Let's solve both parts of the problem step-by-step.

Part (a): Two Hard Disk Drives

1. Given Information:

   • Probability of failure for one hard disk drive in a year = 8% = 0.08.
   • Probability of success (working) for one hard disk drive = 1 - 0.08 = 0.92.

2. Calculating the Probability of Avoiding Catastrophe:

   • For catastrophe to occur, both drives must fail.

   • Probability that both drives fail = $0.08 \times 0.08 = 0.0064$.

   • Probability that at least one drive works (avoiding catastrophe) = $1 - 0.0064 = 0.9936$.

The answer for part (a) is: $\text{Probability} = 0.9936$

Part (b): Four Hard Disk Drives

1. Calculating the Probability of Avoiding Catastrophe:

   • Probability of all four drives failing = $(0.08)^4$.

   Let's compute this:

$(0.08)^4 = 0.00004096$

2. Probability of at Least One Drive Working:
   • Probability of at least one drive working = $1 - (0.08)^4 = 1 - 0.00004096$.

$1 - 0.00004096 = 0.99995904$

Rounding to six decimal places:

$\text{Probability} = 0.999959$

So, the answer for part (b) is: $\text{Probability} = 0.999959$

Would you like more details, or do you have any other questions? Here are some related questions you might find interesting:

1. What would be the probability of avoiding catastrophe with three independent hard drives?
2. How does the probability change if the failure rate increases to 10% per year?
3. What is the probability that at least two of the four hard drives fail?
4. How many hard drives would be needed to achieve a 99.999% reliability rate?
5. How would correlated failures between drives impact the overall probability?

Tip: When dealing with reliability problems, redundancy (using multiple independent backups) significantly increases the probability of avoiding data loss.