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
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 0.40000
1 0.36652
2 0.16792
3 0.05129
4 0.01175
5 0.00215
6 0.00033
7 0.00004
8 0.00000
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Posted by Charlie
on 2004-04-09 13:38:18 |
re: Auto-suggestion
