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?

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.

i wish i could u

step-by-step explanation:

i wish i could u