If the probability of failure of a single drive is 0.000001 per 1,000 hours, what is the probability of failure if exactly:
a. one drive in an array of four fails?
b. two drives in an array of four fails?
a. In order for one failure to occur, three drives must not fail and one drive must fail. If p is the
probability of failure, 1 ? p is the probability of not failing. Therefore, the chance of one failure is p x (1 ?
p) x (1 ? p) x (1 ? p) = p(1 ? p)3. However, as any one drive out of four can fail, there are 4C1 ways = 4!/3!
= 4 of suffering one failed disk out of four.
The probability of failure is 4 p(1 ? p)3 or 0.000004 per 1,000 hours.
b. For two failures to occur two drives must fail and two must not fail. The probability of two failures and
two non?failures is p2(1 ? p)2
In this case, there 4C2 ways = 4 x 3 x 2/2 = 6 ways of suffering two failed disks out of four. The
probability is approximately 6 x 10?12.
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