Question

initial knowledge check
question 25
the workers union at a particular university is quite strong. about 96% 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 4 of the workers interviewed are union members?
round your response to at least three decimal places. (if necessary, consult a list of formulas.)

Explanation:

Step1: Define binomial parameters

We use the binomial probability formula: $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$. Here, $n=5$ (total trials), $k=4$ (successes), $p=0.96$ (union member prob).

Step2: Calculate combination term

Compute $\binom{5}{4}$, which is $\frac{5!}{4!(5-4)!} = 5$.

Step3: Compute success/failure terms

Calculate $p^4 = 0.96^4 = 0.84934656$, and $(1-p)^{1} = 0.04^1 = 0.04$.

Step4: Multiply all terms

Multiply the three values: $5 \times 0.84934656 \times 0.04$.
<Expression>
$5 \times 0.84934656 \times 0.04 = 0.169869312$

Step5: Round to 3 decimals

Round the result to at least 3 decimal places.

Answer:

0.170