, it doesn't break again that day. For poisson he might get it fixed and it break yet again in the same day. $\endgroup$ – JMoravitz Dec 11 '14 at 4:18

Poisson distribution

a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time and/or space if these events occur with a known average rate and independently of the time since the last event.

Binomial distribution

the discrete probability distribution of the number of successes in a sequence of n independent yes/no experiments, each of which yields success with probability p

Emphasis mine

For the Poisson you need a known interval (365 days) and a known failure rate (average failures per day - Note: this can be any number $> 0$). For the Binomial you would need a fixed number of trials (365) and a known failure rate per trial (failure chance on a given day Note: this must be a number $\in [0,1]$).

For the specific question, it is a matter of interpretation and both could be justified here.

The Poisson is more appropriate if it is conceivable that the bike could break on a given day, be repaired and break again (and again etc.). For minor failures this is appropriate.

The Binomial is more appropriate if a failure on a given day takes the bike out for the rest of the day (but not for more than that because it would then reduce the total number of days). That is, a moderate failure.

I know from your earlier question here that this is then combined with a Gamma distributed cost - there is no mention of the time the repair takes. If there were, this would be a fairly typical queuing problem which typically uses Poisson distributions. I must say that it was this that led me towards the Poisson.


It's not incorrect to use a binomial distribution because indeed that is what it would be.

However, it is best modeled as a poison distribution because the calculations are much simpler and the approximation is sufficiently close for large $n$

$$\mathcal{Bin}(n, p) \approx \mathcal {Pois}(np), \mbox{ for }n\to\infty$$


I think the main difference is that the Poisson distribution is used to approximate a very large sample like casuality in the war, fish caught in a big lake, number of traffic accidents, etc. If you review the derivation of Poission distribution, at some step we let $n\rightarrow \infty$ and assume $np\rightarrow \lambda$, which is a constant. At here $n=365$ is a relatively large number, comparing to a typical sample side of $25-100$, so I think using Poisson distribution is justified if $p$ is relatively small.

The wikipedia article on Poission distribution might also helps.


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