Questions: 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.

Solution Steps

We need to find the probability that exactly 3 out of 4 randomly selected workers from a university are members of the workers' union, given that 94% of all workers belong to the union.

Let:

• \( n = 4 \) (the number of trials, or workers interviewed)
• \( x = 3 \) (the number of successes, or union members)
• \( p = 0.94 \) (the probability of success, or being a union member)
• \( q = 1 - p = 0.06 \) (the probability of failure, or not being a union member)

The probability of exactly \( x \) successes in \( n \) trials is given by the formula:

\[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

Substituting the values:

\[ P(X = 3) = \binom{4}{3} \cdot (0.94)^3 \cdot (0.06)^{4-3} \]

The binomial coefficient \( \binom{4}{3} \) is calculated as:

\[ \binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4 \cdot 3!}{3! \cdot 1!} = 4 \]

Now, substituting back into the formula:

\[ P(X = 3) = 4 \cdot (0.94)^3 \cdot (0.06)^1 \]

Calculating \( (0.94)^3 \):

\[ (0.94)^3 \approx 0.830584 \]

Now, substituting this value:

\[ P(X = 3) = 4 \cdot 0.830584 \cdot 0.06 \approx 0.199 \]

Final Answer