## It is generally recognized as wise to back up computer data. Assume that there is a 11% rate of disk drive failure in a year. a. If all of

It is generally recognized as wise to back up computer data. Assume that there is a 11% rate of disk drive failure in a year. a. If all of a computer’s data are stored on a single hard disk drive, what is the probability that the drive will fail during a year? b. If all of a computer’s data are stored on a hard disk drive with a copy stored in a second hard disk drive, what is the probability that both drives will fail during a year? c. If all of a computer’s data is stored on three independent hard disk drives, what is the probability that all three will fail during a year?

## Answers ( )

Answer:a.0.11

b. 0.0121

c. 0.001331

Step-by-step explanation:Given:– The probability of 1 disk fails per year = 0.11

Find:a. If all of a computer’s data are stored on a single hard disk drive, what is the probability that the drive will fail during a year?

Solution:– The probability given is given to be 0.11 for a disk to fail. If there is one hard disk then the failure of data is equivalent.

Hence, P ( disc fails/year) =

0.11b.If all of a computer’s data are stored on a hard disk drive with a copy stored in a second hard disk drive, what is the probability that both drives will fail during a year?Solution:– The given probability of one disk failing in a year does not affect the probability of another disc failing. Hence, the two events are independent.

– Hence. the probability of independent events can be calculated as:

P ( 1st disc & 2nd disc fails / year) = P(disc fail/year)*P(disc fail/year)

P ( 1st disc & 2nd disc fails / year) = 0.11*0.11 =

0.0121c.If all of a computer’s data is stored on three independent hard disk drives, what is the probability that all three will fail during a year?– The given probability of one disk failing in a year does not affect the probability of another disc failing. Hence, the three events are independent.

– Hence. the probability of independent events can be calculated as:

P ( 1st, 2nd & 3rd disc fails / year) = P(disc fail/year)^3

P ( 1st disc & 2nd disc fails / year) = 0.11^3 =

0.001331