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

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,

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

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:

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

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.

b.

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

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

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.

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.