Jerold

2020-11-26

The following two-way contingency table gives the breakdown of a town's population according to party affiliation (A, B, C, or None) and opinion on a property tax issue:
Opinion
$\begin{array}{|cccc|}\hline \text{Affiliation}& \text{Favors}& \text{Opposes}& \text{Undecided}\\ \text{A}& 0.12& 0.09& 0.07\\ \text{B}& 0.16& 0.12& 0.14\\ \text{C}& 0.04& 0.03& 0.06\\ \text{None}& 0.08& 0.06& 0.03\\ \hline\end{array}\phantom{\rule{0ex}{0ex}}$
A person is selected at random. What is the probability that the person is affiliated with parties A or B?

yagombyeR

Step 1
Given information-
We have given the table gives the breakdown of a town's population according to party affiliation (A, B, C, or None) and opinion on a property tax issue.
We have to find the probability that the person is affiliated with parties A or B.$\begin{array}{|ccccc|}\hline \text{Affiliation}& \text{Favors}& \text{Opposes}& \text{Undecided}& \text{Total}\\ \text{A}& 0.12& 0.09& 0.07& 0.28\\ \text{B}& 0.16& 0.12& 0.14& 0.42\\ \text{C}& 0.04& 0.03& 0.06& 0.13\\ \text{None}& 0.08& 0.06& 0.03& 0.17\\ \text{Total}& 0.4& 0.3& 0.3& 1\\ \hline\end{array}\phantom{\rule{0ex}{0ex}}$
Step 2
So,
$P\left(A\right)=0.28,P\left(B\right)=0.42$
Since both are independent events.
$P\left(\text{the person is affiliated with parties A or B}\right)=P\left(A\right)+P\left(B\right)$
$P\left(\text{the person is affiliated with parties A or B}\right)=0.28+0.42=0.70$
Hence , the probability that the person is affiliated with parties A or B is 0.70

Do you have a similar question?