Question

A production facility employs 10 workers on the day shift, 8 workers on the swing shift, and 6 workers on the graveyard shift. A quality control consultant is to select 4 of these workers for in-depth interviews. Suppose the selection is made in such a way that any particular group of 4 workers has the same chance of being selected as does any other group (drawing 4 slips without replacement from among 24).
(a) How many selections result in all 4 workers coming from the day shift? What is the probability that all 4 selected workers will be from the day shift? (Round your answer to four decimal places.)
(b) What is the probability that all 4 selected workers will be from the same shift? (Round your answer to four decimal places.)
(c) What is the probability that at least two different shifts will be represented among the selected workers? (Round your answer to four decimal places.)
(d) What is the probability that at least one of the shifts will be unrepresented in the sample of workers? (Round your answer to four decimal places.)

Answer 1

Answer:

The probability that all 4 selected workers will be from the day shift is, = 0.0198

The probability that all 4  selected workers will be from the same shift is = 0.0278

The probability that at least two different shifts will be represented among the selected workers is = 0.9722

The probability that at least one of the shifts will be unrepresented in the sample of workers is P(A∪B∪C) = 0.5256

Step-by-step explanation:

Given that:

A production facility employs 10 workers on the day shift, 8 workers on the swing shift, and 6 workers on the graveyard shift. A quality control consultant is to select 4 of these workers for in-depth interviews:

The number of selections result in all 4 workers coming from the day shift is :

$(^n _r) = (^{10} _4)$

$=\dfrac{(10!)}{4!(10-4)!}$

= 210

The probability that all 5 selected workers will be from the day shift is,

$\begin{array}{c}\\P\left( {{\rm{all \ 4 \ selected \ workers\ will \ be \ from \ the \ day \ shift}}} \right) = \frac{{\left( \begin{array}{l}\\10\\\\4\\\end{array} \right)}}{{\left( \begin{array}{l}\\24\\\\4\\\end{array} \right)}}\\\\ = \frac{{210}}{{10626}}\\\\ = 0.0198\\\end{array}$

(b) The probability that all 4 selected workers will be from the same shift is calculated as follows:

P( all 4 selected workers will be) $= \dfrac{ (^{10}_4) }{(^{24}_4)}+\dfrac{ (^{8}_4) }{(^{24}_4)} + \dfrac{ (^{6}_4) }{(^{24}_4)}$

where;

$(^{8}_4) } = \dfrac{8!}{4!(8-4)!} = 70$

$(^{6}_4) } = \dfrac{6!}{4!(6-4)!} = 15$

∴ P( all 4 selected workers is ) $=\dfrac{210+70+15}{10626}$

The probability that all 4  selected workers will be from the same shift is = 0.0278

(c) What is the probability that at least two different shifts will be represented among the selected workers?

P ( at least two different shifts will be represented among the selected workers)  $= 1-\dfrac{ (^{10}_4) }{(^{24}_4)}+\dfrac{ (^{8}_4) }{(^{24}_4)} + \dfrac{ (^{6}_4) }{(^{24}_4)}$

$=1 - \dfrac{210+70+15}{10626}$

= 1 - 0.0278

The probability that at least two different shifts will be represented among the selected workers is = 0.9722

(d)What is the probability that at least one of the shifts will be unrepresented in the sample of workers?

The probability that at least one of the shifts will be unrepresented in the sample of workers is:

$P(AUBUC) = \dfrac{(^{6+8}_4)}{(^{24}_4)}+ \dfrac{(^{10+6}_4)}{(^{24}_4)}+ \dfrac{(^{10+8}_4)}{(^{24}_4)}- \dfrac{(^{6}_4)}{(^{24}_4)}-\dfrac{(^{8}_4)}{(^{24}_4)}-\dfrac{(^{10}_4)}{(^{24}_4)}+0$

$P(AUBUC) = \dfrac{(^{14}_4)}{(^{24}_4)}+ \dfrac{(^{16}_4)}{(^{24}_4)}+ \dfrac{(^{18}_4)}{(^{24}_4)}- \dfrac{(^{6}_4)}{(^{24}_4)}-\dfrac{(^{8}_4)}{(^{24}_4)}-\dfrac{(^{10}_4)}{(^{24}_4)}+0$

$P(AUBUC) = \dfrac{1001}{10626}+ \dfrac{1820}{10626}+ \dfrac{3060}{10626}-\dfrac{15}{10626}-\dfrac{70}{10626}-\dfrac{210}{10626} +0$

The probability that at least one of the shifts will be unrepresented in the sample of workers is P(A∪B∪C) = 0.5256