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 working drive? b. If copies of all 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? a. With two hard disk drives, the probability that catastrophe can be avoided is (Round to four decimal places as needed.)

To solve the problem, we first need to understand the probability of failure and success for the hard disk drives. Given: - The probability of failure of a hard disk drive in a year, \( P(F) = 0.10 \). - The probability of success (not failing) of a hard disk drive in a year, \( P(S) = 1 - P(F) = 0.90 \). ### Part a: Two Hard Disk Drives When you have two independent hard disk drives, the probability that both drives fail (which would lead to a catastrophe) is given by: \[ P(\text{both fail}) = P(F) \times P(F) = 0.10 \times 0.10 = 0.01 \] Now, the probability of avoiding catastrophe (at least one drive working) is the complement of both drives failing: \[ P(\text{at least one works}) = 1 - P(\text{both fail}) = 1 - 0.01 = 0.99 \] Thus, the probability that catastrophe can be avoided with two hard disk drives is: \[ \boxed{0.9900} \] ### Part b: Three Hard Disk Drives Now, if you have three independent hard disk drives, the probability that all three drives fail is: \[ P(\text{all fail}) = P(F) \times P(F) \times P(F) = 0.10 \times 0.10 \times 0.10 = 0.001 \] The probability of avoiding catastrophe (at least one drive working) is: \[ P(\text{at least one works}) = 1 - P(\text{all fail}) = 1 - 0.001 = 0.999 \] Thus, the probability that catastrophe can be avoided with three hard disk drives is: \[ \boxed{0.9990} \]