Use the data in the following​ table, which lists​ drive-thru order accuracy at popular fast food chains. Assume that orders are randomly selected from those included in the table.

​Drive-thru **Restaurant**   

**A**

**B**

**C**

**D**

 **Order Accurate**

327327

277277

2492\\49138138

 **Order Not Accurate**
A.39 B. 53 C. 37 D. 14If two orders are​ selected, find the probability that they are both from Restaurant D.

**a.** Assume that the selections are made with replacement. Are the events​ independent?

**b.** Assume that the selections are made without replacement. Are the events​ independent?

Let's first analyze the given data to clarify the problem.

### Data Table:
| **Restaurant** | **A** | **B** | **C** | **D** |
|:--|--:|--:|--:|--:|
| **Order Accurate** | 327 | 277 | 249 | 138 |
| **Order Not Accurate** | 39 | 53 | 37 | 14 |

From this table, we can calculate the total number of orders for each restaurant:
- Restaurant A: \( 327 + 39 = 366 \)
- Restaurant B: \( 277 + 53 = 330 \)
- Restaurant C: \( 249 + 37 = 286 \)
- Restaurant D: \( 138 + 14 = 152 \)

The total number of orders for all restaurants is:
\[
366 + 330 + 286 + 152 = 1134
\]

### Part (a): **Selections Made with Replacement**

When selections are made with replacement, the selections are independent. The probability of each event does not change based on previous events.

To find the probability that both selected orders are from Restaurant D:
1. The probability that the first order is from Restaurant D is:
   \[
   P(\text{1st order from D}) = \frac{152}{1134}
   \]
2. The probability that the second order is also from Restaurant D (since the events are independent):
   \[
   P(\text{2nd order from D}) = \frac{152}{1134}
   \]
3. The probability that both orders are from Restaurant D is:
   \[
   P(\text{both orders from D}) = \left( \frac{152}{1134} \right) \times \left( \frac{152}{1134} \right) = \frac{152^2}{1134^2} = \frac{23104}{1285956} \approx 0.01797
   \]

Thus, the probability that both orders are from Restaurant D when selections are made with replacement is approximately **0.01797**.

### Part (b): **Selections Made without Replacement**

When selections are made without replacement, the selections are dependent, because the probability for the second selection changes depending on the first.

To find the probability that both selected orders are from Restaurant D:
1. The probability that the first order is from Restaurant D is the same:
   \[
   P(\text{1st order from D}) = \frac{152}{1134}
   \]
2. However, if the first order is from Restaurant D, there is now one fewer order from Restaurant D and one fewer total order, so the probability for the second order being from Restaurant D is:
   \[
   P(\text{2nd order from D}) = \frac{151}{1133}
   \]
3. The probability that both orders are from Restaurant D is:
   \[
   P(\text{both orders from D}) = \frac{152}{1134} \times \frac{151}{1133} = \frac{152 \times 151}{1134 \times 1133} = \frac{22952}{1285158} \approx 0.01786
   \]

Thus, the probability that both orders are from Restaurant D when selections are made without replacement is approximately **0.01786**.

### Summary of Answers:
- **a.** The probability of both orders being from Restaurant D when selections are made **with replacement** is approximately **0.01797**. The events are **independent**.
- **b.** The probability of both orders being from Restaurant D when selections are made **without replacement** is approximately **0.01786**. The events are **dependent**.

---

Would you like more details, or do you have any questions?

Here are 5 follow-up questions for further exploration:
1. What is the probability of selecting one accurate and one inaccurate order from Restaurant D with replacement?
2. How does the probability of selecting two orders from Restaurant A compare to Restaurant D?
3. What is the probability of selecting exactly one order from Restaurant B and one from Restaurant C without replacement?
4. How does the probability change if we select three orders instead of two?
5. What is the total probability of selecting accurate orders for all restaurants combined?

**Tip:** When working with probabilities involving replacement, remember that the key factor is whether or not the total number of available outcomes changes after each selection!

Use the data in the following table, which lists drive-thru order accuracy at popular fast food chains. Assume that orders are randomly selected from those included in the table. Drive-thru **Restaurant** A, B, C, D **Order Accurate** 327, 277, 249, 138 **Order Not Accurate** A: 39, B: 53, C: 37, D: 14. If two orders are selected, find the probability that they are both from Restaurant D. a. Assume that the selections are made with replacement. Are the events independent? b. Assume that the selections are made without replacement. Are the events independent?

Find the probability of selecting two orders from Restaurant D based on drive-thru order accuracy data. Explore the difference between independent and dependent events, including with and without replacement cases. Learn about probability and how the selection process affects outcomes.

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