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 three different orders are selected, find the probability that they are all from Wendy’s

Short answer

The probability that all three orders are from Wendy’s is equal to 0.0156.

Step-by-step solution

Given information

The given data shows the number of food orders at drive-thru centers of four fastfood chains.

Multiplication rule of probability

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

$P\left(A\text{and}B\text{and}C\right)=P\left(A\right)\times P\left(B\right)\times P\left(C\right)$

Calculation

The table below shows the subtotals for each category:

	McDonald’s	Burger King	Wendy’s	Taco Bell	Totals
Order Accurate	329	264	249	145	987
Order Not Accurate	33	54	31	13	131
Totals	362	318	280	158	Grand Total=1118

The total number of food orders is equal to 1118.

Define the events.

Let E be the event of selecting a food order from Wendy’s on the first selection.

Let Fbe the event of selecting a food order from Wendy’s on the second selection.

Let G be the event of selecting a food order from Wendy’s on the third selection.

The number of food orders from Wendy’s is equal to 318.

The probability of selecting a food order from Wendy’s on the first selection is equal to:

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

As the remaining total number of orders and the remaining number of orders from Wendy’s will be decreased by 1, the probability of selecting a food order from Wendy’s on the second selection is equal to:

$P\left(F\right)=\frac{279}{1117}$

Further, the remaining total number of orders and the remaining number of orders from Wendy’s will again decrease by 1. Thus, the probability of selecting a food order from Wendy’s on the third selection is equal to:

$P\left(G\right)=\frac{278}{1116}$

The probability of selecting three orders from Wendy’s is given by:

$$\begin{gathered} P\left(Ea\text{nd}F\text{and}G\right)=P\left(E\right)\times P\left(F\right)\times P\left(G\right) \\ =\frac{280}{1118}\times \frac{279}{1117}\times \frac{276}{1116} \\ =0.0156 \end{gathered}$$

Therefore, the probability of selecting three orders from Wendy’s is equal to 0.0156.