combined reliability

Question

I thought I knew the answer to this question, but further reflection is showing some holes in my knowledge; my college math is twenty-five years old and google isn't helping today.

Let's say you have a computer with five disk drives. If any one of the drives crashes, the computer is down. For this model of drives, 99% of them run a year without crashing. So the probability that this computer will be up the entire year is

.99 ^{number of drives}

or

.99 ⁵ ≈ .95

That's correct, right?

Now the obverse observation is that each drive has a 1% chance of failure over the course of a year. How does one express the total chance of failure for the whole computer? The answer should be 5%, because that's what's left over from the 95% given above.

The naive approach would be to do the same thing again:

.01⁵

but that gives 0.0000000001 which obviously isn't right. What am I missing here?

That very small number is the probability that all the drives will die. — André Nicolas, Oct 09 '13 at 15:35
$0.99^5$ is slightly more than $0.95$ but this is not important here — Henry, Oct 09 '13 at 15:36

score 2 · Accepted Answer · answered Oct 09 '13 at 15:35

$0.01^5$ is the probability that ALL 5 drives crash.

To see why, consider that an individual drive failing is unaffected by what happens with the other drives, and the probability of any one of them failing is 0.01. So the probability of all of them failing is the product of each probability.

To calculate the probability of failure, you do exactly as you have done, calculate 1 - Probability of not failing.

score 0 · Answer 2 · answered Oct 09 '13 at 15:53

You could do as David suggests, or you might employ a much longer solution if you want to express this probability in terms of, well, some drives crashing.

So, suppose your computer crashes due to the malfunction of drives. What does it mean? It can mean that

1st drive crashed AND others did not

OR

1st AND 2nd drives crashed AND others did not

OR

... and so on...

OR

1st AND 2nd AND 3d AND 4th AND 5th drives crashed.

So the event $A$ = "the computer crashed" is equivalent to the union of independent events of the form, $A_S$ = "each drive from the set $S$ crashed AND other drives did not crash", where $A_S$ is any nonempty subset of your set $T$ of five drives. Hence $P(A) = \sum_{S\subset T,S\neq \emptyset}P(A_S)$. Now what is the probability of each event $A_S$? $P(A_S) = 0.01^{|S|}\cdot 0.99^{5-|S|}$, where $|S|$ stands for the number of elements in $S$. There will be $2^5 - 1$ addants in the sum (this is the number of subsets of a 5-element set minus 1 which stands for the empty set (if $S$ is empty, none of the drives crashed and hence the computer is working - that's not the case we're dealing with)). Theoretically, if you happen to do all the calculations the right way, you'll get these same $100\cdot(1 - 0.99^5)$ percent, but, as i've said earlier, David's way is certainly easier... ;)

combined reliability

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