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The principle of redundancy is used when system reliability is improved through redundant or backup components. Assume that a student's alarm clock has a \( 13.3 \% \) daily failure rate. Complete parts (a) through (d) below. a. What is the probability that the student's alarm clock will not work on the morning of an important final exam? 0.133 (Round to three decimal places as needed.) b. If the student has two such alarm clocks, what is the probability that they both fail on the morning of an important final exam? (Round to five decimal places as needed.)

Ask by Alexander Curry. in the United States
Feb 14,2025

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Answer

a. The probability the alarm clock fails is 0.133. b. The probability both alarm clocks fail is 0.01769.

Solution

For part (a), the probability the alarm clock fails on any given day is given as 13.3%, or 0.133. For part (b), assuming independence between the two clocks, the probability that both fail is given by multiplying the individual failure probabilities:   Probability = 0.133 × 0.133 = 0.017689. Rounded to five decimal places, this probability is 0.01769.

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The probability that both alarm clocks fail can be calculated by multiplying their individual failure probabilities. Since each clock has a \( 13.3\% \) failure rate (or \( 0.133 \)), the probability that both clocks fail is: \[ P(\text{both fail}) = P(\text{clock 1 fails}) \times P(\text{clock 2 fails}) = 0.133 \times 0.133 = 0.0176889 \] Rounding to five decimal places, the final answer is \( 0.01769 \). c. If the student has two alarm clocks, what is the probability that at least one of them will work on the morning of an important final exam? To find the probability that at least one works, we can use the complement rule. First, find the probability that both fail (which we already calculated as \( 0.01769 \)). Then, subtract that from \( 1 \): \[ P(\text{at least one works}) = 1 - P(\text{both fail}) = 1 - 0.01769 = 0.98231 \] Rounding to five decimal places, the final answer is \( 0.98231 \). d. If the student has three such alarm clocks, what is the probability that at least one will work on the morning of an important final exam? We apply the same logic as before. First, we find the probability that all three fail: \[ P(\text{all fail}) = P(\text{clock 1 fails}) \times P(\text{clock 2 fails}) \times P(\text{clock 3 fails}) = 0.133^3 = 0.00235273 \] Now, subtract this from \( 1 \) to find the probability that at least one works: \[ P(\text{at least one works}) = 1 - P(\text{all fail}) = 1 - 0.00235273 = 0.99764727 \] Rounding to five decimal places, the final answer is \( 0.99765 \).

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