Vehicle insurance claims arrive randomly at a certain office at an average rate of $18$ claims per 9 hour working day (from 08:00 - 17:00). Calculate the probability that this office will receive at least 1 claim in each of the 9 working hours on any given day.

My working is as follows:

$$λ = \frac{18 \ \text{claims}}{9 \ \text{hours}} = 2$$ $$\int_0^1 2 \ e^{-2x} \,dx = 0.8656647...$$

I know that the answer is $0.27$ but I can only get that by raising the integral to 9.

$$(\int_0^1 2 \ e^{-2x} \,dx)^9 = 0.2701653...$$

I have no idea how this answer works if anybody could explain to me how one would traditionally answer it (without the to the 9th power) as well as why raising it to the 9th power works that would be much appreciated.


First, compute the probability that no claim in the $1$ hours. Note that it follows Poisson distribution.

$$\exp(-2) \frac{2^0}{0!}=\exp(-2)$$

Now, the probability that at least one claim in an hour would be


Hence arrival in each the hours are independent. Hence the desired probability is


  • $\begingroup$ So are all probabilities found in an exponential pdf independent? $\endgroup$ – Gregory Oct 4 '18 at 10:40
  • 1
    $\begingroup$ yes, each hour is independent. $\endgroup$ – Siong Thye Goh Oct 4 '18 at 10:42
  • $\begingroup$ Would there have been no way to find the answer without raising the (1-expo(-2)) to the 9? I do understand your methodology though. $\endgroup$ – Gregory Oct 4 '18 at 10:48
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    $\begingroup$ you mean another to compute but somehow equal to $(1-\exp(-2))^9$ in the end? no idea for now. $\endgroup$ – Siong Thye Goh Oct 4 '18 at 10:52
  • $\begingroup$ It's fine. Thank you for the assistance! $\endgroup$ – Gregory Oct 4 '18 at 11:06

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