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.)
Accepted Solution
A:
Answer:0.019 or 1.9%Step-by-step explanation:Let U = Union member N= Non union memberLet the probability of U ; P(U)= 0.94P(N) = 0.06Sample size = 4Outcomes = 2 i.e either Union or Non-UnionTotal possible outcomes= 4² = 16Total 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 NP( 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%