Assume that when human resource managers are randomly selected, 52% say job applicants should follow up within two weeks. If 5 human resource managers are randomly selected, find the probability that at least 3 of them say job applicants should follow up within two weeks.
T he Correct Answer and Explanation is :
To solve this problem, we use the binomial probability distribution, as we are dealing with a fixed number of independent trials (5 HR managers), each with only two possible outcomes (say “yes” or “no” to following up within two weeks).
Given:
- Probability of success (saying “yes”) p=0.52p = 0.52
- Number of trials n=5n = 5
- We want the probability that at least 3 say yes: P(X≥3)P(X \geq 3)
- This is: P(X≥3)=P(X=3)+P(X=4)+P(X=5)P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5)
We use the binomial probability formula: P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k} p^k (1 – p)^{n – k}
Let’s compute:
- P(X = 3):
(53)(0.52)3(0.48)2=10×0.140608×0.2304≈0.324\binom{5}{3} (0.52)^3 (0.48)^2 = 10 \times 0.140608 \times 0.2304 \approx 0.324
- P(X = 4):
(54)(0.52)4(0.48)1=5×0.073725×0.48≈0.177\binom{5}{4} (0.52)^4 (0.48)^1 = 5 \times 0.073725 \times 0.48 \approx 0.177
- P(X = 5):
(55)(0.52)5(0.48)0=1×0.038337×1=0.038\binom{5}{5} (0.52)^5 (0.48)^0 = 1 \times 0.038337 \times 1 = 0.038
Add them up:
P(X≥3)≈0.324+0.177+0.038=0.539P(X \geq 3) \approx 0.324 + 0.177 + 0.038 = \boxed{0.539}
Explanation
This problem involves calculating the probability that at least 3 out of 5 randomly selected human resource (HR) managers believe job applicants should follow up within two weeks. Since we are repeating the same trial (asking an HR manager) a fixed number of times (5), and each trial has only two possible outcomes (either they say “yes” or they don’t), the binomial distribution is appropriate.
The binomial distribution is used to model the number of successes in a fixed number of independent trials, each with the same probability of success. Here, a “success” is an HR manager saying that applicants should follow up within two weeks, and the given success probability is 52% or 0.52.
We are asked to find the probability that at least 3 out of 5 HR managers say “yes.” This is equivalent to finding the sum of the probabilities of exactly 3, 4, or 5 managers saying “yes.” We calculate each of these probabilities using the binomial probability formula: P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k} p^k (1 – p)^{n – k}
This formula calculates the probability of getting exactly kk successes in nn trials, with a success probability pp. We compute the individual probabilities for 3, 4, and 5 successes and then add them up. The result, approximately 0.539, indicates there is about a 53.9% chance that at least three of the five HR managers will recommend following up within two weeks.
