Question

A warehouse employs 22 workers on first shift and 18 workers on second shift. Eight workers are chosen at random to be interviewed about the work environment. Complete parts​ (a) through​ (d). ​(a) Find the probability of choosing all​ first-shift workers.

Answer 1

Answer:

The probability is 0.0042

Step-by-step explanation:

The probability is the division between the number of ways in which the eight workers are chosen from the first shift and the total number of ways in which they can select the eight workers.

So,  the number of ways in which the eight workers are chosen from the first shift is calculate using the following equation:

$nCk=\frac{n!}{k!(n-k)!}$

Where nCk give as the number of ways in which we can select k element from a group of n elements.

Then replacing k by the eight workers that are going to do the interview and n by the 22 workers from the first shift we get:

$22C8=\frac{22!}{8!(22-8)!}= 319,770$

So there are 319,770 ways to select eighth workers and all of them are first shift workers.

At the same way, we can calculate the total number of ways in which they can select the eight workers. Replacing n by 40 because this is the sum of 22 workers of the first shift and 18 workers of the second shift, we get:

$40C8=\frac{40!}{8!(40-8)!}= 76,904,685$

So, There are 76,904,685 ways to select eight workers from a group of 40.

Finally the probability is:

$\frac{22C8}{40C8} = \frac{319,770}{79,904,685} =0.0042$