Question

You have four hard drives. Two of the drives have a failure rate of 1.75% (pf = 0.0175), and the other two have a failure rate of 2.5%. Assume that failures of the hard drives are independent events. (a) What is the probability of failure of your system? (b) Assuming that your quantities of 1.75% and 2.5% drives remain equal (that is, you have n 1.75% and n 2.5% drives), what is the minimum total number of hard drives your system would have to have to exceed a 50% failure rate?

(Note: failure of the system is defined to be the failure of one of the hard drives.)

Answer 1

(a) We know that the system fails if one or more hard drives fail.

We have to find the probability that the system fails

That is we have to find the probability that one or more hard drive fail

Probability that one or more hard drive fail

= 1- Probability that none of the hard drive fail

= 1- (1-0.0175) *(1-0.0175)*(1- 0.025)*(1-0.025)

=1- 0.9176

= 0.0824

Probability that system fails =0.0824

Note : Probability that a hard drive fails = 0.0175 (with failure rate 1.75%)

Probability that hard drive does not fail = 1-0.0175 =0.9825 (with failure rate 1.75%)

probability that a hard drive fails = 0.025  (with failure rate 2.5%)

Probability that hard drive does not fail = 1-0.025 =0.975 (with failure rate 2.5%)

Two hard drives have failure rate 1.75% and other two 2.5%

As failures are independent , we multiply the four probabilities to get, Probability that none of the hard drive fail  = (1-0.0175) *(1-0.0175)*(1- 0.025)*(1-0.025)

(b) Probability of Failure of system with 2+2 =4 hard drives = 0.0824

We need to find n such that

Probability of Failure of system with n+n =2n hard drives at least  0.5

2n = 32.2

Therefore minimum total number of hard drives =32  

Note : log 0.9579 = -0.01866

log 0.5 = -0.30103