Question

on a particular day, a restaurant that is open for lunch and dinner had 114 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	24	51

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 number of customers

The total number of customers is $6 + 33+24 + 51=114$.

Step2: Calculate probability of coming for lunch (part a)

The number of customers who came for lunch is $6 + 33 = 39$. The probability $P(\text{lunch})$ is $\frac{39}{114}\approx0.34$.

Step3: Calculate number of customers who did not order dessert

The number of customers who did not order dessert is $33 + 51=84$.

Step4: Calculate number of customers who came for lunch and did not order dessert

The number of customers who came for lunch and did not order dessert is $33$.

Step5: Calculate probability of coming for lunch or not ordering dessert (part b)

Use the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Here, $A$ is the event of coming for lunch and $B$ is the event of not ordering dessert. $P(A)=\frac{39}{114}$, $P(B)=\frac{84}{114}$ and $P(A\cap B)=\frac{33}{114}$. So $P(A\cup B)=\frac{39 + 84- 33}{114}=\frac{90}{114}\approx0.79$.

Answer:

(a) $0.34$
(b) $0.79$