Mathematics, 20.06.2020 16:57 uhhgray

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.)

Answers: 3

Show answers

Answers

Answer from: rachel2005smith

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) = 0.5257

To solve this question properly, we will need to make use of the concept of combination along with set theory.

What is Combination?

In mathematical concept, Combination is the grouping of subsets from a set without taking the order of selection into consideration.

The formula for calculating combination can be expressed as:

$\mathbf{(^n _r) =\dfrac{n!}{r!(n-r)! }}$

From the parameters given:

Workers employed on the day shift = 10Workers on swing shift = 8Workers on graveyard shift = 6

A quality control consultant is to select 4 of these workers for in-depth interviews:

Using the expression for calculating combination:

(a)

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

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

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

= 210

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

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

$\mathbf{= \dfrac{210}{10626}} \\ \\ \\ \mathbf{= 0.0198}$

(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) $\mathbf{= \dfrac{ \Big(^{10}_4\Big) }{\Big(^{24}_4\Big)}+\dfrac{ \Big(^{8}_4\Big) }{\Big(^{24}_4\Big)} + \dfrac{ \Big(^{6}_4\Big) }{\Big(^{24}_4\Big)}}$

where;

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

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

P( all 4 selected workers is:)

$\mathbf{=\dfrac{210+70+15}{10626}}$

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

(c)

The probability that at least two different shifts will be represented among the selected workers can be computed as:

$= 1-\dfrac{ (^{10}_4) }{(^{24}_4)}+\dfrac{ (^{8}_4) }{(^{24}_4)} + \dfrac{ (^{6}_4) }{(^{24}_4)}$

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

= 1 - 0.0278

= 0.9722

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

(d)

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

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

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