An internet access provider (IAP) owns two servers. Each server has a 50% chance of being "down" independently of the other. Fortunately, only one server is necessary to allow the IAP to provide service to its customers, i. e., only one server is needed to keep the IAP’s system up. Suppose a customer tries to access the internet

Question Continuation

Suppose a customer tries to access the internet on 8 different occasions, which are sufficiently spaced apart in time, so that we may assume that the states of the system corresponding to these 8 occasions are independent. What is the probability that the customer will only be able to access the internet on 2 out of the 8 occasions?

0.311

Explanation:

Given

p = success = 2/8 = ¼ --- Probability that both servers down at any one event

p + q = 1

q = failure = 1 - ¼ = ¾ --- Probability that at most one server down at any one event

The number of ways that the customer will only be able to access the internet on 2 out of the 8 occasions is solved by applying binomial distribution below;

(p + q) ^n where n = 8 and r = 2 is

nCr * p^r * q ^ (n-r)

This becomes

8C2 * (¼)² * (¾)^6

= 8!/(6!2!) * 1/16 * 729/4096

= 0.31146240234375

= 0.311 --- approximated

solution:

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