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]
