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 puzzle solution
by K Sengupta)
Alternatively, let the probability that at least one car was observed in 5 minutes.
Then the probability that one no car was observable in 20 minutes was 1-p, so that the probability that no car would be seen in 20 minutes would correspond to (1-p)^4.
Hence probability of observing at least one car is 1 - (1-p)^4
By the problem:
1 - (1-p)^4 = 609/625
Or, (1-p)^4 = 16/625
Or, 1-p = 2/5
Or, p = 3/5 = 0.6
Thus the required probability is 0.6