According to the 2012 census, 62% of U.S. Households own the home they live in.
If three households are randomly selected, find the probability that at least one owns their home.
SO.
p=.62 1-p=.38 n=3
Now how do I find the probability of at least 1 owns their home?

Question

According to the 2012 census, 62% of U.S. Households own the home they live in. 
If three households are randomly selected, find the probability that at least one owns their home.
SO.
p=.62 1-p=.38 n=3
Now how do I find the probability of at least 1 owns their home?

parthkohli · Answer

Well, if you think about it, the event "owning at least one home" is mutually exclusive with "owning no home". Is it easier to find the probability that no one owns a home?

anonymous · Answer

Thanks I actually found the answer but I couldn't explain it without your comment about understanding it.
I took
P(0)=(3C0)(.62)^0(.38)^3=.054872
P(at least One) = 1-.054872= .945128 or 94.51%
Thanks!

parthkohli · Answer

Yeah, that makes sense. You don't even need to understand binomial probability to find the probability of owning no home.

anonymous · Answer

I was more concerned with how to approach it versus the concept at the moment. ahah.