I am having issues with constructing a probability distribution table of a random variable x.
Here is the question:
According to recent data, 16.1% of motorists are uninsured. Suppose that 2 motorists are selected at random. Let x denote the number of motorists from this sample of two who are uninsured. Construct the probability distribution table of x.
I am really unsure of how to answer this.
The answer is p(0) = .7039, p(1) = .2702, p(2) = .0259
Call the two chosen motorists $A$ and $B$.
The probability that both $A$ and $B$ are uninsured is $0.161^2$.
The probability that $A$ is uninsured and $B$ is insured is $0.161\times(1-0.161)$.
The probability that $B$ is uninsured and $A$ is insured is $0.161\times(1-0.161)$.
The probability that EITHER $A$ is uninsured and $B$ insured OR $B$ is uninsured and $A$ is insured is the sum of those two numbers.
The probability that both are insured is $(1-0.161)^2$.
