Assume that there is a \( 10 \% \) 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 drive? b. If copies of all your computer data are stored on four independent hard disk drives, what is the probability that during a year, you can avoid catastrophe with at least one working drive?

To solve these problems, we will use the concept of complementary probability. The probability of avoiding catastrophe with at least one working drive is equal to \(1\) minus the probability that all drives fail. ### Part a: Two Hard Disk Drives 1. **Probability of failure for one drive**: \(P(\text{failure}) = 0.10\) 2. **Probability of success for one drive**: \(P(\text{success}) = 1 - P(\text{failure}) = 1 - 0.10 = 0.90\) For two independent drives, the probability that both drives fail is: \[ P(\text{both fail}) = P(\text{failure}) \times P(\text{failure}) = 0.10 \times 0.10 = 0.01 \] Now, the probability of avoiding catastrophe with at least one working drive is: \[ P(\text{at least one works}) = 1 - P(\text{both fail}) = 1 - 0.01 = 0.99 \] ### Part b: Four Hard Disk Drives 1. **Probability of failure for one drive**: \(P(\text{failure}) = 0.10\) 2. **Probability of success for one drive**: \(P(\text{success}) = 0.90\) For four independent drives, the probability that all four drives fail is: \[ P(\text{all fail}) = P(\text{failure})^4 = 0.10^4 = 0.0001 \] Now, the probability of avoiding catastrophe with at least one working drive is: \[ P(\text{at least one works}) = 1 - P(\text{all fail}) = 1 - 0.0001 = 0.9999 \] ### Summary of Results - a. The probability of avoiding catastrophe with at least one working drive (2 drives) is \(0.99\). - b. The probability of avoiding catastrophe with at least one working drive (4 drives) is \(0.9999\).