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If the probability of observing a car in 30 minutes on a highway is 0.95, what is the probability of observing a car in 10 minutes (assuming constant default probability)?
Problem source: 140 Google Interview Questions.
If the probability of observing a car in 30 minutes on a highway is 0.95, what is the probability of observing a car in 10 minutes (assuming constant default probability)?
Problem source: 140 Google Interview Questions.
December 15th, 2009 at 1:23 am
Whatever Googler wrote “assuming constant default probability” should have been flunked right then and there! Constant probability often conjures up a uniform distribution, which would lead the careless student to answer 0.3167, which sounds reasonable until she is asked the follow-up question: “OK, then what is the probability of observing a car in 60 minutes?” Naively she answers 1.90, and hilarity ensues.
The only sensible interpretation to that phrase is that the cars (randomly) pass at some constant RATE, so their probability of being observed can be described by a Poisson model. Using this model, the only feasible event that can be modeled is ONE OR MORE cars passing in a 30 minute interval. This gives a rate of 5.9915 cars per hour. So in 10 minutes, the Poisson probabililty of at least one car passing is approximately 0.6316 ( roughly 1-Exp[-1]).
This question is just poorly-worded enough to qualify as a “think again” problem for my stats students next semester….THANKS!
December 16th, 2009 at 4:09 am
Fish-free method (this is the method I had worked out for myself when I saw the problem on Google’s site, before I saw it this evening).
If the probability of observing a car in 30 minutes on a highway is 0.95, then the probability of not observing any cars in 30 minutes is 0.05. This is the probability of not observing any cars in any of the 3 10-minute periods that make up this 30-minute period. Assuming that the probability is the same for one 10-minute period as for any other, then the probability of not observing any cars in 10 minutes is 0.05^(1/3), and the probability of observing any in 10 minutes is 1 – 0.05^(1/3) ~~ 0.6316.
December 16th, 2009 at 5:29 am
Richard’s being a math fundamentalist (not that there’s anything wrong with that!). Look up lim(n->infinity) (1-k/n)^n to see the link back to Poisson.