Mathematics, 22.06.2019 15:20, thefish81

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 15.7% 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? 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? c. what is the probability of not being awakened if the student uses three independent alarm clocks? d. do the second and third alarm clocks result in greatly improved reliability? (a) yes, because you can always be certain that at least one alarm clock will work. (b) no, because the malfunction of both is equally or more likely than the malfunction of one. (c) yes, because total malfunction would not be impossible, but it would be unlikely. (d) no, because total malfunction would still not be unlikely.

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answered: lesly0913

Using the binomial distribution, it is found that:

a) 15.7% probability that the student's alarm clock will not work on the morning of an important final exam.

b) 0.0246 = 2.46% probability that they both fail on the morning of an important final exam.

c) 0.0039 = 0.39% probability of not being awakened if the student uses three independent alarm clocks.

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For each alarm clock, there are only two possible outcomes. Either it works, or it does not. The probability of an alarm working is independent of any other alarm, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by:

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of a success on a single trial.

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Item a:

15.7% probability of the alarm clock falling each day, thus, the same probability on the day of the final exam.

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Item b:

Two clocks, thus n = 2

Each with a 100 - 15.7 = 84.3% probability of working, thus p = 0.843

The probability of both falling is the probability that none works, thus P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.843)^{0}.(0.157)^{2} = 0.0246$

0.0246 = 2.46% probability that they both fail on the morning of an important final exam.

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Item c:

Same as item b, just with 3 clocks, thus n = 3

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.843)^{0}.(0.157)^{3} = 0.0039$

0.0039 = 0.39% probability of not being awakened if the student uses three independent alarm clocks.

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Item d:

Each extra clock, the probability of malfunctions become increasingly smaller, thus very unlikely, which means that the correct option is:

A similar problem is given at link

answered: Reward18

Step-by-step

Given:

alarm clocks that fail at 15.7% on any day.

Solution

Probability of failure of a single clock = 15.7% = 0.157

(a)

probability of failure of a single clock on any given day (final exam or not)

= 15.7% (given)

(b)

probability of failure of two independent alarm clocks on the SAME day

= 0.157^2

= 0.024649 (from independence of events)

(c)

probability of failure of three independent alarm clocks on the SAME day

= 0.157^3

= 0.00387 (from independence of events)

(d)

Since the probability of failure has been reduced from 0.157 to 0.00387, we can conclude that yes, even though malfunction of all three clocks is not impossible, it is unlikely at a probability of 0.00387 (less than 1 %)

answered: Guest

a box 3 3/4 feet long or 4+ feet.

step-by-step explanation:

answered: Guest

step-by-step explanation:

this is not a question