Problem

Use the data in the following table, which lists drive-thru order accuracy at popular fast food chains. Assume that orders are randomly selected from those included in the table. Drive-thru Restaurant \begin{tabular}{|l|c|c|c|c|} \hline & A & B & C & D \\ \hline Order Accurate & 316 & 272 & 249 & 146 \\ \hline Order Not Accurate & 34 & 57 & 39 & 12 \\ \hline \end{tabular} If three different orders are selected, find the probability that they are all from Restaurant D. The probability is (Round to four decimal places as needed.)

Solution

Step 1 :First, calculate the total number of orders by adding all the orders from each restaurant. This gives us a total of 1125 orders.

Step 2 :Next, find the number of orders from Restaurant D, which is 158.

Step 3 :Then, calculate the probability of an order being from Restaurant D by dividing the number of orders from Restaurant D by the total number of orders. This gives us a probability of \(\frac{158}{1125} \approx 0.1404\).

Step 4 :Since we are selecting three orders, and we want all of them to be from Restaurant D, we need to multiply the probability of one order being from Restaurant D by itself three times. This gives us a probability of \(0.1404 \times 0.1404 \times 0.1404 \approx 0.0028\).

Step 5 :Final Answer: The probability that all three orders are from Restaurant D is approximately \(\boxed{0.0028}\).

From Solvely APP
Source: https://solvelyapp.com/problems/36083/

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