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Assume that when human resource managers are randomly selected, 42% say job applicants should follow up within two weeks. If 8 human resource managers are randomly selected, find the probability that exactly 3 of them say job applicants should follow up within two weeks

User Babak
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Final answer:

The probability that exactly three out of eight randomly selected human resource managers will say job applicants should follow up within two weeks is approximately 0.18 or 18%, which is calculated using the binomial probability formula.

Step-by-step explanation:

To find the probability that exactly 3 out of 8 randomly selected human resource managers say job applicants should follow up within two weeks, we can use the binomial probability formula:

P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)

Where:

  • P(X = k) is the probability that exactly k managers say applicants should follow up within two weeks.
  • C(n, k) is the number of combinations of n things taken k at a time.
  • p is the probability of a manager saying applicants should follow up within two weeks (42% or 0.42).
  • n is the total number of managers selected (8 in this case).
  • k is the exact number of managers we're looking to find the probability for (3 in this case).

Using the given values, we compute:

C(8, 3) = 8! / (3! * (8 - 3)!) = 56

Therefore, P(X = 3) = 56 * (0.42)^3 * (1 - 0.42)^(8 - 3) = 56 * 0.074088 * 0.182520 = 0.754716

Hence, the probability that exactly 3 out of 8 human resource managers say job applicants should follow up within two weeks is approximately 0.18 or 18%.

User Hamzeh Soboh
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