Let's say, that we know, that in some period of year there are really common sightings of meteors with an average of 100 meteors per hour. What's the chance of no meteor seen in 5 minutes?

I was trying to compute it like this:

100 meteors per hour = 0.6 minute per 1 meteor on average

Also, falling of meteors is approximately a Poisson process, so I'd choose an exponential distribution to compute the probability of time elapsed before the first event occurs.

So, let's have the time elapsed until the meteor fall $x$. Then

$$ \begin{align} E(X) &= \frac{1}{\lambda} = 0.6\\ \lambda &= \frac{1}{0.6}\\ F(X) &= 1-e^{-\lambda x} = 1 - e^{-\frac{x}{0.6}}\\ F(x > 5) &= F(\infty) - F(5) = 1 - \left( 1 - e^{-\frac{5}{0.6} } \right) = e^{-\frac{5}{0.6}} \approx 0.00024 \end{align} $$

As can be seen, the probability of no meteor being seen in the first 5 minutes is not even one tenth of a percent, which I find to be really strange. Am I doing something wrong?

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    $\begingroup$ If you know that the average number of meteors per hour is 100, you could calculate the average number of meteors in 5 minutes as $100/60\times5$ and then calculate the $P(X=0)$. $\endgroup$ – RScrlli Apr 8 '19 at 21:32
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    $\begingroup$ It coincides with your calculations though. $\endgroup$ – RScrlli Apr 8 '19 at 21:34
  • $\begingroup$ @RScrlli Yes, I've just checked it :-) Well, it seems, that the task is just not very realistic... $\endgroup$ – Eenoku Apr 8 '19 at 21:36

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The average rate is $100/12$ per five-minute interval.

$$P(100/12;0) = \frac{1}{e^{25/3}}$$

  • $\begingroup$ Which is exactly the same number as mine... Well then :-) $\endgroup$ – Eenoku Apr 8 '19 at 21:36

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