Sure! Let's solve the problem step by step using the data provided:
1. Determine the total number of orders for each restaurant:
- For restaurant A:
[tex]\[
\text{Total orders} = 337 (\text{Accurate}) + 39 (\text{Not Accurate}) = 376
\][/tex]
- For restaurant B:
[tex]\[
\text{Total orders} = 273 (\text{Accurate}) + 51 (\text{Not Accurate}) = 324
\][/tex]
- For restaurant C:
[tex]\[
\text{Total orders} = 248 (\text{Accurate}) + 34 (\text{Not Accurate}) = 282
\][/tex]
- For restaurant D:
[tex]\[
\text{Total orders} = 146 (\text{Accurate}) + 11 (\text{Not Accurate}) = 157
\][/tex]
2. Calculate the total number of accurate orders and the total number of orders overall:
- Total accurate orders:
[tex]\[
337 (\text{A}) + 273 (\text{B}) + 248 (\text{C}) + 146 (\text{D}) = 1004
\][/tex]
- Total number of orders overall:
[tex]\[
376 (\text{A}) + 324 (\text{B}) + 282 (\text{C}) + 157 (\text{D}) = 1139
\][/tex]
3. Calculate the probability of selecting an accurate order:
[tex]\[
P(\text{Accurate}) = \frac{\text{Total number of accurate orders}}{\text{Total number of orders}} = \frac{1004}{1139} \approx 0.8815
\][/tex]
4. If selections are made with replacement, the probability of both being accurate is calculated as follows:
[tex]\[
P(\text{Both Accurate}) = P(\text{Accurate}) \times P(\text{Accurate}) = 0.8815 \times 0.8815 \approx 0.7770
\][/tex]
(Rounded to four decimal places as needed)
5. Determine if the events are independent:
Since selections are made with replacement, the outcome of the first selection does not affect the outcome of the second selection. Therefore, the events are independent.
Combining all the results, we have:
- The probability that both orders are accurate is approximately [tex]\(0.7770\)[/tex].
- The events are independent.
Thus, the answer is:
The probability is [tex]\(0.7770\)[/tex]. The events are independent.
