//Comment
This technique is wrong because you are only solving for if all fail on exactly one day in one year. You are asked to solve for the probability that all fail at least once. So solve for the probability that all do not fail in one year and then take 1-P(all do not fail) to get the probability that they will all fail at least once in the year. (See other comments for help with this)
-Ken Pesyna
//End Comment
//As written in a previous comment: Someone tell me if i am thinking about this wrong.
1) .001^k = p[total outage on a given day]
2) 365*0.001^k = p[total outage at least once during the year]
3) we want 365*0.001^k to be less than or equal to 0.001
4) 365*0.001^k = 0.001
5) k=1.8541
//
