Use the data in the following table, which lists drive-thru order accuracy at popular fast food chains. Assume that orders are randomly selected from those included in the table.

[tex]\[
\begin{tabular}{|l|c|c|c|c|}
\hline
& A & B & C & D \\
\hline
Order Accurate & 326 & 265 & 233 & 126 \\
\hline
Order Not Accurate & 34 & 54 & 39 & 10 \\
\hline
\end{tabular}
\][/tex]

1. If one order is selected, find the probability of getting an order that is not accurate or is from Restaurant C. Are the events of selecting an order that is not accurate and selecting an order from Restaurant C disjoint events?

The probability of getting an order from Restaurant [tex][tex]$C$[/tex][/tex] or an order that is not accurate is [tex]\square[/tex] (Round to three decimal places as needed).

2. Are the events of selecting an order from Restaurant [tex][tex]$C$[/tex][/tex] and selecting an inaccurate order disjoint events?

The events [tex]\square[/tex] disjoint because it [tex]\square[/tex] possible to [tex]\square[/tex]



Answer :

To solve this problem, we need to calculate the probability of getting an order that is not accurate or an order from Restaurant C, and determine if these events are disjoint.

### Step-by-Step Solution

1. Sum of Total Orders:
- Restaurant A: 326 (Accurate) + 34 (Not Accurate) = 360 orders
- Restaurant B: 265 (Accurate) + 54 (Not Accurate) = 319 orders
- Restaurant C: 233 (Accurate) + 39 (Not Accurate) = 272 orders
- Restaurant D: 126 (Accurate) + 10 (Not Accurate) = 136 orders
- Total orders: 360 + 319 + 272 + 136 = 1087 orders

2. Total Number of Inaccurate Orders:
- Not Accurate from A: 34 orders
- Not Accurate from B: 54 orders
- Not Accurate from C: 39 orders
- Not Accurate from D: 10 orders
- Total not accurate: 34 + 54 + 39 + 10 = 137 orders

3. Total Orders from Restaurant C:
- Orders from C: 233 (Accurate) + 39 (Not Accurate) = 272 orders

4. Probability Calculations:
- Probability of an order being not accurate:
[tex]\[ \text{P(Not Accurate)} = \frac{\text{Total Not Accurate Orders}}{\text{Total Orders}} = \frac{137}{1087} \approx 0.126 \][/tex]
- Probability of an order being from Restaurant C:
[tex]\[ \text{P(From Restaurant C)} = \frac{\text{Total Orders from C}}{\text{Total Orders}} = \frac{272}{1087} \approx 0.250 \][/tex]
- Probability of an order being not accurate and from Restaurant C:
[tex]\[ \text{P(Not Accurate and From C)} = \frac{\text{Not Accurate Orders from C}}{\text{Total Orders}} = \frac{39}{1087} \approx 0.036 \][/tex]

5. Combined Probability (Not Accurate OR From Restaurant C):
Using the formula for the union of two events:
[tex]\[ \text{P(A or B)} = \text{P(A)} + \text{P(B)} - \text{P(A and B)} \][/tex]
Here, A is "Not Accurate" and B is "From Restaurant C".
[tex]\[ \text{P(Not Accurate or From C)} = \text{P(Not Accurate)} + \text{P(From Restaurant C)} - \text{P(Not Accurate and From C)} \][/tex]
Plugging the values:
[tex]\[ \text{P(Not Accurate or From C)} \approx 0.126 + 0.250 - 0.036 \approx 0.340 \][/tex]

The probability of getting an order from Restaurant C or an order that is not accurate is approximately 0.340 (to three decimal places).

### Determining if the Events are Disjoint

Two events are disjoint if they cannot happen simultaneously. In this context, disjoint events would mean that an order cannot be both from Restaurant C and not accurate at the same time.

Since it is possible for an order to be from Restaurant C and be not accurate (as we have 39 such cases), these events are not disjoint.

### Conclusion

- Probability Calculation:
[tex]\[ \text{The probability of getting an order that is not accurate or is from Restaurant C is } 0.340 \][/tex]
- Event Disjoint Status:
- The events are not disjoint because it is possible to get an order that is both from Restaurant C and not accurate.