Question

Let Event A represent a late arrival and event B represent transportation by car. what would the product of P (A) and P (B) need to equal to show that the events are independent? express the answer as a decimal to the nearest hundredth

150 people were late
250 people used a car to transport
230 people were on time with a car
20 were late with a car

Answer 1

To determine if events A (late arrival) and B (transportation by car) are independent, we need to verify the condition:

\[ P(A \cap B) = P(A) \cdot P(B) \]

1. **Calculating the probabilities:**

Let's find \( P(A) \) (the probability of a late arrival), \( P(B) \) (the probability of being transported by car), and \( P(A \cap B) \) (the probability of being late and using a car).

Given:
- Total number of people = \( 150 \) (late) + \( 250 - 20 \) (on time with a car) + \( 20 \) (late with a car) = \( 150 + 250 = 400 \).

- Number of people transported by car = \( 250 \).
- Number of people who were late with a car = \( 20 \).

2. **Calculating \( P(A) \):**

\[
P(A) = \frac{\text{Number of late people}}{\text{Total number of people}} = \frac{150}{400} = 0.375
\]

3. **Calculating \( P(B) \):**

\[
P(B) = \frac{\text{Number of people using a car}}{\text{Total number of people}} = \frac{250}{400} = 0.625
\]

4. **Calculating \( P(A \cap B) \):**

\[
P(A \cap B) = \frac{\text{Number of late people using a car}}{\text{Total number of people}} = \frac{20}{400} = 0.05
\]

5. **Checking independence:**

Now we need to check if \( P(A \cap B) = P(A) \cdot P(B) \):

\[
P(A) \cdot P(B) = 0.375 \cdot 0.625 = 0.234375
\]

6. **Conclusion:**

Since:

\[
P(A \cap B) = 0.05 \quad \text{and} \quad P(A) \cdot P(B) = 0.234375
\]

These two values are not equal, hence events A and B are not independent.

Thus, for events A and B to be independent, the product \( P(A) \cdot P(B) \) would need to equal \( 0.05 \). Rounding to the nearest hundredth, that is:

\[
\boxed{0.05}
\]