probability of multiple car ownership for a household Given that out of 100 houses in a locality, 20 have no car, 40 have one car, 10 have 2 cars, and 30 have 3 cars. For a randomly selected car on the street, what is the probability that that car's household owns at least one more car. 
Edit* 
This isn't a homework problem. I came across the problem while going through probability questions that were asked for a junior statistician role. 
My logic so far has been this. 
1) 80 houses have cars, of which 40 have more than one. Therefore for a car to be a part of a household with multiple cars, it would be 40/80 = 50%.
2) That being said, I have 90+20= 110 cars being from houses with more than one car; and 40 cars from households with only one car. 
I'm not sure how to factor this into my calculations. Would the probability I am looking for, be just 110/150? Or would it be conditioned on the probability from (1)? Or am I just looking at things the wrong way. 
Its been a while since I've had to work on these sort of problems.
Thank you for any additional insight. 
 A: Your point 2 is correct.  You are randomly picking a car and asking what kind of house it belongs to, not picking a house and asking what cars they have.  Your point 1 would be correct if you picked a house that was guaranteed to have at least one car and asked if they had more than one.  These are two different questions which have different answers.  In many probability problems, the important thing is to understand what question you are asking.
A: I think it is best to think about this in terms of the set of cars you have. As stated in your original post, you have a total of 150 cars. Out of these 150 cars, 90 belong to a household with 3 cars $ (30 \times 3) $, 20 belong to a household with 2 cars $ (10 \times 2) $, and 40 belong to a household with 1 car $ (1 \times 40) $. 
Now you are randomly drawing from this set of 150 cars.
P(car belongs to household with 3 cars) $ = \frac{90}{150} $
P(car belongs to household with 2 cars) $ = \frac{20}{150} $ 
P(car belongs to household with 1 car) $ = \frac{40}{150} $ 
