To answer the question about the probability of two orders being accurate with replacement, we first need to determine the following details:

Step 1: Calculate the total number of orders for each restaurant

For each restaurant, the total number of orders is the sum of accurate and inaccurate orders. The data for each restaurant is as follows:

• Restaurant A:

  • Accurate Orders = 330
  • Not Accurate Orders = 34
  • Total Orders = 330 + 34 = 364

• Restaurant B:

  • Accurate Orders = 278
  • Not Accurate Orders = 59
  • Total Orders = 278 + 59 = 337

• Restaurant C:

  • Accurate Orders = 238
  • Not Accurate Orders = 33
  • Total Orders = 238 + 33 = 271

• Restaurant D:

  • Accurate Orders = 127
  • Not Accurate Orders = 12
  • Total Orders = 127 + 12 = 139

Step 2: Find the probability of selecting an accurate order from each restaurant

The probability of selecting an accurate order from a specific restaurant is given by:

$P(\text{Accurate}) = \frac{\text{Accurate Orders}}{\text{Total Orders}}$

Now, let's compute the probability for each restaurant:

• Restaurant A: $P_A(\text{Accurate}) = \frac{330}{364} \approx 0.9074$

• Restaurant B: $P_B(\text{Accurate}) = \frac{278}{337} \approx 0.8263$

• Restaurant C: $P_C(\text{Accurate}) = \frac{238}{271} \approx 0.8775$

• Restaurant D: $P_D(\text{Accurate}) = \frac{127}{139} \approx 0.9136$

Step 3: Assume selections are made with replacement

The probability that both selected orders are accurate is the product of the probabilities of selecting an accurate order in each selection. If selections are made with replacement, then the probability that both orders are accurate from any restaurant is simply:

$P(\text{Both Accurate}) = P(\text{Accurate}) \times P(\text{Accurate}) = P(\text{Accurate})^2$

Let's calculate the probability for each restaurant.

• Restaurant A: $P_A(\text{Both Accurate}) = 0.9074^2 \approx 0.8234$

• Restaurant B: $P_B(\text{Both Accurate}) = 0.8263^2 \approx 0.6828$

• Restaurant C: $P_C(\text{Both Accurate}) = 0.8775^2 \approx 0.7696$

• Restaurant D: $P_D(\text{Both Accurate}) = 0.9136^2 \approx 0.8345$

Step 4: Determine if the events are independent

For events to be independent, the probability of the combined event (two orders being accurate) must be the product of their individual probabilities. Since selections are being made with replacement, the events are independent because the second selection does not depend on the first.

Final Answer:

• The probability is:

  • For Restaurant A: $P_A(\text{Both Accurate}) = 0.8234$
  • For Restaurant B: $P_B(\text{Both Accurate}) = 0.6828$
  • For Restaurant C: $P_C(\text{Both Accurate}) = 0.7696$
  • For Restaurant D: $P_D(\text{Both Accurate}) = 0.8345$

• The events are independent.

Would you like further explanations on any step, or would you like me to elaborate on any part?