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: Auto-suggestion by Charlie)
If the only deviation from independence is assumed to be that cars will follow the 2-second rule and that there is only one lane in each direction, then a passing spot takes 1 second, leaving 300 finite intervals in a 5-minute 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 1-second intervals is:
p^n * p0^(300-n) * 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
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Posted by Charlie
on 2004-04-10 14:17:42 |
Alternative to Poisson
