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