Given Information
A production facility employs 10 workers on the dayshift, 8 workers on the swing shift, and 6 workers on thegraveyard shift.


Step-1: Part (a)
There are total of 10+8+6 = 24 workers
All 5 workers coming from the day shift. For this, We have to choose 5 workers from 10 workers, hence
Number of ways are
\[\begin{array}{l}{N_a} = \left( {\begin{array}{*{20}{c}}{10}\\5\end{array}} \right)\\\\{N_a} = 252\end{array}\]Number of ways of choosing 5 workers from the entire lot
\[\begin{array}{l}{N_T} = \left( {\begin{array}{*{20}{c}}{24}\\5\end{array}} \right)\\\\{N_T} = 42504\end{array}\]The probability of randomly selecting 5 workers from day shift
\[\begin{array}{l}P = \dfrac{{252}}{{42504}}\\\\P = 0.00593\end{array}\]

Step-2: Part (b)
Number of ways to select 5 swing shift workers are
\[\begin{array}{l}{N_1} = \left( {\begin{array}{*{20}{c}}8\\5\end{array}} \right)\\\\{N_1} = 56\end{array}\]Number of ways to select 5 swing graveyard workers are
\[\begin{array}{l}{N_2} = \left( {\begin{array}{*{20}{c}}6\\5\end{array}} \right)\\\\{N_2} = 6\end{array}\]Total number of ways are

\[\begin{array}{l}{N_b} = {N_a} + {N_1} + {N_2}\\\\{N_b} = 252 + 56 + 6\\\\{N_b} = 314\end{array}\]Required Probability is
\[\begin{array}{l}P = \dfrac{{{N_b}}}{{{N_T}}}\\\\P = \dfrac{{314}}{{42504}}\\\\P = 0.00739\end{array}\]

Step-3: Part (c)
Probability that at least two different shifts will be represented among the selected workers
\[P(\text { at least two shifts represented })=1-P(\text { all from same shift })=1-.00739=.99261 \text { . }\]

Step-4: Part (d)
Probability that at least one of the shifts will be unrepresented in the sample of workers.
Let,
\[\begin{array}{l}{A_1} = {\rm{Day}}\,\,\,\,{\rm{Shift}}\,\,\,{\rm{unrepresented}}\\\\{A_2} = {\rm{Swing}}\,\,\,\,{\rm{Shift}}\,\,\,{\rm{unrepresented}}\\\\{A_3} = {\rm{Graveyard}}\,\,\,\,{\rm{Shift}}\,\,\,{\rm{unrepresented}}\end{array}\]We want to find out the probability of case
\[P\left(A_{1} \cup A_{2} \cup A_{3}\right)\]This means at least one of the case is represented
Number of cases of A$_1$
\[N\left(A_{1}\right)=N(\text { day shift unrepresented })=N(\text { all from swing/graveyard })=\left(\begin{array}{c}8+6 \\5\end{array}\right)=2002\]Number of cases of A$_2$
\[N\left(A_{2}\right)=N(\text { Swing shift unrepresented })=N(\text { all from day/graveyard })=\left(\begin{array}{c}10+6 \\5\end{array}\right)=4368\]Number of cases of A$_3$
\[N\left(A_{3}\right)=N(\text { Graveyard shift unrepresented })=N(\text { all from day/graveyard })=\left(\begin{array}{c}10+8 \\5\end{array}\right)=8568\]

Step-5
Also, \[N\left(A_{1} \cap A_{2}\right)=N(\text { all from graveyard })=6\]and \[N\left(A_{1} \cap A_{3}\right)=56\]and \[N\left(A_{2} \cap A_{3}\right)=252\]

Step-6
Therefore, the required probability is
\[P\left(A_{1} \cup A_{2} \cup A_{3}\right)=\dfrac{2002+4368+8568-6-56-252+0}{42504}=\dfrac{14624}{42504}=.3441\]==============================================================================================================