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?
Answer:
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
