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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) | 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) | ||
Latest revision as of 07:46, 16 September 2008
//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
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
// Comments
I believe k should also be an integer because you cannot have a partial connection between the two cities. You either have the connection or you don't.
-Arie Lyles
