   cars on the road - solution I think this problem can be solved as a geometric distribution: p = probability of seeing a car in one 5-minute time interval q = probability of not seeing a car in one 5-minute time interval where p = 1 – q p(y<=4) = probability of seeing a car in four 5-minute 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 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 Recent Topics Fog Creek Home