Question

a warehouse employs 25 workers on first - shift, 19 workers on second - shift, and 11 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 0.051. (round to three decimal places as needed.)

Explanation:

Step1: Calculate total number of workers

The total number of workers is $25 + 19+11 = 55$ workers.

Step2: Use combination formula for probability

The combination formula is $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n$ is the total number of items and $r$ is the number of items to be chosen. The total number of ways to choose 8 workers out of 55 is $C(55,8)=\frac{55!}{8!(55 - 8)!}$. The number of ways to choose 5 first - shift workers out of 25 and $8 - 5=3$ non - first - shift workers out of $55 - 25 = 30$ is $C(25,5)\times C(30,3)=\frac{25!}{5!(25 - 5)!}\times\frac{30!}{3!(30 - 3)!}$.

Step3: Calculate the probability

The probability $P$ is given by $P=\frac{C(25,5)\times C(30,3)}{C(55,8)}$.
$C(25,5)=\frac{25!}{5!(25 - 5)!}=\frac{25\times24\times23\times22\times21}{5\times4\times3\times2\times1}=53130$
$C(30,3)=\frac{30!}{3!(30 - 3)!}=\frac{30\times29\times28}{3\times2\times1}=4060$
$C(55,8)=\frac{55!}{8!(55 - 8)!}=\frac{55\times54\times53\times52\times51\times50\times49\times48}{8\times7\times6\times5\times4\times3\times2\times1}=190688430$
$P=\frac{53130\times4060}{190688430}\approx0.051$

Answer:

$0.051$