​21% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and ask each to name the reason he or she uses credit cards. Find the probability that the number of college students who say they use credit cards because of the rewards program is​ (a) exactly​ two, (b) more than​ two,
and​ (c) between two and five inclusive. If​ convenient, use technology to find the probabilities.

This problem involves a binomial probability distribution, where:

- \( n = 10 \) (the number of trials, i.e., college students asked),
- \( p = 0.21 \) (the probability that a student uses credit cards because of the rewards program),
- \( q = 1 - p = 0.79 \) (the probability that a student does **not** use credit cards because of the rewards program).

The probability mass function (PMF) for a binomial distribution is:

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

Where:
- \( X \) is the random variable representing the number of students who use credit cards because of the rewards program,
- \( k \) is the number of successes (in this case, the number of students who say they use credit cards for the rewards),
- \( \binom{n}{k} \) is the binomial coefficient (combinations).

### (a) **Probability that exactly two students use credit cards because of the rewards program (i.e., \( P(X = 2) \)):**

We substitute \( k = 2 \), \( n = 10 \), and \( p = 0.21 \) into the binomial formula:

\[
P(X = 2) = \binom{10}{2} (0.21)^2 (0.79)^8
\]

Let’s calculate this.

### (b) **Probability that more than two students use credit cards because of the rewards program (i.e., \( P(X > 2) \)):**

We need to find:

\[
P(X > 2) = 1 - P(X \leq 2)
\]

We can sum the probabilities for \( X = 0, 1, 2 \), and subtract from 1:

\[
P(X > 2) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]
\]

### (c) **Probability that between two and five students use credit cards because of the rewards program (inclusive) (i.e., \( P(2 \leq X \leq 5) \)):**

We need to sum the probabilities for \( X = 2, 3, 4, 5 \):

\[
P(2 \leq X \leq 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
\]

Let me compute these values for you.Here are the results for the probabilities:

- **(a)** The probability that exactly two students use credit cards because of the rewards program is approximately \( 0.301 \) or 30.1%.
- **(b)** The probability that more than two students use credit cards because of the rewards program is approximately \( 0.353 \) or 35.3%.
- **(c)** The probability that between two and five students (inclusive) use credit cards because of the rewards program is approximately \( 0.645 \) or 64.5%.

Would you like more details or have any questions?

Here are 5 related questions to explore:
1. What is the expected number of students using credit cards for rewards out of 10?
2. How does the probability change if 15 students are randomly selected?
3. What is the probability that none of the 10 students use credit cards for rewards?
4. How does increasing the success probability (e.g., to 0.25) affect the results?
5. Can we find the standard deviation for the number of students using credit cards for rewards?

**Tip**: The binomial distribution is useful for modeling the number of successes in fixed trials where the outcome is binary (like success/failure).

21% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and ask each to name the reason he or she uses credit cards. Find the probability that the number of college students who say they use credit cards because of the rewards program is (a) exactly two, (b) more than two, and (c) between two and five inclusive. If convenient, use technology to find the probabilities.

Calculate the binomial probabilities for college students using credit cards for rewards. This problem involves finding the probability that exactly two, more than two, or between two and five students (inclusive) use credit cards due to rewards programs. Suitable for high school students in Grades 11-12, this solution covers key concepts in statistics and probability.

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