Answer:
a) 32.04% probability that overbooking occurs.
b) 40.79% probability that the flight has empty seats.
Step-by-step explanation:
For each booked passenger, there are only two possible outcomes. Either they arrive for the flight, or they do not arrive. The probability of a passenger arriving is independent of other passengers. 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.
Our variable of interest are the 8 reservations that went for the passengers with a 48% probability of arriving.
This means that 
A) Find the probability that overbooking occurs.
12 seats, 8 of which are already occupied. So overbooking occurs if more than 4 of the reservated arrive.







32.04% probability that overbooking occurs.
B) Find the probability that the flight has empty seats.
Less than 4 of the booked passengers arrive.
To make it easier, i will use

From a), P(X > 4) = 0.3204


40.79% probability that the flight has empty seats.