Question

assume that there is a 5% 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 four 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
(round to four decimal places as needed.)
b. with four hard disk drives, the probability that catastrophe can be avoided is
(round to six decimal places as needed.)

Explanation:

Step1: Find the failure rate of a single drive

The failure rate of a single disk - drive is $p = 0.05$, so the success rate (the drive works) is $q=1 - p=1 - 0.05 = 0.95$.

Step2: Calculate the probability of all - drives failing for part a

For two independent hard - disk drives, the probability that both drives fail is $P(\text{both fail})=0.05\times0.05 = 0.0025$ (using the multiplication rule for independent events $P(A\cap B)=P(A)\times P(B)$).

Step3: Calculate the probability of avoiding catastrophe for part a

The probability of avoiding catastrophe (at least one working drive) is the complement of the event that both drives fail. So $P(\text{at least one working}) = 1 - P(\text{both fail})=1 - 0.0025=0.9975$.

Step4: Calculate the probability of all - drives failing for part b

For four independent hard - disk drives, the probability that all four drives fail is $P(\text{all four fail})=0.05\times0.05\times0.05\times0.05=0.05^{4}=0.00000625$.

Step5: Calculate the probability of avoiding catastrophe for part b

The probability of avoiding catastrophe (at least one working drive) is the complement of the event that all four drives fail. So $P(\text{at least one working}) = 1 - P(\text{all four fail})=1 - 0.00000625 = 0.99999375\approx0.999994$.

Answer:

a. $0.9975$
b. $0.999994$