a movie company surveyed 1000 people. 229 people said they went to see the new movie on friday, 256 said they went on saturday. if 24 people saw the movie both nightd, what is the probality that a person chosen at random saw the movie on friday or saturday? round to nearest thousandth

Answer 1

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.