cars on a road  solution
if the probability of observing a car in 20 minutes on a highway is 609/625, what is the probability of observing a car in 5 minutes (assuming constant default probability)?
Solution:
the probability of NOT observing a car in 20 minutes on a highway is 1 609/625 = 16/625.
Assume the probability of not observing a car in 5 minutes
is x/5, and 4 5minute in a 20minute, so
the probability of not observing a car in 20 minutes
is (x/5)^4 = x^4/625 = 16/625 ==> x^4 = 16
==> x=2.
Thus, the probability of not observing a car in 5 minutes
is 2/5, the probability of observing a car in 5 minutes
is 1 2/5 = 3/5
Pan, Wenyu
Sunday, September 22, 2002
I could be quite wrong here, but I'm afraid I did not follow that at all. Why is the simple solution wrong?
Change the problem for a second. If a car goes by every 2 hours, then the odds of a car going by each hour is 1 in 2 (1/2). One car will go by in eight quarterhours, so the odds of a car passing in a quarter hour is 1/(2*4)==1/8.
Now reuse the logic. (Also let's be hockey players for a second and call a 20minute span a period.) If the odds of a car passing in a given period is (609/625) then isn't the odds of a car passing in a quarterperiod (5 minutes) ==(609/(625*4))==(609/2500)?
What am I missing? Why is the negative possibility easier to calculate?
Ran
Wednesday, October 30, 2002
You've got a couple of problems with this.
The probability of two unrelated occurrences is multiplicative not additive (in your example the probability of seeing a car in 40 minutes would be 1218/625 !!).
The original statement doesn't restrict the number of cars that pass in a period to one. If you pick a random 20 minute period the probability that you will see a car pass in that period is 609/625 but you may, in fact, see 34 or 192 cars pass in that period.
The reason you use the negative case is due to the fact that there is one probability of seeing zero cars in a random period so this probability can be multiplied by itself for longer periods. But the probability of seeing a car in this period is, in fact, an infinite sum over the probabilities of seeing a particular number of cars (i.e. probability of 1 car + probability of two cars + etc.) so in order to expand this to cover a longer period you would need to calculate the infinite sum of all the possible product combinations of probabilities.
Your solution would be correct if the statement refered to the probability of seeing one car in the 20 minute period and that it was not possible to see more than one. But those restrictions mean that the probabilities for the 5 minute periods are no longer unrelated.
SM
Sunday, November 3, 2002
I think this problem can be solved as a geometric distribution:
p = probability of seeing a car in one 5minute time interval
q = probability of not seeing a car in one 5minute time interval
where p = 1 – q
p(y<=4) = probability of seeing a car in four 5minutes time intervals (20 minutes)
Find p given that p(y<=4) = 609/625
p(y<=4) = p(1) + p(2) + p(3) + p(4)
= q^0*p + q^1*p + q^2*p + q^3*p
= p*(q^0 + q^1 + q^2 + q^3)
= (1  q)*(q^0 + q^1 + q^2 + q^3)
= (q^0 + q^1 + q^2 + q^3) –
(q^1 + q^2 + q^3 + q^4)
= 1  q^4
= 609/625
1 – q^4 = 609/625
q^4 = 1 – 609/625 = 16/625
q = 2/5
p = 1  q = 1  2/5 = 3/5
CM
Wednesday, November 6, 2002
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