To determine [tex]\( P(A \text{ and } B) \)[/tex], we need to calculate the probability that a place is both a city and in North America. We will follow these steps:
1. Identify the places and their properties:
- Chicago: City and in North America
- Peru: Not a city and not in North America
- Miami: City and in North America
- Canada: Not a city but in North America
- Mexico: Not a city but in North America
2. Count the number of places that are both cities and in North America:
- Chicago is a city and in North America.
- Miami is a city and in North America.
From the table, we see that there are 2 places that satisfy both conditions. Therefore, [tex]\( \text{Number of places that are both cities and in North America} = 2 \)[/tex].
3. Count the total number of places:
- The total number of places listed is 5.
4. Calculate [tex]\( P(A \text{ and } B) \)[/tex]:
- The probability of an event is given by the number of favorable outcomes divided by the total number of possible outcomes.
[tex]\[
P(A \text{ and } B) = \frac{\text{Number of places that are both cities and in North America}}{\text{Total number of places}}
\][/tex]
Substituting the values we counted:
[tex]\[
P(A \text{ and } B) = \frac{2}{5}
\][/tex]
Therefore, the probability [tex]\( P(A \text{ and } B) \)[/tex] is [tex]\( \frac{2}{5} \)[/tex], which is equal to 0.4. However, notice that none of the options exactly match [tex]\( \frac{2}{5} \)[/tex] or 0.4 in their given forms.
It seems the problem might have an issue with provided choices if matching exact values. None of the provided options (A through D) accurately represent [tex]\( \frac{2}{5} \)[/tex], or exactly 0.4.
Given the solution we've reached, the correct probability amongst provided options should ideally have been [tex]\( \frac{2}{5} \)[/tex].
