Questions: Assume that when human resource managers are randomly selected, 45% say job applicants should follow up within two weeks. If 6 human resource managers are randomly selected, find the probability that exactly 3 of them say job applicants should follow up within two weeks. The probability is (Round to four decimal places as needed.)

Assume that when human resource managers are randomly selected, 45% say job applicants should follow up within two weeks. If 6 human resource managers are randomly selected, find the probability that exactly 3 of them say job applicants should follow up within two weeks.

The probability is
(Round to four decimal places as needed.)

Transcript text: Assume that when human resource managers are randomly selected, $45 \%$ say job applicants should follow up withir two weeks. If 6 human resource managers are randomly selected, find the probability that exactly 3 of them say job applicants should follow up within two weeks. The probability is $\square$ (Round to four decimal places as needed.)

Solution

Solution Steps

Step 1: Define the Problem

We need to find the probability that exactly $ x = 3 $ out of $ n = 6 $ human resource managers say that job applicants should follow up within two weeks, given that the probability of success $ p = 0.45 $.

Step 2: Use the Binomial Probability Formula

The probability of exactly $ x $ successes in $ n $ trials is given by the formula:

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

where $ q = 1 - p = 0.55 $.

Step 3: Calculate the Binomial Coefficient

The binomial coefficient $ \binom{n}{x} $ is calculated as:

\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

Step 4: Calculate the Probability

Substituting the values into the formula:

\[ P(X = 3) = 20 \cdot (0.45)^3 \cdot (0.55)^{6-3} \]

Calculating $ (0.45)^3 $ and $ (0.55)^3 $:

\[ (0.45)^3 = 0.091125 \quad \text{and} \quad (0.55)^3 = 0.166375 \]