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 not accurate	39	54	38	20

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

The total number of orders is the sum of all values in the table.

$$314 + 263+242 + 138+39+54+38+20=\sum_{i = 1}^{8}x_{i}=1108$$

Step2: Calculate number of orders from Restaurant A

The number of orders from Restaurant A is \(314 + 39=353\).

Step3: Calculate number of accurate orders

The number of accurate orders is \(314+263 + 242+138=957\).

Step4: Calculate number of accurate orders from Restaurant A

The number of accurate orders from Restaurant A is 314.

Step5: Use the addition - rule for probability

The formula for \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\).
Let \(A\) be the event of getting an order from Restaurant A and \(B\) be the event of getting an accurate order.

$$P(A)=\frac{353}{1108}$$

,

$$P(B)=\frac{957}{1108}$$

,

$$P(A\cap B)=\frac{314}{1108}$$

$$P(A\cup B)=\frac{353 + 957-314}{1108}=\frac{996}{1108}\approx0.899$$

Answer:

0.899