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

Question

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?

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4 weeks 2021-12-27T09:09:52+00:00 1 Answer 0 views 0

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) = 0.11

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?

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 = 0.0121

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?

–  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