Answered

\begin{tabular}{|c|c|c|}
\hline Place & Is a city & Is in North America \\
\hline \multicolumn{3}{|l|}{India} \\
\hline Tokyo & [tex]$\checkmark$[/tex] & \\
\hline Houston & [tex]$\checkmark$[/tex] & [tex]$\checkmark$[/tex] \\
\hline \multicolumn{3}{|l|}{Peru} \\
\hline New York & [tex]$\checkmark$[/tex] & [tex]$\checkmark$[/tex] \\
\hline Tijuana & [tex]$\checkmark$[/tex] & [tex]$\checkmark$[/tex] \\
\hline Canada & & [tex]$\checkmark$[/tex] \\
\hline
\end{tabular}

Let event [tex]$A =$[/tex] The place is a city.
Let event [tex]$B =$[/tex] The place is in North America.

What is [tex]$P(A \text{ and } B)$[/tex]?



Answer :

To find the probability [tex]\( P(A \text{ and } B) \)[/tex], we need to determine the proportion of places that satisfy both conditions: being a city and being in North America.

Let's break down the problem step by step:

1. Count the total number of places:
- By examining the table, we see there are 6 places listed (Tokyo, Houston, New York, Tijuana, and Canada).

2. Identify the places that are both a city and in North America:
- From the table, the places that meet both criteria (city and in North America) are Houston and Tijuana.

3. Count the number of places that satisfy both conditions:
- We have identified 2 places (Houston and Tijuana) that are both a city and in North America.

4. Calculate the probability:
- The probability [tex]\( P(A \text{ and } B) \)[/tex] is calculated as the ratio of the number of places that are both a city and in North America to the total number of places.

Thus, the probability [tex]\( P(A \text{ and } B) \)[/tex] is:

[tex]\[ P(A \text{ and } B) = \frac{\text{Number of places that are both a city and in North America}}{\text{Total number of places}} = \frac{2}{6} = 0.3333333333333333 \][/tex]

So, the probability [tex]\( P(A \text{ and } B) \)[/tex] is approximately [tex]\( 0.3333 \)[/tex] or [tex]\( \frac{1}{3} \)[/tex].