Assume the lifespan of hard disks are independent and identically distributed.

Let the probability of lifespan of a hard disk be P(t)

Let the CDF of P(t) be C(t)

CDF of minimum of 2 i.i.d. random variable is 1-(1-C(t))^2

And then differentiate the above w.r.t. t, we have

PDF of RAID-0 is 2*(1-C(t))*P(t)

So, for any distribution of lifespan P(t), the RAID-0 comparing to single hard disk is always 2*(1-C(t)) times more likely to fail.

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To illustrate the difference, let further assume P(t) is Normal distribution with mean = 5 years and s.d. = 1 years

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