Sorry, dude... For A and B not mutually exclusive, P(A or B) = P(A) + P(B). No, the individual drives are not more likely to fail, but the "effective drive", the array, is twice as likely to fail. Just probability...Originally posted by: JKing76
But the likelyhood of a failure occuring does not increase.
Originally posted by: cleverhandle
Sorry, dude... For A and B not mutually exclusive, P(A or B) = P(A) + P(B). No, the individual drives are not more likely to fail, but the "effective drive", the array, is twice as likely to fail. Just probability...Originally posted by: JKing76
But the likelyhood of a failure occuring does not increase.