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)?

(The mention of cars in this puzzle, made me think of all the poor dogs who are run over every day by cars. How sad. Someone should take up some donations for these dogs.)

As usual, Ady is correct. The odds of seeing one or more cars in time period T, is 1 minus the odds of not seeing any cars in time period T. The probability of not seeing any car in  20 minutes is the product of the probabilities not seeing a car in four successive 5 minute intervals: 1 - p^4, if 'p' is the probability of non-seeing in 5 minutes.  If 1 - p^4 = 609/625, then p^4 = 16 / 625, so p = 2/5.  The probability of seeing one or more cars in 5 minutes is then 1 - p, or 3/5.

If the problem had asked for the odds of seeing exactly one car in 5 minutes, the problem would have been a lot more difficult.  

Posted by Penny on 2004-04-09 12:01:02