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 3 of the workers interviewed are union members? Round your response to at least three decimal places. (If necessary, consult a list of formulas.)

Step 1 ：This problem is about calculating a binomial probability. The binomial probability formula is: \(P(X=k) = C(n, k) * (p^k) * ((1-p)^{n-k})\), where:

Step 2 ：- n is the number of trials (in this case, the number of workers interviewed)

Step 3 ：- k is the number of 'successes' (in this case, the number of union members)

Step 4 ：- p is the probability of success on a single trial (in this case, the probability that a worker is a union member)

Step 5 ：- C(n, k) is the binomial coefficient, which gives the number of ways to choose k successes out of n trials

Step 6 ：In this case, we want to find the probability that exactly 3 out of 4 workers interviewed are union members. So, n=4, k=3, and p=0.94.

Step 7 ：Substituting these values into the formula, we get: \(P(X=3) = C(4, 3) * (0.94^3) * ((1-0.94)^{4-3})\)

Step 8 ：Calculating the above expression, we get a probability of approximately 0.199

Step 9 ：Final Answer: The probability that exactly 3 of the workers interviewed are union members is approximately \(\boxed{0.199}\)