assume that there is a 6% rate of disk - drive failure in a year.\na. if all your computer data is stored on…

assume that there is a 6% rate of disk - drive failure in a year.\na. 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?\nb. 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?\na. with two hard disk drives, the probability that catastrophe can be avoided is \n(round to four decimal places as needed.)

assume that there is a 6% rate of disk - drive failure in a year.\na. 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?\nb. 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?\na. with two hard disk drives, the probability that catastrophe can be avoided is \n(round to four decimal places as needed.)

Answer

Explanation:

Step1: Find the failure - probability of a single drive

The rate of disk - drive failure is $p = 0.06$, so the probability of a drive working is $q=1 - p=1 - 0.06 = 0.94$.

Step2: Calculate the probability of both drives failing (for two - drive case)

Since the drives are independent, the probability that both drives fail is $P(\text{both fail})=p\times p = 0.06\times0.06=0.0036$.

Step3: Calculate the probability of avoiding catastrophe (for two - drive case)

The probability of avoiding catastrophe (at least one working drive) is the complement of both drives failing. So $P(\text{at least one working}) = 1 - P(\text{both fail})=1 - 0.0036 = 0.9964$.

Step4: Calculate the probability of all three drives failing (for three - drive case)

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

Step5: Calculate the probability of avoiding catastrophe (for three - drive case)

The probability of avoiding catastrophe (at least one working drive) is the complement of all three drives failing. So $P(\text{at least one working})=1 - P(\text{all three fail})=1 - 0.000216 = 0.999784$.

Answer:

a. $0.9964$ b. $0.9998$ (rounded to four decimal places)