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 two orders are selected, find the probability that they are both accurate.

a. Assume that the selections are made with replacement. Are the events independent?

b. Assume that the selections are made without replacement. Are the events independent?

Frequencies of food orders of two types, accurate and inaccurate, are tabulated for four different fast-food chains.

Consider two events, A and B.

Theprobability that events A and B both will occur simultaneouslyis given by the following:

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

Here, $P\left(B|A\right)$represents the probability of B given A has already occurred.

• If selections are made with replacement, events are independent.
• If selections are made without replacement, events are not independent.

The following table shows all the subtotals and the grand total:

The total number of food orders is equal to 1,118.

The number of accurate food orders is equal to 987.

The probability of selecting an inaccurate food order,

$P\left(\text{accurate}\text{order}\right)=\frac{987}{1118}$

.

Let E be the event of selecting an accurate order on the first try.

Let F be the event of selecting an accurate order on the second try.

a.

As selections are made with replacement, the total number of orders and the number of accurate orders remains the same for both tries.

$$\begin{gathered} P\left(E\right)=P\left(F|E\right) \\ =\frac{987}{1118} \end{gathered}$$

The probability that both the orders are inaccurate is given as follows:

$$\begin{gathered} P\left(E\text{and}F\right)=P\left(E\right)\times P\left(F|E\right) \\ =\frac{987}{1118}\times \frac{987}{1118} \\ =0.7793 \end{gathered}$$

Therefore, the probability of selecting both accurate orders with replacement is equal to0.779.

As the selections are made with replacement, they are independent.

b.

As selections are made without replacement, the total number of orders and the number of accurate orders will decrease by one for the second try.

$P\left(F|E\right)=\frac{986}{1117}$

The probability that both the orders selected are inaccurate,

$$\begin{gathered} P\left(E\text{and}F\right)=P\left(E\right)\times P\left(F|E\right) \\ =\frac{987}{1118}\times \frac{986}{1117} \\ =0.7792 \end{gathered}$$

Therefore, the probability of selecting both accurate orders without replacement is equal to 0.779.

As the selections are made without replacement, they are not independent.