Question

a warehouse employs 27 workers on first - shift, 17 workers on second - shift, and 12 workers on third - shift. eight workers are chosen at random to be interviewed about the work environment. find the probability of choosing exactly five first - shift workers. the probability of choosing exactly five first - shift workers is (round to three decimal places as needed.)

Explanation:

Step1: Calculate total number of workers

The total number of workers is \(27 + 17+12 + 8=64\) workers.

Step2: Use combination formula for probability

The number of ways to choose 8 workers out of 64 is \(C(64,8)=\frac{64!}{8!(64 - 8)!}\). The number of ways to choose 5 first - shift workers out of 27 first - shift workers and \(8 - 5=3\) non - first - shift workers out of \(64 - 27 = 37\) non - first - shift workers is \(C(27,5)\times C(37,3)=\frac{27!}{5!(27 - 5)!}\times\frac{37!}{3!(37 - 3)!}\).

Step3: Calculate the probability

The probability \(P=\frac{C(27,5)\times C(37,3)}{C(64,8)}\)
\[

$$\begin{align*} C(27,5)&=\frac{27!}{5!(27 - 5)!}=\frac{27\times26\times25\times24\times23}{5\times4\times3\times2\times1}=80730\\ C(37,3)&=\frac{37!}{3!(37 - 3)!}=\frac{37\times36\times35}{3\times2\times1}=7770\\ C(64,8)&=\frac{64!}{8!(64 - 8)!}=\frac{64\times63\times62\times61\times60\times59\times58\times57}{8\times7\times6\times5\times4\times3\times2\times1}=44674665270 \end{align*}$$

\]
\[

$$\begin{align*} P&=\frac{80730\times7770}{44674665270}\\ &=\frac{627272100}{44674665270}\approx0.014 \end{align*}$$

\]

Answer:

0.014