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 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)

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 re: Auto-suggestion | Comment 4 of 20 |
(In reply to Auto-suggestion by Penny)

As has been noted, that the solution depends on the independence of successive time intervals, once that assumption has been made, the conversion of a probability of seeing at least one car to the probability of seeing exactly one car just involves fitting an appropriate Poisson distribution to the probability.

The probability of observing n occurrences when t occurrences are expected in an interval is t^n*e^-t/n!

If the probability of seeing exactly zero cars in 5 minutes is 2/5, then

2/5 = e^-t

or -t = ln(2/5)

or t = ln(5/2) = .9162907318741551

Then the probability of observing exactly 1 would be

t*e^-t = .366516292749662

The probabilities of different numbers of cars in 5 minutes then becomes:

`0                      0.400001                      0.366522                      0.167923                      0.051294                      0.011755                      0.002156                      0.000337                      0.000048                      0.00000`

 Posted by Charlie on 2004-04-09 13:38:18

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