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.

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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!