Question

on a particular day, a restaurant that is open for lunch and dinner had 119 customers. each customer came in for one meal. an employee recorded at which meal each customer came in and whether the customer ordered dessert. the data are summarized in the table below.

dessert	no dessert
dinner	37	52

suppose a customer from that day is chosen at random. answer each part. do not round intermediate computations, and round your answers to the nearest hundredth.
(a) what is the probability that the customer came for lunch?
(b) what is the probability that the customer came for lunch or did not order dessert?

Explanation:

Step1: Calculate total lunch - goers

The number of customers who came for lunch is the sum of those who ordered dessert and those who did not at lunch. So, \(21 + 9=30\). The total number of customers is \(119\).

Step2: Calculate probability for part (a)

The probability \(P(\text{lunch})\) is the number of lunch - goers divided by the total number of customers. So, \(P(\text{lunch})=\frac{30}{119}\approx0.25\).

Step3: Calculate number of non - dessert - orderers

The number of customers who did not order dessert is \(9 + 52 = 61\).

Step4: Calculate number of lunch - goers and non - dessert - orderers

The number of customers who came for lunch and did not order dessert is \(9\).

Step5: Calculate probability for part (b)

Use the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Here, \(A\) is the event that the customer came for lunch and \(B\) is the event that the customer did not order dessert. \(P(A)=\frac{30}{119}\), \(P(B)=\frac{61}{119}\), and \(P(A\cap B)=\frac{9}{119}\). Then \(P(A\cup B)=\frac{30 + 61-9}{119}=\frac{82}{119}\approx0.69\).

Answer:

(a) \(0.25\)
(b) \(0.69\)