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 both of them are not 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.

Let A and B be two events.

Theprobability of occurrence of events A and B simultaneouslyhas the following notation:

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

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 inaccurate food orders is equal to 131.

The probability of selecting an inaccurate food order is as follows:

$P\left(\text{inaccurate}\text{order}\right)=\frac{131}{1118}$

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

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

a.

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

$$\begin{gathered} P\left(E\right)=P\left(F|E\right) \\ =\frac{131}{1118} \end{gathered}$$

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

$$\begin{gathered} P\left(EandF\right)=P\left(E\right)\times P\left(F|E\right) \\ =\frac{131}{1118}\times \frac{131}{1118} \\ =0.0137 \end{gathered}$$

Therefore, the probability of selecting both inaccurate orders with replacement is equal to 0.0137.

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 inaccurate orders will decrease by one for the second try.

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

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

$$\begin{gathered} P\left(E\text{and}F\right)=P\left(E\right)\times P\left(F|E\right) \\ =\frac{131}{1118}\times \frac{130}{1117} \\ =0.0136 \end{gathered}$$

Therefore, the probability of selecting both inaccurate orders without replacement is equal to 0.0136.

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