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 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?

Short answer

a. The probability that both orders are from Burger King selected with replacement is equal to 0.0809.

Yes, the events are independent.

b. The probability that both orders are from Burger King selected without replacement is equal to 0.0807.

No, the events are not independent.

Step-by-step solution

Given information

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

Multiplication rule of probability

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 $P\left(B|A\right)$.

Independence of events

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

Calculation

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

	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.

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.