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I have recently tackled this question: If the probability of observing a car in 30 minutes on a highway is 0.95, what is the probability of observing a car in 10 minutes (assuming constant default probability)?

While I fully understand the answer, I was wondering if the proposed answered is a good one? What if you've been asked to calculate the probability of seeing a car in 12 or 13 minutes? This will present a problem with the proposed strategy no?

Is there a way to calculate the probability for any given interval?

Another question would be what does it mean "assuming constant default probability" wouldn't that question follow under the poisson distribution?

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The accepted answer gives a good approach. If we model the arrival of cars as a Poisson distribution, the fact that the probability of observing at least one event in $30$ minutes is $0.95$ tells us the probability of observing no events is $0.05$. This gives us that the parameter $\lambda$ of the distribution satisfies $e^{-\lambda}=0.05$ or $\lambda =-\log(0.05) \approx 2.996$ The expected rate of cars to see is then about $\frac {2.996}{30}$ per minute and the expected number to be seen in $12$ minutes is about $\frac {12 \cdot 2.996}{30}=1.198.$ The chance of seeing none in $12$ minutes is then $e^{-1.198}\approx 0.302$

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    thanks for the answer. Can you elaborate why e^-mu is 0.05? thanks! – 2017-02-24
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    That is a property of the Poisson distribution. The chance of no events is $e^{-\mu}$ where $\mu$ is the mean number. – 2017-02-25
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    Thank you for following up. I'm aware that this falls under the Poisson distribution but the equation is not e^-mu (I added the equation in my post). What I'm struggling with is why e^-mu is 0.05 and not (e^-mu*mu^x)/x!=0.05. Thanks again!!! – 2017-02-25
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    Because if we set $x=0$, which corresponds to zero events, we get $e^{-\mu}$ (or $ \lambda$, whichever you use as the parameter). That is why if you know the probability of zero events, you can get the parameter this way. – 2017-02-25
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    ahhhh...got it. Thank you very much! – 2017-02-25