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
(round to three decimal places as needed.)

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 orders

Sum the non - accurate orders: \(35 + 56+33 + 17 = 141\)

Step3: Calculate number of orders from Restaurant C

Sum the orders from Restaurant C: \(231+33 = 264\)

Step4: Calculate number of non - accurate orders from Restaurant C

The number of non - accurate orders from Restaurant C is 33.

Step5: Use the addition rule for probability

The formula for \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Here, let \(A\) be the event of getting a non - accurate order and \(B\) be the event of getting an order from Restaurant C.
\[P(A)=\frac{141}{1105}\], \[P(B)=\frac{264}{1105}\], \[P(A\cap B)=\frac{33}{1105}\]
\[P(A\cup B)=\frac{141 + 264-33}{1105}=\frac{372}{1105}\approx0.337\]

Answer:

0.337