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