Question

The workers' union at a particular university is quite strong. About 94% of all workers employed by the university belong to the workers' union. Recently, the workers went on strike, and now a local TV station plans to interview 4 workers (chosen at random) at the university to get their opinions on the strike. What is the probability that exactly 2 of the workers interviewed are union members? Round your response to at least three decimal places. (If necessary, consult a list of formulas.)

Answer 1

0.019 or 1.9%

Explanation:

Let U = Union member

N= Non union member

Let the probability of U ; P(U)= 0.94

P(N) = 0.06

Sample size = 4

Outcomes = 2 i.e either Union or Non-Union

Total possible outcomes= 4²

= 16

Total Possible outcomes is shown below

1st member 2nd member 3rd member 4th member

U U U U

U U U N

U U N U

U U N N

U N U U

U N U N

U N N U

U N N N

N U U U

N U U N

N U N U

N U N N

N N U U

N N U N

N N N U

N N N N

P( exactly two (2) workers are union members )

= P(UUNN) + P (UNUN) + P (UNNU) + P (NUUN) + P (NUNU) + P (NNUU)

= 6 x P(UUNN)

= 6 x (0.94 x0.94 x0.06 x0.06)

= 0.01908576

≈ 0.019 or 1.9%