To find the probability that a randomly selected person is male or is carpooling, we use the principle of inclusion and exclusion in probability. The formula for this is:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Where:

• $A$ is the event that the person is male.
• $B$ is the event that the person is carpooling.

Step 1: Calculate $P(A)$

The probability that a person is male: $P(A) = \frac{\text{Number of males}}{\text{Total number of people}} = \frac{110}{200} = 0.55$

Step 2: Calculate $P(B)$

The probability that a person is carpooling: $P(B) = \frac{\text{Number of people carpooling}}{\text{Total number of people}} = \frac{63}{200} = 0.315$

Step 3: Calculate $P(A \cap B)$

The probability that a person is both male and carpooling: $P(A \cap B) = \frac{\text{Number of males carpooling}}{\text{Total number of people}} = \frac{28}{200} = 0.14$

Step 4: Calculate $P(A \cup B)$

Using the principle of inclusion and exclusion: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.55 + 0.315 - 0.14 = 0.725$

Therefore, the probability that a randomly selected person is male or is carpooling is 0.725 or 72.5%.

Do you want more details or have any questions?

Here are some related questions:

1. What is the probability that a randomly selected person is female or rides the bus?
2. What is the probability that a randomly selected person is male and rides the train?
3. What is the probability that a randomly selected person is female or rides the train?
4. What is the probability that a randomly selected person rides the bus or carpools?
5. What is the probability that a randomly selected person is female and rides the bus?
6. What is the probability that a randomly selected person is male and does not carpool?
7. What is the probability that a randomly selected person is female and does not ride the train?
8. What is the probability that a randomly selected person rides the train and carpools?

Tip: When solving probability problems involving "or" conditions, always remember to subtract the overlap (the intersection of the two events) to avoid double counting.