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
a b c d
order accurate 321 261 242 127
order not accurate 32 59 37 17

if one order is selected, find the probability of getting an order from restaurant a or an order that is accurate. are the events of selecting an order from restaurant a and selecting an accurate order disjoint events?

the probability of getting an order from restaurant a or an order that is accurate is
(round to three decimal places as needed.)

Explanation:

Step1: Calculate total number of orders

First, we sum up all the values in the table. For accurate orders: \(321 + 261 + 242 + 127 = 951\). For not accurate orders: \(32 + 59 + 37 + 17 = 145\). Total orders \(N=951 + 145 = 1096\).

Step2: Calculate number of orders from Restaurant A or accurate

Let \(A\) be the event of selecting from Restaurant A, \(B\) be the event of selecting an accurate order. We use the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\).

• Number of orders from A: \(321 + 32 = 353\), so \(P(A)=\frac{353}{1096}\).
• Number of accurate orders: \(951\), so \(P(B)=\frac{951}{1096}\).
• Number of orders from A and accurate: \(321\), so \(P(A\cap B)=\frac{321}{1096}\).

Now, \(P(A\cup B)=\frac{353 + 951 - 321}{1096}=\frac{983}{1096}\).

Step3: Compute the value

Calculate \(\frac{983}{1096}\approx0.897\) (rounded to three decimal places).

Answer:

\(0.897\)