Mathematics, 05.10.2019 17:30 mawawakaiii

Assume that there is a 5% 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 three independent hard disk drives, what is the probability that during a year, you can avoid catastrophe with at least one working drive?

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Probability of survival of one disk, p=1-0.05=0.95
If there are n drives, and failures are independent (e.g. mechanical wear and tear, and not a lightning strike or a current spike), then
Probability of survival = p^n
(a) for two disks, P(2) = 0.95^2=0.9025
(b)for three disks P(3) = 0.95^3 = 0.8574

Answer from: dinarussell74

The probability that during a year, you can avoid catastrophe with at least one working drive is 0.9975

The probability that during a year, you can avoid catastrophe with at least one working drive is 0.999875

Step-by-step explanation:

There is a 5% rate of disk drive failure in a year

So, probability of failure q = 0.05

Since the sum of probabilities = 1

So, probability of success p= 1-0.05

=0.95

Part a : all your computer data is stored on a hard disk drive with a copy stored on a second hard disk drive

So, n =2

We are required to find what is the probability that during a year, you can avoid catastrophe with at least one working drive

So, he probability that during a year, you can avoid catastrophe with at least one working drive :

$1-(^2C_0 \times p^0\times q^2)$

$1-(\frac{2!}{0!(2-0)!}\times (0.95)^0\times(0.05)^2)$

$1-(1\times0.0025)$

0.9975

Hence the probability that during a year, you can avoid catastrophe with at least one working drive is 0.9975

Part b : If copies of all your computer data are stored on three three independent hard disk drives, what is the probability that during a year, you can avoid catastrophe with at least one working drive?

Since there are 3 copies so, n =3

So, the probability that during a year, you can avoid catastrophe with at least one working drive:

$1-(^3C_0 \times p^0\times q^2)$

$1-(\frac{3!}{0!(3-0)!} \times (0.95)^0\times(0.05)^3)$

$1-(1\times0.000125)$

0.999875

Hence the probability that during a year, you can avoid catastrophe with at least one working drive is 0.999875

Answer from: Quest

ryan had a head start of 10 meters.

Answer from: Quest

The answer is something i don’t know

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