Questions: Assume that when human resource managers are randomly selected, 47% say job applicants should follow up within two weeks. If 7 human resource managers are randomly selected, find the probability that at least 3 of them say job applicants should follow up within two weeks. The probability is (Round to four decimal places as needed.)

Solution Steps

We are tasked with finding the probability that at least 3 out of 7 human resource managers say job applicants should follow up within two weeks, given that the probability of a manager saying this is \( p = 0.47 \).

To find the probability of at least 3 successes, we first calculate the probabilities of getting exactly 0, 1, and 2 successes using the binomial probability formula:

\[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

where \( n = 7 \) and \( q = 1 - p = 0.53 \).

• For \( x = 0 \): \[ P(X = 0) = \binom{7}{0} \cdot (0.47)^0 \cdot (0.53)^7 = 0.0117 \]

• For \( x = 1 \): \[ P(X = 1) = \binom{7}{1} \cdot (0.47)^1 \cdot (0.53)^6 = 0.0729 \]

• For \( x = 2 \): \[ P(X = 2) = \binom{7}{2} \cdot (0.47)^2 \cdot (0.53)^5 = 0.194 \]

The probability of getting at least 3 successes is given by:

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

Substituting the calculated probabilities:

\[ P(X \geq 3) = 1 - (0.0117 + 0.0729 + 0.194) = 1 - 0.2786 = 0.7214 \]

Final Answer