A data center has 10,000 disk drives. Suppose that a disk drive fails in a given day with
Probability 10^-3.
(a) Find the probability that there are no failures in a given day.
(b) Find the probability that there are fewer than 10 failures in two days.
Answers
Step-by-step explanation:
Given
Number of disc, n = 10000
Probability that a disc fail in a day, p = 1/1000
Calculation
Probability that a disc do not fail on a day
q = 1-p = 1-1/1000 = 999/1000
(a) Probability of no disc failure on a day
= C(10000, 0)q¹⁰⁰⁰⁰
= 1*(999/1000)¹⁰⁰⁰⁰
(b) Let us assume that a disc failed on day 1 is replaced with a new one before day 2. The failures can occur in the following ways (0, 0), (0, 1),(1, 0),(0, 2),(2, 0), .....(4, 5),(5,4) where the two number represent failure on first and second day respectively. Their respective probabilities are
P(0,0) = C(10000, 0)q¹⁰⁰⁰⁰*C(10000, 0)q¹⁰⁰⁰⁰
P(0,1) = C(10000, 0)q¹⁰⁰⁰⁰*C(10000, 1)pq⁹⁹⁹⁹
P(1,0) = C(10000, 1)q⁹⁹⁹⁹*C(10000, 0)q¹⁰⁰⁰⁰
···
····
···
P(4,5) = C(10000, 4)q⁹⁹⁹⁶*C(10000, 5)q⁹⁹⁹⁵
P(5,4) = C(10000, 5)q⁹⁹⁹⁵*C(10000, 4)q⁹⁹⁹⁶
The required probability is the sum of the above probabilities.
P = P(0,0)+P(0,1)+P(1,0)+.........+P(4,5)+P(5,4)