Here we have a situation where the probability of a driver wearing seat belts is known and remains constant throughout the experiment of stopping 20 drivers.
The drivers stopped are assumed to be random and independent.
These conditions are suitable for modelling using he binomial distribution, where
where n=number of drivers stopped (sample size = 20)
x=number of drivers wearing seatbelts (4)
p=probability that a driver wears seatbelts (0.35), and
C(n,x)=binomial coefficient of x objects chosen from n = n!/(x!(n-x)!)
So the probability of finding 4 drivers wearing seatbelts out of a sample of 20
P(4;20;0.35)
=C(20,4)*(0.35)^4*(0.65)^16
= 4845*0.0150061*0.0010153
= 0.07382