Answers to "cars on the road"
1(1609/625)^(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 20minute 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 5min 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
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