A consumer organization estimates that over a 1-year period 17% of cars will need to be repaired once, 7% will need repairs twice, and 4% will require three or more repairs. If you own 2 cars, what is the probability that

a) neither will need repair?
b) both will need repair?
c) at least one car will need repair?

I'm studying for a test! I understand how you would find the probability for 1 car chosen at random, but not for 2 cars.

Thanks for the help!!

You are given three probabilities corresponding to P(X=1, 2 or 3), where X is the number of times a particular car will need a repair. The probability that one car will need a repair is independent of the probability that another car will need a repair. Armed with this knowledge you can set up some equations:

1. P(Neither Car Will Need Repair). If neither car will need a repair, that means that both cars are fine right? So that means that the P(0)=0.72 (be sure you understand how I got this). We established that the probability that one car needs repair is independent of the other so the probability that they BOTH will not need to be repaired is (0.72)(0.72)=0.5184 chance that neither car will need to be repaired.

I will leave the remaining two to you, but part of the problem in solving these is not only understanding some of the properties of probability, but understanding WHAT the question is asking. Questions one and two are "AND" statements, and question three is an "OR" statement.