If the probability of observing a car (read: at least one car) in 20 minutes on a highway is 609/625, what is the probability of observing a car (read: at least one) in 5 minutes (assuming constant default probability)?
(In reply to
re: Autosuggestion by Charlie)
If the only deviation from independence is assumed to be that cars will follow the 2second rule and that there is only one lane in each direction, then a passing spot takes 1 second, leaving 300 finite intervals in a 5minute time span.
The probability of a given second having no car pass is then the 300th root of .4, and the probability that this second does have a car pass is 1 minus that. Let the former be called p0 and the latter be called p. Then the binomial distribution applies, where the probability of n cars passing during the 300 1second intervals is:
p^n * p0^(300n) * Combin(300,n)
Tabulated for different values of n:
0 0.400000000000
1 0.367076588845
2 0.167870089173
3 0.051008643198
4 0.011585523713
5 0.002098034981
6 0.000315517149
7 0.000039459216
8 0
which are close to the Poisson approximation.
Edited on April 10, 2004, 2:18 pm

Posted by Charlie
on 20040410 14:17:42 