Answer:
A consumer organization estimates that over a 1-year period 16% of cars will need to be repaired once, 9% will need repairs twice, and 3% will require three or more repairs..
We have the cars that do not need repair =[tex]100-(16+9+3)=72[/tex]%
(16+9+3=28% cars need repair of some sort)
What is the probability that :
a) neither will need repair?
=> [tex]0.72\times0.72=0.5184[/tex]
b) both will need repair?
=> [tex]0.28\times0.28=0.0784[/tex]
c) at least one car will need repair?
=> [tex]1-(0.72\times0.72)[/tex]
=> [tex]1-0.5184=0.4816[/tex]
