Redundant array of inexpensive discs(RAID) is a technologythat uses multiple har

Question

Redundant array of inexpensive discs(RAID) is a technologythat uses multiple hard drives to increase the speed of datatransfer and provide instant data backup. Suppose that theprobability of any hard drive failing in a day is .001 and thedrive failures are independent. A) suppose you implement a RAID 0 scheme that uses two harddrives each containing a mirror image of the other. what is theprobability of data loss? Assume that data loss occurs if bothdrives fail within same day? B) suppose you implement a RAID 1 scheme that splits thedata over two hard drives. what is the probability of dataloss? Assume that data loss occurs if at least one drive failswithin the same day? Redundant array of inexpensive discs(RAID) is a technologythat uses multiple hard drives to increase the speed of datatransfer and provide instant data backup. Suppose that theprobability of any hard drive failing in a day is .001 and thedrive failures are independent. A) suppose you implement a RAID 0 scheme that uses two harddrives each containing a mirror image of the other. what is theprobability of data loss? Assume that data loss occurs if bothdrives fail within same day? B) suppose you implement a RAID 1 scheme that splits thedata over two hard drives. what is the probability of dataloss? Assume that data loss occurs if at least one drive failswithin the same day?

Explanation / Answer

a.) If the events are independent of each other then the probability oftwo drives failing is .001x.001 which is .000001. b.) Since their are only two events here crash or no crashthe probability of crashing is .001 and not crashing is1-.001=.999 Here is all the possible outcomes and their probabilities, if wehave listed all possibilities and done our math correctly then theprobability of all events should add up to 1. First DriveCrashes            Second Drive Doesn't Crash    All datalost   .001x.999=.000999 First DriveCrashes            Second DriveCrashes            All data lost    .001x.001=.000001 First Drive Doesn't Crash   Second Drive DoesCrash        All datalost     .999x.001=.000999 First Drive Doesn't Crash Second Drive Doesn'tCrash     No data lost    .999x.999=.998001 Total                                                                                                                       = 1 So you can see that you have two ways to solve this. You caneither add up all the ways of getting at least one drive to crashwhich involves adding up 3 different probabilities which gets you.000999+.000001+.000999= .001999. Or you can look it from theopposite perspective. If you know that their are only twooutcomes crash or don't crash and that together the probability ofthose two outcomes add up to be 1 then you could just figure outthe probability that they don't crash and subtract it from 1. Most often in at least problems the best way to solve is to to take1-(the opposite event). In this case we are asking what isthe probability that at least one will crash. Well the bestway to answer this is to say what is one minus the probability thatthey both will not crash (the opposite event) since their is onlyone way they both will not crash. This would be 1-(.999x.999)=.001999

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