A data center has 10,000 disk drives. Suppose that a disk drive fails

1.	A data center has 10,000 disk drives. Suppose that a disk drive fails in a given day withProbability 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.
Answer» 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)q¹⁰⁰⁰⁰C(10000, 0)q¹⁰⁰⁰⁰ P(0,1) = C(10000, 0)q¹⁰⁰⁰⁰C(10000, 1)pq⁹⁹⁹⁹ P(1,0) = C(10000, 1)q⁹⁹⁹⁹C(10000, 0)q¹⁰⁰⁰⁰ ··· ···· ··· P(4,5) = C(10000, 4)q⁹⁹⁹⁶C(10000, 5)q⁹⁹⁹⁵ P(5,4) = C(10000, 5)q⁹⁹⁹⁵C(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)

A data center has 10,000 disk drives. Suppose that a disk drive fails in a given day withProbability 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.

Answer»

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)