Question

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
order accurate 339 266 234 150
order not accurate 40 57 30 13
if one order is selected, find the probability of getting an order that is not accurate or is from restaurant c. are the events of selecting an order that is not accurate and selecting an order from restaurant c disjoint events?
the probability of getting an order from restaurant c or an order that is not accurate is
(round to three decimal places as needed )

Explanation:

Step1: Calculate total number of orders

First, we sum up all the orders. For each restaurant, add the accurate and not accurate orders.
Total for A: \(339 + 40 = 379\)
Total for B: \(266 + 57 = 323\)
Total for C: \(234 + 30 = 264\)
Total for D: \(150 + 13 = 163\)
Now total orders \(N = 379 + 323 + 264 + 163\)
\(N = 379+323 = 702\); \(702 + 264 = 966\); \(966+163 = 1129\)

Step2: Calculate number of "not accurate" orders

Sum the "Order Not Accurate" row: \(40 + 57 + 30 + 13 = 140\)

Step3: Calculate number of orders from Restaurant C

Total orders from C: \(234 + 30 = 264\)

Step4: Calculate number of orders that are "not accurate" and from C

From the table, "Order Not Accurate" and "Restaurant C" is \(30\)

Step5: Apply the formula for \(P(A \cup B)=P(A)+P(B)-P(A \cap B)\)

Let \(A\) be "not accurate", \(B\) be "from C".
\(P(A)=\frac{140}{1129}\), \(P(B)=\frac{264}{1129}\), \(P(A \cap B)=\frac{30}{1129}\)
So \(P(A \cup B)=\frac{140 + 264 - 30}{1129}=\frac{374}{1129}\)

Step6: Compute the value

\(\frac{374}{1129}\approx0.331\) (rounded to three decimal places)

Answer:

\(0.331\)