Let:
• A, be the event of Bob choosing an used vehicle
,
• B, be the event of Bob choosing a car
Now, we know that:
![P(A\cup B)=P(A)+P(B)-P(A\cap B)](https://img.qammunity.org/2023/formulas/mathematics/college/fmis6wxz6gpakvw9xm1kdfvr92kx8wdfo2.png)
Out of the 41 vehicles, 18 are used. Therefore,
![P(A)=(18)/(41)](https://img.qammunity.org/2023/formulas/mathematics/college/2xcz1lcu196uwv4yhs47spy6tm6r9p0yct.png)
Out of the 41 vehicles, 19 are cars. This way,
![P(B)=(19)/(41)](https://img.qammunity.org/2023/formulas/mathematics/college/erj3s341hlnc9cfugb7c8eytcaeklfam28.png)
Out of the 41 vehicles, 8 are used cars. Thereby,
![P(A\cap B)=(8)/(41)](https://img.qammunity.org/2023/formulas/mathematics/college/fnvdwl57ifh67ivoy78thv14u3v6384943.png)
This way, we'll have that
![\begin{gathered} P(A\cup B)=(18)/(41)+(19)/(41)-(8)/(41) \\ \\ \Rightarrow P(A\cup B)=(29)/(41) \\ \\ \rightarrow P(A\cup B)=0.71 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qggvan27axhatpxvpym3vfdwx6f2bd6e09.png)
Therefore, the probability that the vehicle that Bob chooses is used or is a car is aproximately 0.71