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
Autosuggestion 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

Posted by Charlie
on 20040409 13:38:18 