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	35	56	33	17

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 0.337
(round to three decimal places as needed.)

are the events of selecting an order from restaurant c and selecting an inaccurate order disjoint events?

the events drop - down because it drop - down possible to drop - down

Explanation:

Step1: Calculate total number of orders

First, find the sum of all values in the table.
\[321 + 267+231 + 145+35 + 56+33 + 17=1105\]

Step2: Calculate number of non - accurate or from Restaurant C orders

Number of non - accurate orders: \(35 + 56+33 + 17 = 141\)
Number of orders from Restaurant C: \(231+33 = 264\)
Number of non - accurate orders from Restaurant C: \(33\)
Using the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), we have \(141 + 264-33=372\)

Step3: Calculate the probability

The probability \(P=\frac{372}{1105}\approx0.337\)
For the disjoint - events part:
Two events are disjoint if they cannot occur at the same time. Since there are non - accurate orders from Restaurant C (33 non - accurate orders from C), it is possible to have an order that is both non - accurate and from Restaurant C. So the events are not disjoint.

Answer:

The probability of getting an order that is not accurate or is from Restaurant C is approximately \(0.337\). The events of selecting an order from Restaurant C and selecting an inaccurate order are not disjoint because it is possible to have an order that is both from Restaurant C and inaccurate.