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 6, 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 9, 2005
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