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 survey:
- \( |A| = 229 \) (people who saw the movie on Friday)
- \( |B| = 256 \) (people who saw the movie on Saturday)
- \( |A \cap B| = 24 \) (people who saw the movie on both nights)
Using the formula for the union of two sets, we have:
\[ |A \cup B| = |A| + |B| - |A \cap B| \]
Substituting the values:
\[ |A \cup B| = 229 + 256 - 24 = 461 \]
Now, the probability \( P(A \cup B) \) that a randomly chosen person saw the movie on either Friday or Saturday can be calculated as follows:
\[ P(A \cup B) = \frac{|A \cup B|}{\text{Total number of people}} = \frac{461}{1000} \]
Calculating this gives:
\[ P(A \cup B) = 0.461 \]
Rounding to the nearest thousandth, we get:
\[ \boxed{0.461} \]
Thus, the probability that a randomly chosen person saw the movie on either Friday or Saturday is approximately 0.461.