Hooli is a large international corporation that makes the world a better place t
ID: 3044398 • Letter: H
Question
Hooli is a large international corporation that makes the world a better place through ubiq- uitous cloud storage. Hooli has many data centers throughout the world. The data center closest to UTD has 10,000 hard drives. Each of these hard drives has a probability of 2 x 10-4 of failing in any given day o) I 4or mare hoand drive al in n givenr oo move data from their UTD data center to another data center. What is the probability that Hooli initiates a data transfer in a given day? (b) What is the probability of having t days in a 7 day week where data transfers occur? Evaluate this for t = 3. (c) What is the probability that no data transfers are required in the month of February?Explanation / Answer
Back-up Theory
Let X = Number of hard drives failing in a day at the data center. Then,
X ~ B(n, p), ………………………………………………………………………………..(1)
where n = number of hard drives available at the data center and p = probability of a hard drive failing in a day
We are given n = 104 and p = 2 x 10- 4.
Since n is very large and p is very small, Binomial probabilities can be approximated by Poisson(), where = np.
Thus, X ~ Poisson(2)……………………………………………………………………..(2)
If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and p = probability of one success, then
probability mass function (pmf) of X is given by
p(x) = P(X = x) = (nCx)(px)(1 - p)n – x, x = 0, 1, 2, ……. , n ……………………………..(3)
[The above probability can also be directly obtained using Excel Function of Binomial Distribution] ……………………………………………………………………………….(3a)
If a random variable X ~ Poisson(), i.e., X has Poisson Distribution with mean then
probability mass function (pmf) of X is given by P(X = x) = e – .x/(x!) …………..(4)
where x = 0, 1, 2, ……. ,
Values of p(x) for various values of and x can be obtained by using Excel Function….. (4a)
Part (a)
Probability the data center would go for a data transfer
= P(X 4)
= 1 – 0.9473
= 0.0527 ANSWER
Part (b)
Let Y = Number of days in a 7day-week where data transfer occur. Then, Y ~ B(7, p), where
p = Probability the data center would go for a data transfer = 0.0527 [from answer of Part (a)].
So, probability of t days in a week having data transfer = P(Y = t)
= (7Ct)(0.0527t)(0.9473)7 – t [vide (3) above] ANSWER 1
Substituting t = 3 in the above formula, answer is 0.0041 ANSWER 2
Part (c)
Assuming normal year, probability of no data transfer in February = P(Z = 0), where
Z = Number of days in February where data transfer occur.
We know Z ~ B(28, 0.0527)
So, the required probability = (28C0)(0.05270)(0.9473)28 [vide (3) above]
= 0.2196 ANSWER
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