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
Order Not Accurate	33	54	31	13

Fast Food Drive-Thru Accuracy If three different orders are selected, find the probability that they are all not accurate.

Short answer

The probability that all three orders are inaccurate is equal to 0.00158.

Step-by-step solution

Given information

The given data shows the number of accurate and inaccurate food orders at drive-thru centers of four fast food chains.

Multiplication rule of probability

For three events A, B, and C to occur simultaneously, 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.

Let E be the event of selecting an inaccurate food order on the first selection.

Let F be the event of selecting an inaccurate food order on the second selection.

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

The number of inaccurate food orders is equal to 131.

The probability of selecting an inaccurate food order on the first selection is equal to:

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

The remaining total number of orders and the number of inaccurate orders will decrease by 1.

The probability of selecting an inaccurate food order on the second selection is equal to:

$P\left(F\right)=\frac{130}{1117}$

For the third selection, the remaining total number of orders and the number of inaccurate orders will further decrease by 1.

The probability of selecting an inaccurate food order on the third selection is equal to:

$P\left(G\right)=\frac{129}{1116}$

The probability of selecting three inaccurate orders is given by:

$$\begin{gathered} P\left(E\text{and}F\text{and}G\right)=P\left(E\right)\times P\left(F\right)\times P\left(G\right) \\ =\frac{131}{1118}\times \frac{130}{1117}\times \frac{129}{1116} \\ =0.00158 \end{gathered}$$

Therefore, the probability of selecting three inaccurate orders is equal to 0.00158.