Answer:
a) 0.214 = 21.4% probability that at most 4 of the calls involve a fax message
b) 0.118 = 11.8% probability that exactly 4 of the calls involve a fax message
c) 0.904 = 90.4% probability that at least 4 of the calls involve a fax message
d) 0.786 = 78.6% probability that more than 4 of the calls involve a fax message
Step-by-step explanation:
For each call, there are only two possible outcomes. Either it involves a fax message, or it does not. The probability of a call involving a fax message is independent of other calls. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.
25% of the incoming calls involve fax messages
This means that 
25 incoming calls.
This means that 
a. What is the probability that at most 4 of the calls involve a fax message?
.
In which







0.214 = 21.4% probability that at most 4 of the calls involve a fax message
b. What is the probability that exactly 4 of the calls involve a fax message?

0.118 = 11.8% probability that exactly 4 of the calls involve a fax message.
c. What is the probability that at least 4 of the calls involve a fax message?
Either less than 4 calls involve fax messages, or at least 4 do. The sum of the probabilities of these events is 1. So

We want
. Then

In which








0.904 = 90.4% probability that at least 4 of the calls involve a fax message.
d. What is the probability that more than 4 of the calls involve a fax message?
Very similar to c.

From a), 
Then

0.786 = 78.6% probability that more than 4 of the calls involve a fax message