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
re(2): Autosuggestion by ThoughtProvoker)
The Poisson distribution is the standard one for arrivals of independent events such as telephone calls. If we do assume Poisson for this problem then the probability that, in time t, k cars have passed by is given by (Lt)^k*exp(Lt)/k! where L is to be determined from the data of the problem. The probability that at least one car has passed in time T is clearly 1exp(LT). Hence 20L is determined by exp(20L) = 16/625, so that L = (1/20)log(16/625) and then using this L we compute the probability that at least one car has passed by in 5 minutes as 1exp(5L) which works out to be 1 minus the 4th root of 16/625 or 3/5 (surprise!).

Posted by Richard
on 20040409 14:28:47 