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on a particular day, a restaurant that is open for lunch and dinner had…

Question

on a particular day, a restaurant that is open for lunch and dinner had 111 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. \
\

dessertno dessert\
---------\
lunch164\
dinner2467\

\
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. \
(if necessary, consult a list of formulas.) \
\
(a) what is the probability that the customer did not order dessert? \
\
(b) what is the probability that the customer came for lunch or did not order dessert? \

Explanation:

Part (a) Step 1: Find total customers who did not order dessert

To find the number of customers who did not order dessert, we add the number of lunch customers who didn't order dessert and dinner customers who didn't order dessert. From the table, lunch no dessert is 4 and dinner no dessert is 67. So total no dessert is \(4 + 67 = 71\).

Part (a) Step 2: Calculate the probability

The total number of customers is 111. The probability that a customer did not order dessert is the number of customers who did not order dessert divided by the total number of customers. So probability \(P(\text{no dessert})=\frac{71}{111}\approx0.64\) (rounded to nearest hundredth).

Part (b) Step 1: Recall the formula for "or" probability

The formula for \(P(A \text{ or } B)=P(A)+P(B)-P(A \text{ and } B)\). Let \(A\) be the event that the customer came for lunch and \(B\) be the event that the customer did not order dessert.

Part (b) Step 2: Find \(P(A)\), \(P(B)\), and \(P(A \text{ and } B)\)

  • \(P(A)\): Number of lunch customers is \(16 + 4 = 20\), so \(P(A)=\frac{20}{111}\).
  • \(P(B)\): We already found \(P(B)=\frac{71}{111}\) from part (a).
  • \(P(A \text{ and } B)\): Number of customers who came for lunch and did not order dessert is 4, so \(P(A \text{ and } B)=\frac{4}{111}\).

Part (b) Step 3: Calculate \(P(A \text{ or } B)\)

Using the formula: \(P(A \text{ or } B)=\frac{20}{111}+\frac{71}{111}-\frac{4}{111}=\frac{20 + 71 - 4}{111}=\frac{87}{111}\approx0.78\) (rounded to nearest hundredth).

Answer:

(a) \(\boxed{0.64}\)
(b) \(\boxed{0.78}\)