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At a concession stand, the combo purchases for one week are recorded in the table below:

\begin{tabular}{|l|c|c|c|}
\hline & Hamburger & Pizza & \begin{tabular}{c}
Deli \\
Sandwich
\end{tabular} \\
\hline French Fries & 83 & 67 & 37 \\
\hline Peanuts & 2 & 5 & 14 \\
\hline Popcorn & 19 & 29 & 3 \\
\hline
\end{tabular}

If we choose a customer at random, what is the probability that they have purchased French fries and a deli sandwich?

[tex]\[
P(\text{French Fries and Deli Sandwich}) = \frac{1}{[?]}
\][/tex]

Give your answer in simplest form.


Sagot :

To determine the probability that a randomly chosen customer purchased french fries and a deli sandwich, let's follow a systematic approach to solve the problem.

### Step 1: Identify the Event
The event in question is "purchasing french fries and a deli sandwich."

From the table, the number of customers who purchased french fries along with a deli sandwich is 37.

### Step 2: Calculate the Total Number of Purchases
We need to find the total number of all purchases recorded.

1. Total French Fries Purchases:
- Hamburger with French Fries: 83
- Pizza with French Fries: 67
- Deli Sandwich with French Fries: 37
- Total French Fries: [tex]\( 83 + 67 + 37 = 187 \)[/tex]

2. Total Peanuts Purchases:
- Hamburger with Peanuts: 2
- Pizza with Peanuts: 5
- Deli Sandwich with Peanuts: 14
- Total Peanuts: [tex]\( 2 + 5 + 14 = 21 \)[/tex]

3. Total Popcorn Purchases:
- Hamburger with Popcorn: 19
- Pizza with Popcorn: 29
- Deli Sandwich with Popcorn: 3
- Total Popcorn: [tex]\( 19 + 29 + 3 = 51 \)[/tex]

4. Total Purchases:
Sum of all purchases:
[tex]\[ 187 \text{ (French Fries)} + 21 \text{ (Peanuts)} + 51 \text{ (Popcorn)} = 259 \][/tex]

### Step 3: Calculate the Probability
The probability of choosing a customer who purchased french fries and a deli sandwich is the ratio of the number of customers who made that purchase to the total number of purchases.

[tex]\[ P(\text{French Fries and Deli Sandwich}) = \frac{\text{Number of French Fries and Deli Sandwich Purchases}}{\text{Total Purchases}} = \frac{37}{259} \][/tex]

To give the answer in simplest form, the fraction [tex]\(\frac{37}{259}\)[/tex] does not reduce any further as 37 is a prime number and does not divide 259 exactly.

Thus, the probability is:
[tex]\[ P(\text{French Fries and Deli Sandwich}) = \frac{1}{7} \][/tex]

Therefore, the final answer is:

[tex]\[ P(\text{French Fries and Deli Sandwich}) = \frac{1}{7} \][/tex]