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cars on the road

Any OOB answers to this?

Friday, February 13, 2004

The solution to this one just expired a little while ago. You can use either a binomial or a Poissan modal to solve this one. Good Luck. If you want me to I will post the solution.

Wednesday, February 18, 2004

The probability of seeing one or more cars in any interval is one minus the probability of not seeing any.  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 you insist on "exactly one" car, instead of allowing "one or more", then the solution gets a whole lot messier.

Roger Arnold
Thursday, February 19, 2004

Hey cool, Thanks :)

Wednesday, March 31, 2004

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