Fog Creek Software
Discussion Board




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 06, 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 09, 2005

*  Recent Topics

*  Fog Creek Home