Question

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?

Answer 1

Answer:

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 2

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