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You purchased 11 hard drives, but 3 of them are defective. Now, you decide to use...

You purchased 11 hard drives, but 3 of them are defective. Now, you decide to use the drives in a configuration which creates an exact copy (or mirror) of a set of data on two or more disks. What is the chance that you are not going to lose any data prematurely if you use two, three, and four of the drives i.e. what’s the probability for each of those setups?

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Answer #1

Total number of hard drives=11
Number of defective hard drives=3
Number of undefective hard drives=11-3=8

a)
2 hard drives are used. We ll lose data if we use both the defective hard drives simultaneously.
So the probability of not losing the data is
1-益 = 52/55 = 0.945

b)
3 hard drives are used. We ll lose data if we use all the defective hard drives simultaneously.
So the probability of not losing the data is
1- - 164/165 -0.0939

c)
If we use 4 hard drives simultaneously then in nohow we'll be losing data as there is only 3 defective hard drives.
Hence the probability of not losing data is 1.

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