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

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
$$\begin{array}{|c|c|c|}\hline \text{Affiliation}&\text{Favors}&\text{Opposes}&\text{Undecided}\\\hline \text{A} &0.12&0.09&0.07\\ \hline \text{B}&0.16&0.12&0.14 \\ \hline \text{C}&0.04&0.03&0.06 \\ \hline \text{None}&0.08&0.06&0.03 \\ \hline \end{array}\\$$
A person is selected at random. What is the probability that the person is affiliated with parties A or B?

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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. $$\begin{array}{|c|c|c|}\hline \text{Affiliation}&\text{Favors}&\text{Opposes}&\text{Undecided}&\text{Total}\\\hline \text{A} &0.12&0.09&0.07&0.28\\ \hline \text{B}&0.16&0.12&0.14&0.42 \\ \hline \text{C}&0.04&0.03&0.06&0.13\\ \hline \text{None}&0.08&0.06&0.03&0.17\\ \hline \text{Total}&0.4&0.3&0.3&1 \\ \hline \end{array}\\$$
Step 2
So,
$$P(A)=0.28, P(B)=0.42$$
Since both are independent events.
$$P(\text{the person is affiliated with parties A or B})=P(A)+P(B)$$
$$P(\text{the person is affiliated with parties A or B})=0.28+0.42=0.70$$
Hence , the probability that the person is affiliated with parties A or B is 0.70