Let's first compute the total number of orders from all restaurants and then proceed to answer the questions.

The total number of orders for each restaurant is the sum of accurate and not accurate orders:

• Restaurant A: $327 + 40 = 367$
• Restaurant B: $276 + 58 = 334$
• Restaurant C: $235 + 38 = 273$
• Restaurant D: $120 + 12 = 132$

The total number of orders across all restaurants is:

$367 + 334 + 273 + 132 = 1106$

Part 1: Selections with Replacement

When the selections are made with replacement, the probability of selecting an order from Restaurant D on each draw is:

$P(\text{Order from D}) = \frac{132}{1106}$

Since the selections are made with replacement, the probability of selecting two orders from Restaurant D is:

$P(\text{Two orders from D}) = \left( \frac{132}{1106} \right)^2$

Let's compute this:

$P(\text{Two orders from D}) = \frac{132^2}{1106^2}$

$P(\text{Two orders from D}) \approx \frac{17424}{1223236} \approx 0.01424$

So, the probability is approximately 0.0142.

Since the selections are made with replacement, the events are independent because the outcome of the first selection does not affect the probability of the second selection.

Part 2: Selections without Replacement

When the selections are made without replacement, the probability of selecting an order from Restaurant D on the first draw is:

$P(\text{First order from D}) = \frac{132}{1106}$

After selecting one order from Restaurant D, there are only 131 orders left from Restaurant D and 1105 total orders remaining. So, the probability of selecting a second order from Restaurant D is:

$P(\text{Second order from D}) = \frac{131}{1105}$

Thus, the probability of selecting two orders from Restaurant D without replacement is:

$P(\text{Two orders from D without replacement}) = \frac{132}{1106} \times \frac{131}{1105}$

Let's compute this:

$P(\text{Two orders from D without replacement}) \approx \frac{132 \times 131}{1106 \times 1105} \approx \frac{17292}{1223230} \approx 0.01414$

So, the probability is approximately 0.0141.

Since the selections are made without replacement, the events are not independent because the first selection affects the probability of the second selection.

Summary:

• Part 1 (with replacement): The probability is 0.0142 and the events are independent.
• Part 2 (without replacement): The probability is 0.0141 and the events are not independent.

Let me know if you need any further clarification or details.

Here are some related questions:

1. What is the probability of getting one accurate order from Restaurant D?
2. How does the probability change if we select three orders instead of two?
3. What is the probability of selecting one accurate and one not accurate order from Restaurant A?
4. How do probabilities compare across different restaurants for two orders?
5. How would the probability change if Restaurant D had more orders?

Tip: When calculating probabilities with or without replacement, always consider whether the first event affects the second event, as this determines independence.