Questions: The table below shows the results of a survey that asked 2870 people whether they are involved in any type of charity work. A person is selected at random from the sample. Complete parts (a) through (d). Frequently Occasionally Not at all Total ---------------------------------------------------- Male 227 455 791 1473 Female 206 450 741 1397 Total 433 905 1532 2870 (Round to the nearest thousandth as needed.) (c) Find the probability that the person is male or frequently involved in charity work. P(being male or being frequently involved) = 0.5850 (Round to the nearest thousandth as needed.) (d) Find the probability that the person is female or not frequently involved in charity work. P(being female or not being frequently involved) = (Round to the nearest thousandth as needed.)

Solution Steps

We are given a table with survey results showing the involvement of males and females in charity work. We need to find the probability of certain events based on this data.

Let's define the events for clarity:

• \( M \): The person is male.
• \( F \): The person is female.
• \( Fq \): The person is frequently involved in charity work.
• \( O \): The person is occasionally involved in charity work.
• \( N \): The person is not at all involved in charity work.

We need to find the probability that the person is male or frequently involved in charity work: \[ P(M \cup Fq) \]

Using the formula for the union of two events: \[ P(M \cup Fq) = P(M) + P(Fq) - P(M \cap Fq) \]

From the table:

• \( P(M) = \frac{1473}{2870} \)
• \( P(Fq) = \frac{433}{2870} \)
• \( P(M \cap Fq) = \frac{227}{2870} \)

Substituting these values: \[ P(M \cup Fq) = \frac{1473}{2870} + \frac{433}{2870} - \frac{227}{2870} \] \[ P(M \cup Fq) = \frac{1473 + 433 - 227}{2870} \] \[ P(M \cup Fq) = \frac{1679}{2870} \] \[ P(M \cup Fq) \approx 0.5850 \]

We need to find the probability that the person is female or not frequently involved in charity work: \[ P(F \cup (O \cup N)) \]

Using the formula for the union of two events: \[ P(F \cup (O \cup N)) = P(F) + P(O \cup N) - P(F \cap (O \cup N)) \]

From the table:

• \( P(F) = \frac{1397}{2870} \)
• \( P(O \cup N) = \frac{905 + 1532}{2870} = \frac{2437}{2870} \)
• \( P(F \cap (O \cup N)) = \frac{450 + 741}{2870} = \frac{1191}{2870} \)

Substituting these values: \[ P(F \cup (O \cup N)) = \frac{1397}{2870} + \frac{2437}{2870} - \frac{1191}{2870} \] \[ P(F \cup (O \cup N)) = \frac{1397 + 2437 - 1191}{2870} \] \[ P(F \cup (O \cup N)) = \frac{2643}{2870} \] \[ P(F \cup (O \cup N)) \approx 0.9209 \]

Final Answer