The Old techInterview Forum - cars on road

cars on 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)?

Abyss
Wednesday, June 22, 2005

let x be the probability of seeing the car in 5 mins. we assume constant default probabilty so for every 5 mintue interval the distribution is the same

probability of seeing the car in 20 mins =

probability of seeing the car in first 5 mins or
probability of not seeing the car in first 5 minsinterval but in the next 5 mins (10 minute) or
probablity of not seeing the car both in the first and second 5 mins interval and seeing it in the third 5 mins(15 minute)
or
probabilty of not seeing it in first, second and third 5 minute and seeing it in the fourth 5 minute interval(20 mins)

so
x + (1-x)x + (1-x)(1-x)x + (1-x)(1-x)(1-x)x = 609/625

solving the above cubic equation for values x < 609/625

Abyss
Wednesday, June 22, 2005

3/5. You have correct equation, but there is a way to get to the answer without using a pen.

DK
Wednesday, June 22, 2005

yeh i see it.that is by taking the converse right

probability of not seeing the car in 20 mins = (probability of not seeing the car in 5 mins)^ 4

as we can assume constant default probability

probability of not seeing tha car in 5 mins = (1 - (609/625))^ (1/4) = 2/5

so probability of seeing the car in 5 mins = 1 - (2/5) = 3/5

Abyss
Wednesday, June 22, 2005