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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.

\begin{tabular}{|l|c|c|}
\cline { 2 - 3 }
\multicolumn{1}{c|}{} & Dessert & No dessert \\
\hline
Lunch & 21 & 9 \\
\hline
Dinner & 37 & 52 \\
\hline
\end{tabular}

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 and did not order dessert?
[tex]\square[/tex]

(b) What is the probability that the customer came for lunch or did not order dessert?
[tex]\square[/tex]

Sagot :

GinnyAnsweridn
Let's break down the given problem and follow through step-by-step to find the required probabilities.

### Part (a)
Probability that the customer came for lunch and did not order dessert:

From the given data,
- Total number of customers: 119
- Number of customers who came for lunch and did not order dessert: 9

The probability is calculated by dividing the number of customers who came for lunch and did not order dessert by the total number of customers.

[tex]\[ P(\text{Lunch and No dessert}) = \frac{\text{Number of Lunch with No dessert}}{\text{Total number of customers}} \][/tex]

Plugging in the values:

[tex]\[ P(\text{Lunch and No dessert}) = \frac{9}{119} \approx 0.07563025210084033 \][/tex]

Rounding to the nearest hundredth:

[tex]\[ P(\text{Lunch and No dessert}) \approx 0.08 \][/tex]

### Part (b)
Probability that the customer came for lunch or did not order dessert:

To find this probability, we will use the principle of inclusion and exclusion. We first need to calculate:
- The number of customers who came for lunch (lunch with dessert + lunch with no dessert)
- The number of customers who did not order dessert (lunch with no dessert + dinner with no dessert)
- The overlapping customers who fit both criteria (lunch with no dessert).

From the given data:
- Number of customers who came for lunch (with or without dessert) = 21 + 9 = 30
- Number of customers who did not order dessert (lunch with no dessert + dinner with no dessert) = 9 + 52 = 61
- Number of customers who came for lunch and did not order dessert = 9

The probability that a customer came for lunch or did not order dessert is calculated as:

[tex]\[ P(\text{Lunch or No dessert}) = \frac{\text{Number of Lunch customers} + \text{Number of No dessert customers} - \text{Number of overlapping customers}}{\text{Total number of customers}} \][/tex]

Plugging in the values:

[tex]\[ P(\text{Lunch or No dessert}) = \frac{30 + 61 - 9}{119} \approx 0.6890756302521008 \][/tex]

Rounding to the nearest hundredth:

[tex]\[ P(\text{Lunch or No dessert}) \approx 0.69 \][/tex]

Therefore, the answers to the given problems are:
(a) The probability that the customer came for lunch and did not order dessert is approximately 0.08.
(b) The probability that the customer came for lunch or did not order dessert is approximately 0.69.
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