anonymous
  • anonymous
Statisticssssss -_______- Assume that there is a %8 rate of disk drive failure in a year. a. If all 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? b. If copies of all 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? Round to four decimal places.
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
This is a binomial distribution.
anonymous
  • anonymous
\[ X\sim B(n,p) \implies \Pr(X=k) = \binom nk p^{k}(1-p)^{n-k} \]
anonymous
  • anonymous
Yeah, I can.

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anonymous
  • anonymous
Do you know what \(\binom n k\) is?
anonymous
  • anonymous
n choose k
anonymous
  • anonymous
Ok. So if you have two drives, how many ways are there for one to remain?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
What is the probability for 1 to remain?
anonymous
  • anonymous
Not quite.
anonymous
  • anonymous
Okay suppose we give them a letter. There is dive A and drive B.
anonymous
  • anonymous
What is the probability that drive A fails and the probability that drive B doesn't fail?
anonymous
  • anonymous
"Assume that there is a %8 rate of disk drive failure in a year."
anonymous
  • anonymous
What is the probability that drive A fails?
anonymous
  • anonymous
That is the simplest I can get it.
anonymous
  • anonymous
92% is that it doesn't fail.
anonymous
  • anonymous
Now, does the failure of drive A affect the failure of drive B? Is it independent?
anonymous
  • anonymous
Well it is independent.
anonymous
  • anonymous
They are independent because they don't affect each other. Drive A will fail or not fail with the same probability regardless of whether drive B fails.
anonymous
  • anonymous
What is the probability of two independent events happening? Do you remember any formula?
anonymous
  • anonymous
This isn't conditional probability.
anonymous
  • anonymous
What is the probability of BOTH independent events happening.
anonymous
  • anonymous
\[ \Pr(AB) = \Pr(A) \times \Pr(B) \]
anonymous
  • anonymous
When \(A,B\) are independent.
anonymous
  • anonymous
IN this case \(\Pr(A)= 0.08\) and \(\Pr(B) = 0.92 = 1-0.08\)
anonymous
  • anonymous
And \(A\) is the event drive \(A\) fails and \(B\) is the event \(B\) doesn't fail.
anonymous
  • anonymous
We have 0.0736 that A fails and B doesn't fail.
anonymous
  • anonymous
Next we need the probability that A doesn't fail and B fails.
anonymous
  • anonymous
Let's keep it simple for now.
anonymous
  • anonymous
Okay
anonymous
  • anonymous
What is the probability that A doesn't fail and B fails?
anonymous
  • anonymous
probability that A doesn't fail is .92
anonymous
  • anonymous
probability that B does is .08
anonymous
  • anonymous
yeah but for both?
anonymous
  • anonymous
IT's the same as before
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
Now one last one we need to do. What is the probability that A doesn't fail and B doesn't fail.
anonymous
  • anonymous
Okay
anonymous
  • anonymous
So we have probabilities for: A fails, B doesn't A doesn't, B fails A doesn't, B doesn't
anonymous
  • anonymous
There are the 3 ways in which at least one drive doesn't fail.
anonymous
  • anonymous
They are mutually exclusive events, so the probability of either of them happening is just the sum of them all
anonymous
  • anonymous
So if we add up all of the probabilities we have calculated we will have the probability that at least one doesn't fail.
anonymous
  • anonymous
That is the answer to a)
anonymous
  • anonymous
No. 0.0736 was just the probability that A fails and B doesn't
anonymous
  • anonymous
The only way to do it faster is to understand it.
anonymous
  • anonymous
There isn't one simple formula for every problem.
anonymous
  • anonymous
Where would you start?
anonymous
  • anonymous
There are a couple formula you should know by now about probability.
anonymous
  • anonymous
Okay good luck.

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