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Cars on the road (Posted on 2004-04-09) Difficulty: 2 of 5
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)?

See The Solution Submitted by SilverKnight    
Rating: 2.5000 (4 votes)

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Alternative to Poisson | Comment 17 of 20 |
(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
  Posted by Charlie on 2004-04-10 14:17:42

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