\begin{tabular}{|c|c|c|c|c|}
\hline \multicolumn{5}{|c|}{ Network Preference by Floor } \\
\hline & 1st Floor & 2nd Floor & 3rd Floor & Total \\
\hline Internet Only & 13 & 15 & 11 & 39 \\
\hline Cable Only & 14 & 16 & 6 & 36 \\
\hline Internet \& Cable & 11 & 9 & 22 & 42 \\
\hline Total & 38 & 40 & 39 & 117 \\
\hline
\end{tabular}

The two-way table shows network preferences for an apartment building. Based on the table, if we define [tex]$A$[/tex] to be Internet Only and [tex]$B$[/tex] to be 3rd Floor connections, what is [tex]$P(A \text{ and } B)$[/tex]?

A. 0.094
B. 0.952
C. 0.205



Answer :

To determine the probability [tex]\( P(A \text{ and } B) \)[/tex] where [tex]\( A \)[/tex] is the event that a person prefers Internet Only and [tex]\( B \)[/tex] is the event that a person is on the 3rd floor, follow these steps:

1. Identify the total number of people in the surveyed group:
The total number of people is given by the sum of all entries in the table, which is 117.

2. Determine the number of people who satisfy both conditions [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
From the table, the number of people who prefer Internet Only on the 3rd floor is 11.

3. Calculate the probability [tex]\( P(A \text{ and } B) \)[/tex]:
The probability of both events occurring is given by the ratio of the number of favorable outcomes to the total number of outcomes. In this case, it is the number of people who prefer Internet Only on the 3rd floor divided by the total number of people.

Therefore, [tex]\( P(A \text{ and } B) = \frac{11}{117} \)[/tex].

Upon evaluating [tex]\( \frac{11}{117} \)[/tex], we obtain:

[tex]\[ P(A \text{ and } B) \approx 0.09401709401709402 \][/tex]

Hence, the probability [tex]\( P(A \text{ and } B) \)[/tex] is approximately 0.094, which matches with option:

[tex]\[ P(A \text{ and } B) = 0.094 \][/tex]

Thus, the correct answer is:
[tex]\[ 0.094 \][/tex]