Let's solve each part of the problem step by step.

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Part a) Probability that a randomly chosen person has a speeding ticket:

The total number of people surveyed is $612$, and the total number of people who received a speeding ticket is $250$.

$P(\text{Speeding Ticket}) = \frac{\text{Number of people with speeding tickets}}{\text{Total number of people surveyed}}$

$P(\text{Speeding Ticket}) = \frac{250}{612}$

$P(\text{Speeding Ticket}) \approx 0.408$

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Part b) Probability that a randomly chosen person has a red car or got a speeding ticket:

To calculate this, we use the formula for the union of two events:

$P(\text{Red Car or Speeding Ticket}) = P(\text{Red Car}) + P(\text{Speeding Ticket}) - P(\text{Red Car and Speeding Ticket})$

Step 1: Find $P(\text{Red Car})$

The total number of people with red cars is $315$, and the total surveyed is $612$.

$P(\text{Red Car}) = \frac{315}{612} \approx 0.515$

Step 2: Use $P(\text{Speeding Ticket})$

We already calculated this in part (a) as $0.408$.

Step 3: Find $P(\text{Red Car and Speeding Ticket})$

The number of people with red cars and speeding tickets is $144$.

$P(\text{Red Car and Speeding Ticket}) = \frac{144}{612} \approx 0.235$

Step 4: Calculate $P(\text{Red Car or Speeding Ticket})$

Substitute the values:

$P(\text{Red Car or Speeding Ticket}) = 0.515 + 0.408 - 0.235$

$P(\text{Red Car or Speeding Ticket}) \approx 0.688$

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Final Answers:

a) Probability that a randomly chosen person has a speeding ticket:

$\boxed{0.408}$

b) Probability that a randomly chosen person has a red car or got a speeding ticket:

$\boxed{0.688}$

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Would you like additional clarification or details? Here are related questions for exploration:

1. What is the probability that a randomly chosen person has a red car and no speeding ticket?
2. What is the probability of not having a red car or a speeding ticket?
3. How does the probability of having a speeding ticket differ between red cars and non-red cars?
4. What is the conditional probability of having a speeding ticket given the person has a red car?
5. How would the results change if the number of red cars increased by 10%?

Tip: To better understand probabilities involving "or," remember the inclusion-exclusion principle to avoid overcounting overlapping events.