A parking spot is unoccupied 1/3 of the time... But, you find it empty for nine consecutive days in a row.

Find the probability that it will be empty on the tenth day.

Read more: http://www.businessinsider.com/8-mind-bending-interview-questions-that-google-asks-its-engineers-2012-7?op=1#ixzz2zVa8Qb8w

The answer is 2/3 but I can't figure out how that is possible. It should be 1/3 instead.

  • $\begingroup$ answer should be 1/3 $\endgroup$ Apr 9, 2016 at 12:37

4 Answers 4


After actually reading through that article (a painful experience considering that the author clearly doesn't know enough about the subject material to write precise explanations or complete descriptions of the questions), it is obvious that he got the wording mixed up. He would have meant either "the space is occupied $1/3$ of the time" or "the answer is still $1/3$."

But I would take this a step further: there is an underlying assumption being made that the probability of the space being occupied on any given day is independent of whether it was occupied on a previous day. The expected value of the proportion of days on which it is unoccupied may be $p$, but if individual days are not independent of each other (which is a valid model, since for example, weekends might be different than weekdays), then the conditional probability is no longer memoryless and you might use a Markov chain model, for instance.


It must be a typo, since the source says "still 2/3." It should say "still 1/3." The interview question is testing whether you fall for the gambler's falacy.


The question was taken from this site, which says

What is the probability of finding the parking lot empty?

The probability of finding the parking slot occupied is 1/3. You find it empty for 9 consecutive days. Find the probability that it will be empty on the 10th day.

Rather than the source which says

A parking spot is unoccupied 1/3 of the time...

So the site didn't get the answer wrong so much as the question itself.

  • $\begingroup$ Nicely researched! $\endgroup$
    – hardmath
    Apr 21, 2014 at 11:19
  • $\begingroup$ Which of course raises the question if you're just going to straight lift material from another site, why bother editing it? Ctrl+C, Ctrl+V. $\endgroup$
    – Kyle Hale
    Apr 21, 2014 at 18:40

The answer is 2/3

The problem is defined as follows: if the probability of being occupied for one day is 1/3, then P(C) = 1/3, thus the probability of not being occupied is P(~C) = 2/3, let us make it easier by naming if P(F) = P(~C) {probability of being free}

The probability of being not occupied (free) for n consequence days is P(Fn) = (2/3)^n

i.e. Probability of being free for 2 consequence days = 4/9, for 10 consequence days = (2/3)^10, and for 9 days = (2/3)^9

Now the problem asks about the probability of being free for the 10th day GIVEN that it is free for 9 consequence days. that is:

P(F_10 | F_9) = P(F_10)/P(F_9) = (2/3)^10 / (2/3)^9 = 2/3

Please let me know if you find a mistake in my answer.


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