Answer :
Let:
• A, be the event of Bob choosing an used vehicle
,• B, be the event of Bob choosing a car
Now, we know that:
[tex]P(A\cup B)=P(A)+P(B)-P(A\cap B)[/tex]Out of the 41 vehicles, 18 are used. Therefore,
[tex]P(A)=\frac{18}{41}[/tex]Out of the 41 vehicles, 19 are cars. This way,
[tex]P(B)=\frac{19}{41}[/tex]Out of the 41 vehicles, 8 are used cars. Thereby,
[tex]P(A\cap B)=\frac{8}{41}[/tex]This way, we'll have that
[tex]\begin{gathered} P(A\cup B)=\frac{18}{41}+\frac{19}{41}-\frac{8}{41} \\ \\ \Rightarrow P(A\cup B)=\frac{29}{41} \\ \\ \rightarrow P(A\cup B)=0.71 \end{gathered}[/tex]Therefore, the probability that the vehicle that Bob chooses is used or is a car is aproximately 0.71