In the early hours of Easter Friday, a news agency reported that 300 vehicles per hour entered KwaZulu-Natal province in South Africa for the 2023 Easter holiday.

Calculate the probability that no vehicles enter KwaZulu-Natal in a given minute? Answers can be uploaded or typed in the textbox provided. Show all working. Round final answers off to 2 decimal places. Use a period (.) instead of a comma (,) to indicate the decimal place.

To solve this problem, we need to calculate the probability that no vehicles enter KwaZulu-Natal in a given minute, given that 300 vehicles per hour enter the province. This problem can be modeled using the Poisson distribution, which is suitable for counting events that occur independently over a fixed period of time.

### Step-by-Step Solution:

1. **Understanding the Poisson Distribution:**

   The Poisson distribution formula is given by:

   \[
   P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}
   \]

   where:
   - \(P(X = k)\) is the probability of \(k\) events occurring in a fixed interval.
   - \(\lambda\) is the average number of events in that interval.
   - \(e\) is the base of the natural logarithm (approximately 2.71828).
   - \(k\) is the number of events we are interested in (in this case, \(k = 0\) for no vehicles entering).

2. **Determine \(\lambda\) for One Minute:**

   Given that 300 vehicles enter per hour, we need to find the average number of vehicles entering per minute. Since there are 60 minutes in an hour:

   \[
   \lambda = \frac{300 \text{ vehicles/hour}}{60 \text{ minutes/hour}} = 5 \text{ vehicles/minute}
   \]

3. **Calculate the Probability of No Vehicles Entering (k = 0):**

   We want to find the probability that no vehicles enter in a given minute, which corresponds to \(k = 0\). Plugging in the values into the Poisson formula:

   \[
   P(X = 0) = \frac{5^0 e^{-5}}{0!}
   \]

   Simplifying further:

   \[
   P(X = 0) = \frac{1 \cdot e^{-5}}{1} = e^{-5}
   \]

4. **Compute \(e^{-5}\):**

   Using the approximation \(e^{-5} \approx 0.006737947\):

   \[
   P(X = 0) \approx 0.006737947
   \]

5. **Round the Answer to Two Decimal Places:**

   \[
   P(X = 0) \approx 0.01
   \]

### Conclusion:

The probability that no vehicles enter KwaZulu-Natal in a given minute is approximately **0.01**.

---

Would you like more details or have any questions? Here are 5 related questions that might help deepen your understanding:

1. What is the probability that exactly 2 vehicles enter KwaZulu-Natal in a given minute?
2. How would the probability change if the rate was 600 vehicles per hour?
3. What is the expected number of vehicles entering in a 30-minute interval?
4. How would you use the Poisson distribution to model arrivals over different time intervals?
5. Can you explain the difference between the Poisson and Binomial distributions?

**Tip:** The Poisson distribution is widely used for modeling count-based data and is especially useful when events occur independently over time or space.

In the early hours of Easter Friday, a news agency reported that 300 vehicles per hour entered KwaZulu-Natal province in South Africa for the 2023 Easter holiday. Calculate the probability that no vehicles enter KwaZulu-Natal in a given minute? Answers can be uploaded or typed in the textbox provided. Show all working. Round final answers off to 2 decimal places. Use a period (.) instead of a comma (,) to indicate the decimal place.

Calculate the probability that no vehicles enter KwaZulu-Natal in a given minute using Poisson distribution, given that 300 vehicles enter per hour. This step-by-step guide covers key probability concepts and formulas, suitable for advanced high school or college students.

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