I am reading the book Database Systems the Complete Book 2nd Edition. A slightly modified Question 13.4.5 states:

Suppose we use three disks as a mirrored group; i.e., all three hold identical data. If the yearly probability of failure for one disk is F, and it takes H hours to restore a disk, what is the yearly probability data loss?

My answer is $(F*(F*H/365*24)^2)*6$

I would like to know if I am right. I will explain the reasoning behind my answer:

In order for data loss to occur, all 3 disks must fail within a time period of H hours.

The probability that the first disk fails is $F$. The probability that the second and third disks fail within $H$ hours of the first disk failing is $(F*H/365*24)^2$.

Thus the probability of all disks failing within a time period of H hours is $F*((F*H/365*24)^2)*6$.

The reason we multiply by 6 is because there are 6 ways in which this event could happen:

123 132 213 231 312 321

I have trouble with probability when dealing with events that can occur in multiple ways. So the only part of the reasoning that I am unsure about is the part where I multiply by 6. Are all 3 disks failing one event, or are they 6 different events?


Let us look at the probability of the "123" event happening. That is, Disk 1 fails, and then Disk 2, and then Disk 3, and all of these happen within a span on $H$ hours. Your claim is that the probability of this is $F ∗ (\frac{FH}{365∗24})^2$. But the $\frac{FH}{365∗24}$ term only denotes the probability of a disk failing in the $H$ hours after the first disk fails. In particular, it does not consider the ordering constraint of "Disk 2 fails before Disk 3 fails".

Instead, what $F ∗ (\frac{FH}{365∗24})^2$ denotes is the probability that Disk 1 fails, and in the ensuing $H$ hours, both Disk 2 and Disk 3 fail. So this captures both the events "123" and "132".

Similarly, the probability of either the event "213" or "231" happening would be $F ∗ (\frac{FH}{365∗24})^2$. And similarly for the events "312" and "321" combined.

So the final answer should be $(F ∗ (\frac{FH}{365∗24})^2) * 3$, and not $(F ∗ (\frac{FH}{365∗24})^2) * 6$.


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