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Redundancy in Computer Hard Drives Assume that there is a \(3 \%\) rate of disk drive failures in a year (based on data from various sources including lifehacker.com). a. If all of 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? Express the result with four decimal places. b. If copies of all of 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? Express the result with six decimal places. What is wrong with using the usual round-off rule for probabilities in this case?

Short Answer

Expert verified
a: 0.9991, b: 0.999973, because rounding off to four decimal places does not show the incremental benefit of the third drive.

Step by step solution

01

Define the Failure Probability

The probability that a hard disk drive fails in a year is given as 3%, or 0.03. Denote this probability as P(fail) = 0.03.
02

Define the Success Probability

The probability that a hard disk drive does not fail is the complement of the failure probability. Therefore, P(success) = 1 - P(fail) = 1 - 0.03 = 0.97.
03

Calculate for Two Drives (Part a)

To find the probability that at least one of the two drives does not fail, first find the probability that both drives fail: P(both fail) = P(fail) * P(fail) = 0.03 * 0.03 = 0.0009. Then, the probability that at least one drive works is: P(at least one works) = 1 - P(both fail) = 1 - 0.0009 = 0.9991.
04

Answer Part a

The probability that during a year you can avoid catastrophe with at least one working drive is 0.9991.
05

Calculate for Three Drives (Part b)

To find the probability that at least one of the three drives does not fail, calculate the probability that all three drives fail: P(all fail) = P(fail) * P(fail) * P(fail) = 0.03 * 0.03 * 0.03 = 0.000027. Then, the probability that at least one drive works is: P(at least one works) = 1 - P(all fail) = 1 - 0.000027 = 0.999973.
06

Answer Part b

The probability that during a year you can avoid catastrophe with at least one working drive is 0.999973.
07

Discuss Rounding Rule Issue

Using the usual round-off rule to four decimal places would yield the same result for both cases (0.9991 for two drives and 0.9990 for three drives), which would not distinguish the difference. Hence, more precision is required in part b to capture the small difference.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

probability of failure
In probability theory, the probability of failure is the chance that a given event will not succeed. For example, if you have a hard disk drive that fails 3% of the time in a year, this percentage represents its probability of failure. Mathematically, this can be expressed as: \(P(fail) = 0.03\).Understanding the probability of failure helps us plan for and mitigate potential problems. In our exercise, this rate (3%) indicates that each independent hard disk drive has a 3% chance of failing within a year.
complementary probability
Complementary probability refers to the chance of the opposite outcome of a given event. If you know the probability of an event occurring, the complementary probability is simply 1 minus that probability. In the case of our hard disk drives, if there's a 0.03 probability that the drive will fail, then the complementary probability of it not failing (or succeeding) is: \(P(success) = 1 - P(fail) = 1 - 0.03 = 0.97\).This concept is crucial because it allows us to determine the likelihood of all possible outcomes. For instance, if you have two independent hard drives, calculating the probability of at least one success involves using the complementary concept to first find the probability of both failing and then subtracting this from 1.
independent events
Events are considered independent if the outcome of one does not affect the other. In our example, each hard disk drive's failure is independent of the others. This means the failure of one drive does not change the probability of the other drive failing.To calculate the overall probability for multiple independent events, you multiply the probabilities of each event. For instance, with two drives each having a 0.03 probability of failure, the combined probability of both failing is: \( P(both \, fail) = P(fail) * P(fail) = 0.03 * 0.03 = 0.0009 \).Understanding independence is key in accurately calculating the combined probabilities of several events, as seen in our exercise.
rounding rules
Rounding rules in probability can significantly impact the results depending on the required precision. In our exercise, rounding to four decimal places for the probability of avoiding catastrophe with two drives might result in 0.9991. For three drives, a more precise calculation is essential as rounding incorrectly can obscure important differences.For example, the probability of avoiding catastrophe with three drives is calculated to be 0.999973. When rounding to four decimal places, you lose precision, making both two and three drives appear to have the same probability (0.9991). This loss of detail can be problematic when precise differentiation is required. It's important to determine the appropriate level of precision to avoid misleading results.

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