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Sagot :
Answer:
0.9936 = 99.36% probability that during a year, you can avoid catastrophe with at least one working drive
Step-by-step explanation:
For each disk, there are only two possible outcomes. Either it works, or it does not. The probability of a disk working is independent of any other disk, which means that the binomial probability 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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Assume that there is a 8% rate of disk drive failure in a year.
So 100 - 8 = 92% probability of working, which means that [tex]p = 0.92[/tex]
Two disks are used:
This means that [tex]n = 2[/tex]
What is the probability that during a year, you can avoid catastrophe with at least one working drive?
This is:
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{2,0}.(0.92)^{0}.(0.08)^{2} = 0.0064[/tex]
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0064 = 0.9936[/tex]
0.9936 = 99.36% probability that during a year, you can avoid catastrophe with at least one working drive
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