Answer to Question #143155 in Statistics and Probability for Vladimyr Lubin

Question

Assume that there is a ​10% rate of disk drive failure in a year.
a. If all your computer data is stored on a hard disk drive with a copy stored on a second hard disk​ drive, what is the probability that during a​ year, you can avoid catastrophe with at least one working​ drive?
b. If copies of all your computer data are stored on 4 independent hard disk​ drives, what is the probability that during a​ year, you can avoid catastrophe with at least one working​ drive?
a. With two hard disk​ drives, the probability that catastrophe can be avoided is

Answer 1

Probability of failure of the disk is given P_f = 0.1

Hence Probability of non failure or success of the disk P_s = 1 - P_f = 1 - 0.1 = 0.9

a) Probability of avoiding the catastrophe with atleast 1 working drive out of 2 is given by,

Probability that the first drive is working and the Probability that the second drive is not working or the probability that the second drive is working and the first drive is not working or both the drives are working = P_s*P_f + P_f*P_s + P_s*P_s = 0.9*0.1 + 0.1*0.9 + 0.9*0.9 = 0.99

or we can say that there is 99% Probability of avoiding the catastrophe with atleast 1 working drive.

b) Probability of avoiding the catastrophe with atleast 1 working drive out of 4 = 1 - probability that none of the disks are working

= 1 - (0.1*0.1*0.1*0.1) = 1 - 0.0001 = 0.9999

= 99.99% Probability.

c) With two hard disk​ drives, the probability that catastrophe can be avoided is that atleast 1 drive is working, means

Probability that the first drive is working and the Probability that the second drive is not working or the probability that the second drive is working and the first drive is not working or both the drives are working = P_s*P_f + P_f*P_s + P_s*P_s = 0.9*0.1 + 0.1*0.9 + 0.9*0.9 = 0.99

or we can say that there is 99% Probability of avoiding the catastrophe with atleast 1 working drive.