To solve this problem, let’s go through each part in detail:

Step 1: Total Accurate Orders and Total Orders

The table shows the drive-thru order accuracy data across four restaurants (A, B, C, D).

Accurate Orders:

• Restaurant A: 311
• Restaurant B: 273
• Restaurant C: 248
• Restaurant D: 127

Total accurate orders = 311 + 273 + 248 + 127 = 959

Not Accurate Orders:

• Restaurant A: 38
• Restaurant B: 52
• Restaurant C: 32
• Restaurant D: 17

Total not accurate orders = 38 + 52 + 32 + 17 = 139

Total Orders:

Total orders = Total accurate orders + Total not accurate orders
Total orders = 959 + 139 = 1098

Part (a): Probability of Selecting Two Accurate Orders with Replacement

Since selections are made with replacement, the probability of each selection being accurate does not change.

1. Calculate Probability of Selecting an Accurate Order Once: $P(\text{Accurate}) = \frac{\text{Total Accurate Orders}}{\text{Total Orders}} = \frac{959}{1098}$

2. Calculate Probability of Both Orders Being Accurate: Since the selections are made with replacement, the probability of two orders both being accurate is: $P(\text{Both Accurate}) = P(\text{Accurate}) \times P(\text{Accurate}) = \left(\frac{959}{1098}\right)^2$

3. Calculate the Result: $P(\text{Both Accurate}) = \left(\frac{959}{1098}\right)^2 \approx 0.7647$

So, the probability that both orders are accurate is approximately 0.7647.

Part (b): Independence of Events

Since the selections are made with replacement, each selection is independent of the previous one. This means the outcome of one selection does not affect the outcome of the next.

Thus, the events are independent.

Summary of Answers:

1. The probability that both selected orders are accurate is 0.7647.
2. The events are independent.

Would you like more details on any part of the solution?

Related Questions

1. What would be the probability if the selections were made without replacement?
2. How would the probability change if the number of accurate orders was different?
3. What is the probability of selecting one accurate and one not accurate order with replacement?
4. How does independence affect the probability calculation in this scenario?
5. How would the result change if a fifth restaurant was added with additional data?

Tip:

When events are independent, each event's probability remains unaffected by previous outcomes, which simplifies calculations in probability problems involving replacement.