 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Cars on the road (Posted on 2004-04-09) 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) Comments: ( Back to comment list | You must be logged in to post comments.) 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       0.4000000000001       0.3670765888452       0.1678700891733       0.0510086431984       0.0115855237135       0.0020980349816       0.0003155171497       0.0000394592168       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 Please log in:

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