cars on the road Split the 20 minute observation period into 4 intervals of 5 minutes. Assume a constant default probability of p. Now, repeatedly apply the following two formulas: probability(x or y) = probability(x) + probability (y) - probability(x and y) probability(x and y) = probability(x) x probability(y) Where x (or y) is the observation of a car in 5 minute period. This will give: 4p - 6p^2 + 4p^3 - p^4 = 609/625 Which be can be rearranged as: 1 - (1 - p)^4 = 609/625 Rseolving for p gives: p = 3/5 Sajid Umerji Thursday, December 30, 2004 Oops, I was just watching the new year's fireworks and it just occured to me that there is a simpler solution to this puzzle :-) As before, split the the 20 minute observation period into 4 intervals of 5 minutes and let p be the constant default probability of observing a car in a 5 minute period. The probability that one does not observe a car in 20 minutes is 1 - 609/625 = 16/625 Now, the probability of one does not observe a car in a 5 minute period is (1 - p) And therefore the probability that one does not observe a car in any of the 4 consecutive 5 minute periods (i.e. in 20 minutes) is (1 - p)^4 Therefore (1 - p)^4 = 16/625 Resolving for p gives P = 3/5 Sajid Umerji Friday, December 31, 2004 I think the answe of p=3/5 is only valid if the question is changed to "If the probability of observing AT LEAST a car in 20 mins..." If the question is "If the probability of observing ONLY a car in 20 min...", the answer p = [609/625]/4 Ken Thursday, January 6, 2005 I think conventional "logic puzzles" interpret "a car" to mean "at least one but perhaps more" cars. If you want exactly one, I think the problem is expected to say exactly one. WanFactory Wednesday, March 9, 2005   Fog Creek Home