Question

Redundancy in Computer Hard Drives Assume that there is a 3% rate of disk drive failures in a year (based on data from various sources including lifehacker.com).

a. If all of 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? Express the result with four decimal places.

b. If copies of all of 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? Express the result with six decimal places. What is wrong with using the usual round-off rule for probabilities in this case?

Short answer

a. The probability that out of two drives, at least one will work is equal to 0.9991.

b. The probability that out of three drives, at least one will work is equal to 0.999973.

If the probability value is rounded off, the answer will be equal to one, indicating that it is certain that at least one disk drive will be working out of the three drives, which is wrong.

Step-by-step solution

Given information

The probability that a disk drive will fail/not work in a specific year is 3%.

Define the probability of “at least one”

The probability that an event occurs at least once is calculated by deducting the probability of no appearances for the event from 1.

Mathematically, if A is an event,

$P\left(Aoccurringatleastonce\right)=1-P\left(Anotoccurring\right)$

Compute the probability that at least one drive works out of two

a.

Let A be the event of a disk failure.

Then, the probability of a disk failure is:

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

The probability that out of two disk drives, none will work is given as:

$$\begin{gathered} P\left(\text{both}\text{will}\text{fail}\right)=P\left(A\right)\times P\left(A\right) \\ =0.03\times 0.03 \\ =0.0009 \end{gathered}$$

The probability that out of the two disk drives, at least one will work is computed as follows:

$$\begin{gathered} P\left(\text{at}\text{least}\text{one}\text{will}\text{work}\right)=1-P\left(\text{both}\text{will}\text{fail}\right) \\ =1-0.0009 \\ =0.9991 \end{gathered}$$

Therefore, the probability that out of the two disk drives, at least one will work in a year is equal to 0.9991.

Compute the probability that at least one drive works out of three

b.

The probability that out of three disk drives, none will work is given as:

$$\begin{gathered} P\left(all3\text{will}\text{fail}\right)=P\left(A\right)\times P\left(A\right)\times P\left(A\right) \\ =0.03\times 0.03\times 0.03 \\ =0.000027 \end{gathered}$$

The probability that out of the three disk drives, at least one will work is computed as follows:

$$\begin{gathered} P\left(\text{at}\text{least}\text{one}\text{will}\text{work}\right)=1-P\left(\text{all}\text{3}\text{will}\text{fail}\right) \\ =1-0.000009 \\ =0.999973 \end{gathered}$$

Therefore, the probability that out of the three disk drives, at least one will work in a year is equal to 0.999973.

Error in rounding off

If the probability in part (b) is rounded off, the answer will come out to be equal to 1, indicating that the event of at least one working disk drive out of the three is a certain event.

However, this can lead to inaccurate conclusions as the probability is high but not exactly equal to one.