R. H. Bruskin Associates Market Research found that 40% of Americans do not think that having a college education is important to succeed in the business world. If a random sample of 5 Americans is selected, find these probabilities. a) Exactly 2 people will agree with that statement. b) At most 3 people will agree with that statement. c) At least 2 people will agree with that statement. d) Fewer than 3 people will agree with that statement.

This problem involves a binomial probability distribution, where the probability of an individual agreeing with the statement ("success") is \( p = 0.40 \), and the probability of not agreeing ("failure") is \( 1 - p = 0.60 \). The number of trials \( n = 5 \), and we are asked to find different probabilities related to the number of successes in these 5 trials.

The binomial probability formula is:

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

Where:
- \( n \) is the number of trials (5 in this case),
- \( k \) is the number of successes,
- \( p \) is the probability of success (0.40),
- \( \binom{n}{k} \) is the binomial coefficient, which represents the number of ways to choose \( k \) successes from \( n \) trials.

### a) Exactly 2 people will agree with that statement.

We need to find \( P(X = 2) \), where \( X \) is the number of people agreeing.

\[
P(X = 2) = \binom{5}{2} (0.40)^2 (0.60)^3
\]

First, calculate the binomial coefficient \( \binom{5}{2} = \frac{5!}{2!(5-2)!} = 10 \).

\[
P(X = 2) = 10 \times (0.40)^2 \times (0.60)^3
= 10 \times 0.16 \times 0.216 = 0.3456
\]

So, the probability that exactly 2 people will agree is **0.3456**.

### b) At most 3 people will agree with that statement.

"At most 3" means we need to find \( P(X \leq 3) \), which is the sum of the probabilities for 0, 1, 2, and 3 people agreeing:

\[
P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
\]

We'll calculate each term:

- \( P(X = 0) = \binom{5}{0} (0.40)^0 (0.60)^5 = 1 \times 1 \times 0.07776 = 0.07776 \)
- \( P(X = 1) = \binom{5}{1} (0.40)^1 (0.60)^4 = 5 \times 0.40 \times 0.1296 = 0.2592 \)
- \( P(X = 2) = 0.3456 \) (from part a)
- \( P(X = 3) = \binom{5}{3} (0.40)^3 (0.60)^2 = 10 \times 0.064 \times 0.36 = 0.2304 \)

Now, sum them up:

\[
P(X \leq 3) = 0.07776 + 0.2592 + 0.3456 + 0.2304 = 0.91296
\]

So, the probability that at most 3 people will agree is **0.91296**.

### c) At least 2 people will agree with that statement.

"At least 2" means \( P(X \geq 2) \), which is the complement of \( P(X < 2) \) (i.e., \( P(X = 0) + P(X = 1) \)):

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

We already have \( P(X = 0) = 0.07776 \) and \( P(X = 1) = 0.2592 \). So:

\[
P(X \geq 2) = 1 - (0.07776 + 0.2592) = 1 - 0.33696 = 0.66304
\]

Thus, the probability that at least 2 people will agree is **0.66304**.

### d) Fewer than 3 people will agree with that statement.

"Fewer than 3" means \( P(X < 3) \), which is the sum of \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \):

\[
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
\]

We already have all these probabilities:

\[
P(X < 3) = 0.07776 + 0.2592 + 0.3456 = 0.68256
\]

So, the probability that fewer than 3 people will agree is **0.68256**.

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Let me know if you'd like further details or have any questions!

Here are 5 related questions:
1. How do you calculate binomial probabilities for larger sample sizes?
2. What is the role of the binomial coefficient in this calculation?
3. How would you compute the expected value and variance of this binomial distribution?
4. What if the probability of agreeing was higher (e.g., 0.70)? How would that change the probabilities?
5. How can we use technology or statistical tables to compute binomial probabilities efficiently?

**Tip**: When solving binomial problems, it's helpful to remember that the sum of all probabilities for a given number of trials must equal 1.

Explore how to calculate binomial probabilities for a random sample of 5 Americans, where 40% believe college education is not important for business success. Learn to find the probability that exactly 2, at most 3, at least 2, or fewer than 3 people agree. Suitable for Grades 11-12 or early college-level statistics.

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