Question

In Exercises 9–20, use the data in the following table, which lists drive-thru order accuracy at popular fast food chains (data from a QSR Drive-Thru Study). Assume that orders are randomly selected from those included in the table.

	McDonald’s	Burger King	Wendy’s	Taco Bell
Order Accurate	329	264	249	145
OrderNotAccurate	33	54	31	13

Fast Food Drive-Thru Accuracy If one order is selected, find the probability of getting an order from McDonald’s or Wendy’s or an order that is not accurate.

Short answer

The probability of getting an order from McDonald’s or Wendy’s or an inaccurate order is equal to 0.634.

Step-by-step solution

Given information

The number of food orders at drive-thru centres of four fast-food chains are provided.

Some orders are accurate, while others are inaccurate.

Addition Rule of Probability

For three events, A, B, and C, to either occur individually or together, the following probability is calculated:

$P\left(AorBorC\right)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(AandB\right)-P\left(AandC\right)-P\left(BandC\right)$

Calculation

The subtotals for each chain are tabulated below:

	McDonald’s	Burger King	Wendy’s	Taco Bell	Totals
Order Accurate	329	264	249	145	987
OrderNotAccurate	33	54	31	13	131
Totals	362	318	280	158	1118

The total number of food orders is equal to 1118.

Let E be the event of selecting a food order from McDonald’s.

Let F be the event of selecting a food order from Wendy’s.

Let G be the event of selecting an inaccurate food order.

The number of food orders from McDonald’s is equal to 362.

The probability of selecting a food order from McDonald’s,

$P\left(E\right)=\frac{362}{1118}$

.

The number of food orders from Wendy’s is equal to 280.

The probability of selecting a food order from Wendy’s,

$P\left(F\right)=\frac{280}{1118}$

The number of inaccurate food orders is equal to 131.

The probability of selecting an inaccurate food order,

$P\left(G\right)=\frac{131}{1118}$

The number of orders from McDonald’s and Wendy’s is equal to 0. Thus,

$P\left(E\text{and}F\right)=\frac{0}{1118}$

The number of inaccurate orders from McDonald’s is equal to 33. Thus,

$P\left(E\text{and}G\right)=\frac{33}{1118}$

The number of inaccurate orders from Wendy’s is equal to 31. Thus,

$P\left(F\text{and}G\right)=\frac{31}{1118}$

The probability of getting an order from McDonald’s or Wendy’s or an inaccurate order is calculated as follows:

$$\begin{gathered} P\left(E\text{or}F\text{or}G\right)=P\left(E\right)+P\left(F\right)+P\left(G\right)-P\left(E\text{and}F\right)-P\left(E\text{and}G\right)-P\left(F\text{and}G\right) \\ =\frac{362}{1118}+\frac{280}{1118}+\frac{131}{1118}-\frac{0}{1118}-\frac{33}{1118}-\frac{31}{1118} \\ =\frac{709}{1118} \\ =0.6342 \end{gathered}$$

Therefore, the probability of getting an order from McDonald’s or Wendy’s or an inaccurate order is equal to 0.634.