0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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 $(F ∗ (\frac{FH}{365∗24})^2) * 6$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.