The owner of a small deli is trying to decide whether to discontinue selling magazines. He suspects that only 7.9% of his customers buy a magazine and he thinks that he might be able to use the display space to sell something more profitable. Before making a final decision, he decides that for one day he will keep track of the number of customers that buy a magazine. Assuming his suspicion that 7.9% of his customers buy a magazine is correct, what is the probability that exactly 3 out of the first 10 customers buy a magazine?


This problem involves a **binomial distribution**, as it deals with repeated independent trials (each customer deciding to buy or not buy a magazine) with two possible outcomes: success (buying a magazine) or failure (not buying a magazine). 

In a binomial distribution, the probability of exactly \( k \) successes in \( n \) independent trials is given by the formula:

\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k}
\]

Where:
- \( P(X = k) \) is the probability of getting exactly \( k \) successes,
- \( \binom{n}{k} \) is the binomial coefficient, which counts the number of ways to choose \( k \) successes from \( n \) trials,
- \( p \) is the probability of success on each trial,
- \( n \) is the total number of trials, and
- \( k \) is the number of successes.

In this case:
- \( n = 10 \) (since he is observing the first 10 customers),
- \( k = 3 \) (the problem asks for the probability that exactly 3 customers buy a magazine),
- \( p = 0.079 \) (since 7.9% of customers buy a magazine).

We can now calculate the probability.

### Step 1: Calculate the binomial coefficient \( \binom{10}{3} \):
\[
\binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120
\]

### Step 2: Apply the binomial formula:
\[
P(X = 3) = \binom{10}{3} (0.079)^3 (1 - 0.079)^{10-3}
\]
\[
P(X = 3) = 120 \times (0.079)^3 \times (0.921)^7
\]

Let me calculate this.The probability that exactly 3 out of the first 10 customers will buy a magazine is approximately \( 0.0333 \), or about 3.33%.

Would you like to see more details or have any questions about this process?

Here are 5 related questions to further explore the topic:

1. What is the probability that no customers buy a magazine?
2. How would the probability change if the deli owner suspects that 10% of customers buy a magazine?
3. What is the probability that fewer than 2 customers buy a magazine?
4. How can we calculate the expected number of customers who buy a magazine in this scenario?
5. What is the standard deviation of the number of customers buying a magazine?

**Tip:** The sum of probabilities for all possible outcomes (0 through 10 customers buying a magazine) must always equal 1 for a binomial distribution.

The owner of a small deli is trying to decide whether to discontinue selling magazines. He suspects that only 7.9% of his customers buy a magazine and he thinks that he might be able to use the display space to sell something more profitable. Before making a final decision, he decides that for one day he will keep track of the number of customers that buy a magazine. Assuming his suspicion that 7.9% of his customers buy a magazine is correct, what is the probability that exactly 3 out of the first 10 customers buy a magazine?

Learn how to calculate the probability that exactly 3 out of the first 10 customers will buy a magazine using the binomial distribution. This problem assumes a success rate of 7.9% and provides a step-by-step solution with key formulas. Suitable for high school students in Grades 10-12.

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