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8 July, 22:38

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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  1. 9 July, 02:05
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    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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