Motor vehicles sold to individuals are classified as either cars or light trucks (including SUVs) and as either domestic or imported. In a recent year, 69% of vehicles sold were light trucks, 78% were domestic, and 55% were domestic light trucks.
Let A be the event that a vehicle is a car and B the event that it's imported. Write each of the following events in set notation and give its probability.
(a) The vehicle is a light truck.
(b) The vehicle is an imported car.

Event that vehicle is a car is given by A. Since, the vehicles can only be classified as cars or light trucks, the event that a vehicle will be light truck will be compliment of A.

i.e. Event that vehicle is a Light Truck = [tex]A^{c}[/tex]

It is given that in recent year 69% of vehicles sold were light trucks, 78% were domestic, and 55% were domestic light trucks.

So, from here we can say that, if a vehicle is randomly selected from all the vehicles the probability that it would be a light truck will be:

P(Vehicle will be Light Truck) = P([tex]A^{c}[/tex]) = 69% = 0.69

Part b)

Event that vehicle is imported is given by B. We need to find the probability the a randomly chosen vehicle is an imported car i.e. we have to find probability of occurrence of events A and B together, which will be denoted as: P(A and B)

Since, P([tex]A^{c}[/tex]) = 69% = 0.69

P(A) = 1 - P([tex]A^{c}[/tex]) = 1 - 0.69 = 0.31

It is also given that 78% of vehicles were domestic. This means, the percentage/probability of imported vehicles is:

P(B) = 1 - 0.78 = 0.22

55% of vehicles were domestic light trucks. This can be expressed as:

[tex]P(A^{c}\&B^{c})[/tex], the compliment of this event will give us P(A or B).

i.e.

P(A or B) = 1 - [tex]P(A^{c} \&B^{c})[/tex] = 1 - 0.55 = 0.45

According to the addition rule of probabilities:

P(A or B) = P(A) + P(B) - P(A and B)

Substituting the calculated values gives us:

0.45 = 0.31 + 0.22 - P(A and B)

P(A and B) = 0.08

This means, the probability that vehicle is imported car is 0.08 or 8%.