Answer: 0.09
Explanation:
Let O and S denotes the event of having more than one car and having a sports car respectively.
By considering the given information, we have
P(O)=0.85 ; P(S)=0.23 and P(O∩S)= 0.17
Now, [tex]P(O\cup S)=P(O)+P(S)-P(O\cap S)[/tex]
[tex]=0.85+0.23-0.17=0.91[/tex]
Now, the probability that a customer selected at random insures exactly one car and it is not a sports car will be :-
[tex]P(\overline{O}\cap \overline{S})= P(\overline{O\cup S})\\\\=1-P(O\cup S)\\\\1-0.91=0.09[/tex]
Hence, the probability that a customer selected at random insures exactly one car and it is not a sports car = 0.09
