Take P,The probability of not getting connected on any one day = (0.001)$ ^k $
(1-(.001$ ^k $))^365 Serves the condition "atlest"
Therefore 1-(1-(.001$ ^k $))^365 Tells us the proability of total outrage happening "atleast" once a year.
The above expression should be lesser or eual to 0.001....so solve the equation from thereon!!
why 1-.001^k I tought that would give probability of being all on. ///-Andrew Hermann
