Question

Assume that there is a 8% 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.)

Answer 1

The probability of avoiding a hard disk failure with two independent drives is 99.36%, and with three independent drives, it is 99.9488%, after rounding to four decimal places.

Step-by-step explanation:

When considering the probability of avoiding a hard disk failure with multiple independent hard disk drives, we can use complementary events to calculate the probability of at least one drive working. With a failure rate of 8%, the probability that one drive will not fail is therefore 1 - 0.08 = 0.92 or 92%.

For two drives, the probability that both fail is 0.08 * 0.08 = 0.0064 or 0.64%. Thus, the probability of at least one working drive is 1 - 0.0064 = 0.9936, or 99.36%, after rounding to four decimal places.

For three drives, the probability they all fail is 0.08 * 0.08 * 0.08 = 0.000512 or 0.0512%. Hence, the probability of at least one working drive is 1 - 0.000512 = 0.999488, or 99.9488%, rounded to four decimal places.