Redundancy. Exercises 25 and 26 involve redundancy.

Redundancy in Computer Hard Drives It is generally recognized that it is wise to back up computer data. Assume that there is a 3% rate of disk drive failure in a year (based on data from various sources, including lifehacker.com).

a. If you store all of your computer data on a single hard disk drive, what is the probability that the drive will fail during a year? continued 158 CHAPTER 4 Probability

b. If all of your computer data are stored on a hard disk drive with a copy stored on a second hard disk drive, what is the probability that both drives will fail during a year?

c. If copies of all of your computer data are stored on three independent hard disk drives, what is the probability that all three will fail during a year?

d. Describe the improved reliability that is gained with backup drives

The probability of failure for a hard disk drive in a year is equal to 3%.

For a single event, say A, the probability of an event A is defined as:

$P\left(A\right)=\frac{n\left(A\right)}{\text{Total}\text{counts}}$

According to the multiplication rule of probability:

For a group of events (say, A, B, and C), the probability that they will occur together can be calculated as follows:

$P\left(A\text{and}B\text{and}C\right)=P\left(A\right)\times P\left(B\right)\times P\left(C\right)$

a.

Let A be the event of failure of a hard disk drive in a year.

The probability of failure for a single disk drive in a year is given as:

$$\begin{gathered} P\left(A\right)=\frac{3}{100} \\ =0.03 \end{gathered}$$

Therefore, the probability of failure of a single disk drive in a year if all the data is stored in it is equal to 0.03.

b.

Let Bbe the event of failure of the second disk drive in a year.

The probability of failure of the second disk drive in a year is given by:

$$\begin{gathered} P\left(B\right)=\frac{3}{100} \\ =0.03 \end{gathered}$$

The probability of failure of both the first and the second disk drives in a year is given as:

$$\begin{gathered} P\left(A\text{and}B\right)=P\left(A\right)\times P\left(B\right) \\ =0.03\times 0.03 \\ =0.0009 \end{gathered}$$

Therefore, the probability of failure of both the first and the second disk drives in a year is equal to 0.0009.

c.

Let C be the event of failure of the third disk drive in a year.

The probability of failure of the third disk drive in a year is given by:

$$\begin{gathered} P\left(C\right)=\frac{3}{100} \\ =0.03 \end{gathered}$$

The probability of failure of all three (1^st_,2^nd, and 3^rd) disk drives in a year is given as:

$$\begin{gathered} P\left(A\text{and}B\text{and}C\right)=P\left(A\right)\times P\left(B\right)\times P\left(C\right) \\ =0.03\times 0.03\times 0.03 \\ =0.000027 \end{gathered}$$

Therefore, the probability of failure of all three disk drives in a year is equal to 0.000027.

d.

The failure rate of a disk drive when data is stored in only one drive is 3 out of 100.

As data is stored in two disk drives, the total failure rate (of both drives) decreases from 3 out of 100 to 9 out of 10000. Thus, there is a significant improvement in the failure rate.

As data is stored in three disk drives, the total failure rate (of all three drives) decreases from 9 out of 10,000 to 27 out of 1,000,000. Thus, there is an even bigger improvement.

It can be concluded that there is a significant improvement in the total failure rate as the number of disk drives in which data is stored increases.