for 0 <= x = 1? Which can be proven (similarly) by noting that f(x) = (1-x)^r – 1 + rx, then f’(x) >= 0 for all 0 <= x = 0 -> f(x) >= 0 (by mean value theorem).

With the above, x = C^{-n}, r = C^{-m+n}, and you recover your result (only if C^n >= C^m which is to say r>=1).

]]>sum(n * P(there are exactly n events))

i.e. since the number of events is discrete, it is

0 * P(exactly 0 events) + 1 * P(exactly 1 event) + 2 * P(exactly 2 events) + …

This is greater than

1 * P(exactly 1 event) + 1 * P(exactly 2 events) + …

which is equal to P(>=1 event)

]]>If you have 100 events which occur, each with probability 10^-9, the average number of them which are occuring at any point in time is 100 * 10^-9 = 10^-7; so you immediately have that the probability that one or more is occuring is less than or equal to 10^-7.

I believe you are computing the *expected value* here—for example, if I flip a fair coin three times, I expect to see heads 3 * 0.5 = 1.5 times. The *probability* of one or more heads is computed by .

This post was mentioned on Twitter by donsbot: Lee Pike’s post on one-in-a-billion critical system failures, http://leepike.wordpress.com/2010/01/24/10-9/…

]]>No logarithms required. (Also, using this argument you don’t need to make the assumption that failures are independent of each other, which you implicitly do.)

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