Answer:
The probability is 0.4207
Explanation:
The probability of a home-based computer having access to on-line services is p = 0.2 (data from the exercise)
Then, the probability of a home-based computer not having access to on-line services is p = 1 - 0.2 = 0.8
We are going to use this probability (p = 0.8) to solve the exercise.
Let's define the random variable X
X : ''Number of home-based computers not having access to on-line services''
X can be modeled as a binomial random variable
X ~ Bi(p,n)
X ~Bi(0.8,25)
Where p is the success probability and n is the number of Bernoulli independent experiments we are taking place.
We are going to count ''a success'' as a computer not having access to on-line services.
The binomial probability function is :

Where P(X=x) is the probability of the random variable X to assume the value x
nCx is the combinatorial number define as

p is the success probability and n the number of Bernoulli independent experiments taking place.
In our exercise,

We are looking for :




Finally, the probability of finding that more than 20 of 25 home-based computers do not have access to on-line services is 0.4207