Let's break down the problem step by step: ### **Given:** - Probability that a randomly selected HR manager says applicants should follow up within two weeks, \( p = 0.51 \) (since 51%) - Number of HR managers selected, \( n = 5 \) - We need the probability that **exactly 2** of them say applicants should follow up within two weeks. This is a **binomial probability** problem. --- ## **Step 1: Identify the Binomial Probability Formula** The probability of exactly \( k \) successes in \( n \) independent Bernoulli trials is: \[ P(X = k) = {n \choose k} p^k (1-p)^{n-k} \] where: - \( n = 5 \) - \( k = 2 \) - \( p = 0.51 \) - \( 1-p = 0.49 \) --- ## **Step 2: Plug in the Numbers** \[ P(X = 2) = {5 \choose 2} (0.51)^2 (0.49)^{3} \] First, calculate the binomial coefficient: \[ {5 \choose 2} = \frac{5!}{2! \cdot 3!} = \frac{5 \times 4}{2 \times 1} = 10 \] --- ## **Step 3: Calculate Each Component** - \( (0.51)^2 = 0.2601 \) - \( (0.49)^3 = 0.49 \times 0.49 \times 0.49 = 0.117649 \) - Binomial coefficient = 10 --- ## **Step 4: Multiply Everything Together** \[ P(X=2) = 10 \times 0.2601 \times 0.117649 \] Let's do \( 0.2601 \times 0.117649 \): \[ 0.2601 \times 0.117649 = 0.030608 \] Now multiply by 10: \[ 0.030608 \times 10 = 0.30608 \] --- ## **Step 5: Final Answer** **The probability that exactly 2 out of 5 HR managers say job applicants should follow up within two weeks is:** \[ \boxed{0.306} \] (or, rounded to three decimal places, **0.306**, or **30.6%**)