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 5 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.

To solve this problem, we can use the binomial probability formula. The binomial probability formula is used when there are exactly two mutually exclusive outcomes of a trial: "success" (a worker is a union member) and "failure" (a worker is not a union member).

Given:
- The probability of success (a worker being a union member) is \( p = 0.94 \).
- The probability of failure (a worker not being a union member) is \( q = 1 - p = 0.06 \).
- The number of trials (workers interviewed) is \( n = 5 \).
- The number of successes (union members among the interviewed workers) is \( k = 3 \).

The binomial probability formula is:

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

Where:
- \(\binom{n}{k}\) is the binomial coefficient, calculated as \(\frac{n!}{k!(n-k)!}\).
- \(p^k\) is the probability of having exactly \(k\) successes.
- \(q^{n-k}\) is the probability of having exactly \(n-k\) failures.

Let's calculate:

1. **Calculate the binomial coefficient**:

\[
\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10
\]

2. **Calculate \(p^k\) and \(q^{n-k}\)**:

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

\[
q^{2} = (0.06)^2 = 0.0036
\]

3. **Calculate the binomial probability**:

\[
P(X = 3) = 10 \times 0.830584 \times 0.0036 \approx 0.0299
\]

Therefore, the probability that exactly 3 of the workers interviewed are union members is approximately **0.030** (rounded to three decimal places).

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Do you want more details or have any other questions? Here are some related questions:

1. What is the probability that exactly 4 workers interviewed are union members?
2. How does the probability change if the number of workers interviewed increases to 10?
3. What is the probability that none of the workers interviewed are union members?
4. How would the probability change if the proportion of union members was 80% instead of 94%?
5. How does this probability relate to the expected number of union members among the interviewed workers?
6. What is the cumulative probability of having at least 3 union members among the interviewed workers?
7. How would the probability change if the interview group size was decreased to 3 workers?
8. Can you use a normal approximation for this problem, and under what conditions would it be appropriate?

**Tip:** When dealing with binomial probabilities, always ensure that you correctly identify the "success" event and use the appropriate values for \(p\) and \(q\).

Learn how to calculate the probability of exactly 3 union members among 5 workers interviewed at a university. Use the binomial probability formula with a success probability of 0.94. Get step-by-step solutions and explanations.

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