cars on the road  solution
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 5minute 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 06, 2002
That's a good solution, but to nitpick a bit  probability is too high to assume there is always exactly one (or zero) car within 20 min, which makes final result to be "p>=3/5"
Although it's probably too complicated for usual technical interview to get that far into math.
DK
Tuesday, December 17, 2002
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