Question

assume that there is a 11% 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 0.9879 (round to four decimal places as needed.)
b. with three hard disk drives, the probability that catastrophe can be avoided is (round to six decimal places as needed.)

Explanation:

Step1: Find failure probability

The rate of disk - drive failure is $11\%=0.11$, so the probability of a drive failing is $p = 0.11$, and the probability of a drive working is $q=1 - 0.11 = 0.89$.

Step2: Use complementary probability for three - drive case

The probability of all three drives failing is $P(\text{all fail})=p\times p\times p=(0.11)^3$.
The probability of avoiding catastrophe (at least one working drive) is the complement of all drives failing. So $P(\text{at least one working}) = 1-(0.11)^3$.

Step3: Calculate the result

$(0.11)^3=0.11\times0.11\times0.11 = 0.001331$. Then $1-(0.11)^3=1 - 0.001331=0.998669$.

Answer:

$0.998669$