Question

assume that there is a 6% 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 three 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 0.9964 (round to four decimal places as needed.) b. with three 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 in a year is $p = 0.06$, so the success - rate (the drive works) is $q=1 - p=1 - 0.06 = 0.94$.

Step2: Find the probability of all drives failing for three drives

Since the drives are independent, the probability that all three drives fail is $P(\text{all fail})=p\times p\times p=p^{3}$. Substituting $p = 0.06$, we get $P(\text{all fail})=(0.06)^{3}=0.06\times0.06\times0.06 = 0.000216$.

Step3: Find the probability of at least one drive working

The probability of at least one drive working is the complement of the event that all drives fail. Let $A$ be the event that at least one drive works. Then $P(A)=1 - P(\text{all fail})$. Substituting the value of $P(\text{all fail}) = 0.000216$, we get $P(A)=1 - 0.000216=0.999784$.

Answer:

$0.999784$