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 from Burger King.

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?

The number of food orders at drive-thru centers of four fast-food chains are provided under two kinds, accurate and inaccurate.

Consider two events A and B.

The rule says that theprobability of events A and B occurring togetheris given by the following equation:

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

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

• Events are said to be independent if they are selected with replacement.
• Events are said to be dependent if they are selected without replacement.

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

The total number of food orders is equal to 1118.

The number of food orders from Burger King is equal to 318.

The probability of selecting a food order from Burger King,

$P\left(\text{order}\text{from}\text{Burger}\text{King}\right)=\frac{318}{1118}$

Let E be the event of selecting an order from Burger King on the first try.

Let F be the event of selecting an order from Burger King on the second try.

a.

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

$$\begin{gathered} P\left(E\right)=P\left(F|E\right) \\ =\frac{318}{1118} \end{gathered}$$

The probability that both the orders are from Burger King,

$$\begin{gathered} P\left(E\text{and}F\right)=P\left(E\right)\times P\left(F|E\right) \\ =\frac{318}{1118}\times \frac{318}{1118} \\ =0.0809 \end{gathered}$$

Therefore, the probability of selecting both orders from Burger King with replacement is equal to 0.0809.

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 orders from Burger King will decrease by one for the second try.

$P\left(F|E\right)=\frac{317}{1117}$

The probability that both the orders selected are from Burger King,

$$\begin{gathered} P\left(E\text{and}F\right)=P\left(E\right)\times P\left(F|E\right) \\ =\frac{318}{1118}\times \frac{317}{1117} \\ =0.0807 \end{gathered}$$

Therefore, the probability of selecting both orders from Burger King without replacement is equal to 0.0807.

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