we want to find the number of hours (or rather days) until only 4.25% of the drives fail.
that means for an exponential distribution we have
P(X < x) = 1 - e^(-Lx)
L = 1/mean = 1/12500
and we need to find x, so that P(X <x) = 0.0425 (that is 4.25% converted into a probability).
so, we get
0.0425 = 1 - e^(-x/12500)
0.0425 - 1 = -e^(-x/12500)
-0.9575 = -e^(-x/12500)
0.9575 = e^(‐x/12500)
ln(0.9575) = -x/12500
-x = ln(0.9575)×12500
x = -ln(0.9575)×12500 = - -0.043429558...×12500 =
= 0.043429558...×12500 = 542.8694741... hours
after 542.8694741... hours = 22.61956142... days
it is expected that 4.25% of the hard drives fail.
you left out to what decimal this should be rounded. and if you need hours or days. in case of doubt use hours (as the only other time information is given in hours).
please round yourself as you need.