This is a binomial probability problem because we are dealing with a fixed number of trials (9 human resource managers), two possible outcomes (either the manager says job applicants should follow up within two weeks or they don't), and a constant probability of success (43%).

The binomial probability formula is given by:

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

Where:

• $n$ is the number of trials (in this case, 9 human resource managers),
• $k$ is the number of successes (we want exactly 7 managers to say "follow up within two weeks"),
• $p$ is the probability of success on a single trial (43% or 0.43),
• $\binom{n}{k}$ is the binomial coefficient, which represents the number of ways to choose $k$ successes out of $n$ trials and is calculated as:

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$

Step-by-Step Solution:

1. Identify the values: $n = 9, \quad k = 7, \quad p = 0.43, \quad 1 - p = 0.57$

2. Compute the binomial coefficient: $\binom{9}{7} = \frac{9!}{7!(9-7)!} = \frac{9 \times 8}{2 \times 1} = 36$

3. Calculate the probability: $P(X = 7) = 36 \times (0.43)^7 \times (0.57)^2$

Let's compute this step by step.

The probability that exactly 7 out of 9 randomly selected human resource managers say job applicants should follow up within two weeks is approximately $0.0318$ or 3.18%.

Would you like further clarification on any part of the solution?

Here are 5 related questions you might want to explore:

1. How would the probability change if we wanted to find the probability of 8 managers agreeing instead of 7?
2. What is the probability that at least 7 managers say job applicants should follow up within two weeks?
3. How does the probability change if the probability of success was 50% instead of 43%?
4. What is the expected number of managers who will say job applicants should follow up within two weeks?
5. How do we calculate the variance and standard deviation for this binomial distribution?

Tip: For binomial distributions, the mean can be found using $\mu = np$, and the variance using $\sigma^2 = np(1-p)$.