To find the probability that a randomly chosen person saw the movie on either Friday or Saturday, we can use the principle of inclusion-exclusion.
Let:
- \( A \) be the event that a person saw the movie on Friday.
- \( B \) be the event that a person saw the movie on Saturday.
From the data provided:
- \( |A| = 229 \) (number of people who went on Friday)
- \( |B| = 256 \) (number of people who went on Saturday)
- \( |A \cap B| = 24 \) (number of people who went both nights)
We want to find \( |A \cup B| \), the number of people who saw the movie on at least one of the two nights, which is given by the formula:
\[ |A \cup B| = |A| + |B| - |A \cap B| \]
Substituting in the values:
\[ |A \cup B| = 229 + 256 - 24 \] \[ |A \cup B| = 461 \]
Now, to find the probability \( P(A \cup B) \) that a randomly chosen person saw the movie on either night, we divide the number of people who saw the movie by the total number of people surveyed:
\[ P(A \cup B) = \frac{|A \cup B|}{\text{Total number of people}} = \frac{461}{1000} \]
Thus, the probability is:
\[ P(A \cup B) = 0.461 \]
Therefore, the probability that a person chosen at random saw the movie on Friday or Saturday is 0.461.
(Note: The response "0.437" is incorrect based on the calculations above.)