The Old techInterview Forum - Answers to "cars on the road"

Answers to "cars on the road"

1-(1-609/625)^(1/(20/5))=92%

Daniel
Tuesday, April 08, 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 04, 2003