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sladkih [1.3K]
4 years ago
8

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?
Mathematics
1 answer:
lukranit [14]4 years ago
3 0

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) = 0.11

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

     

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