Answers to "cars on the road" 1-(1-609/625)^(1/(20/5))=92% Daniel Tuesday, April 8, 2003 Break down the 20 minutes into four 5 minute periods. Assume they are independent. The probability of no cars in a 20-minute period is the probability of zero cars in each period to the fourth degree. But this number we have, and it is 16/625. Hey - - - This is (2/5) to the fourth degree. Therefore the probability of no car in each 5-min period is 0.4, and the number you're looking for is 0.6. Sandro Perugio Tuesday, April 15, 2003 609/625 cars in 20 minutes reads: In a 20 min period an estimated 609 cars will pass this point of the possible 625. So in a 1 min 609/20 = 30 cars/min Now in 5 mins 30*5min = 120  and (625/20) *5 = 156 In a 5min period an estimated 120 cars will pass this point of the possible 156 cars  so the ratio is 120:156 vrkelley Thursday, June 19, 2003 How does the probablility changes depends on time. Even if you obsert for 5 mins or 1 Hr. The probability to observe a car is same. 609/625 ravi Friday, July 4, 2003   Fog Creek Home