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12 July, 00:43

Your office parking lot has a probability of being occupied of 1/3. You happen to find it unoccupied for nine consecutive days. What are the chances that you find it empty on the 10th day as well?

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  1. 12 July, 01:10
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    0.4

    Step-by-step explanation:

    Let X be the random variable that represents the number of consecutive days in which the parking lot is occupied before it is unoccupied. Then the variable X is a geometric random variable with probability of success p = 2/3, with probability function f (x) = [ (2/3) ^x] (1/3)

    Then the probability of finding him unoccupied after the nine days he has been found unoccupied is:

    P (X> = 10 | X> = 9) = P (X> = 10) / P (X> = 9). For a geometric aeatory variable:

    P (X> = 10) = 1 - P (X <10) = 0.00002

    P (X> = 9) = 1 - P (X <9) = 0.00005

    Thus, P (X> = 10 | X> = 9) = P (X> = 10) / P (X> = 9) = 0.00002 / 0.00005 = 0.4.
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