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There are \({\rm{10}}\) employees on the day shift, \({\rm{8}}\) employees on the swing shift, and 6 employees on the graveyard shift, for a total of \({\rm{24}}\) employees.

An unordered subset is referred to as a combination. The number of combinations of size \({\rm{k}}\) for \({\rm{n}}\) persons in a group is indicated as \({{\rm{C}}_{{\rm{k,n}}}}\)

\(\left( {\begin{aligned}{\rm{n}}\\{\rm{k}}\end{aligned}} \right)\)

which we interpret as \({\rm{k}}\)is chosen from \({\rm{n}}\) items. The proposal is this:

\({{\rm{C}}_{{\rm{k,n}}}}{\rm{ = }}\left( {\begin{aligned}{\rm{n}}\\{\rm{k}}\end{aligned}} \right){\rm{ = }}\frac{{{{\rm{P}}_{{\rm{k,n}}}}}}{{{\rm{k!}}}}{\rm{ = }}\frac{{{\rm{n!}}}}{{{\rm{k!(n - k)!}}}}\)

So, if we have \({\rm{n = 10}}\)employees on the day shift and we remove \({\rm{5}}\)of them, we'll have

\({{\rm{C}}_{{\rm{5,10}}}}{\rm{ = }}\left( {\begin{aligned}{{\rm{10}}}\\{\rm{5}}\end{aligned}} \right){\rm{ = }}\frac{{{\rm{10!}}}}{{{\rm{5!(10 - 5)!}}}}{\rm{ = 252}}\)

The chance of picking \({\rm{5}}\)day shift employees at random may be estimated using the formula below.

\({\rm{P(A) = }}\frac{{{\rm{\# \;favorable outcomes in\;A}}}}{{{\rm{\# \;outcomes in the sample space\;}}}}\)

When \({\rm{\# }}\) positive outcomes in A have been computed and are \({\rm{252}}\), and we utilize combinations to estimate how many ways we can take \({\rm{5}}\)workers out of \({\rm{24}}\)in the sample space, which is

\({{\rm{C}}_{{\rm{5,24}}}}{\rm{ = }}\left( {\begin{aligned}{{\rm{24}}}\\{\rm{5}}\end{aligned}} \right){\rm{ = }}\frac{{{\rm{24!}}}}{{{\rm{5!(24 - 5)!}}}}{\rm{ = 42504}}{\rm{.}}\)

We can calculate the probability using these two.

\({\rm{P(A) = }}\frac{{{\rm{252}}}}{{{\rm{42504}}}}{\rm{ = 0}}{\rm{.0059}}\)

\({{\rm{C}}_{{\rm{5,10}}}}{\rm{ = }}\left( {\begin{aligned}{{\rm{10}}}\\{\rm{5}}\end{aligned}} \right){\rm{ = }}\frac{{{\rm{10!}}}}{{{\rm{5!(10 - 5)!}}}}{\rm{ = 252}}\)

selections.

Similarly, because the day shift has \({\rm{n = 8}}\)employees and we take \({\rm{5}}\) of them, there are

\({{\rm{C}}_{{\rm{5,8}}}}{\rm{ = }}\left( {\begin{aligned}{\rm{8}}\\{\rm{5}}\end{aligned}} \right){\rm{ = }}\frac{{{\rm{8!}}}}{{{\rm{5!(8 - 5)!}}}}{\rm{ = 56}}\)

selections.

And, because there are \({\rm{n = 6}}\)employees on the night shift, and we take \({\rm{5}}\) of them, \({{\rm{C}}_{{\rm{5,6}}}}{\rm{ = }}\left( {\begin{aligned}{\rm{6}}\\{\rm{5}}\end{aligned}} \right){\rm{ = }}\frac{{{\rm{6!}}}}{{{\rm{5!(6 - 5)!}}}}{\rm{ = 6}}\)

there are \({\rm{252 + 56 + 6 = 314}}\) choices (favourable outcomes) to choose \({\rm{5}}\) people from the same shift.

We utilize combinations to figure out how many ways we can take 5 employees out of 24 in the sample space for # outcomes.

\({{\rm{C}}_{{\rm{5,24}}}}{\rm{ = }}\left( {\begin{aligned}{{\rm{24}}}\\{\rm{5}}\end{aligned}} \right){\rm{ = }}\frac{{{\rm{24!}}}}{{{\rm{5!(24 - 5)!}}}}{\rm{ = 42504}}\)

We can calculate the probability using these two.

\({\rm{P(A) = }}\frac{{{\rm{314}}}}{{{\rm{42504}}}}{\rm{ = 0}}{\rm{.0074}}\)

\({\rm{P(A) = 1 - P}}\left( {{\rm{A'}}} \right)\)

Keep in mind that this applies to any event A. It has been computed in the situation that everyone is from the same shift (b).

We now have

\({\rm{P(A) = 1 - P}}\left( {{\rm{A'}}} \right){\rm{ = 1 - 0}}{\rm{.0074 = 0}}{\rm{.9926}}{\rm{.}}\)

\(\begin{aligned}{{A}_{1}}&=\{~day\text{ }shift\text{ }has\text{ }no\text{ }representative~\} \\{{A}_{2}}&=\{~swing\text{ }shift\text{ }has\text{ }no\text{ }representative~\} \\{{A}_{3}}&=\{~graveyard\text{ }shift\text{ }has\text{ }no\text{ }representative~\}, \\\end{aligned}\)

There is no day shift representation, no swing shift representative, and no graveyard shift representative.

\(P\left( {{A}_{1}}\grave{E}{{A}_{2}}\grave{E}{{A}_{3}} \right).\)

It means we're looking for a way to figure out what the chances are of anything happening.

\(\begin{aligned}P(A\grave{E}B\grave{E}C)=P(A)+P(B)+P(C)-P(AB)-P(AC)-P(BC) \\+P(ABC) \\\end{aligned}\)

Due to the fact that this is not a disjoint union, we will apply the following proposition:

The following is true for each of the three occurrences \({\rm{A,B}}\) and \({\rm{C}}\)

start with the probability of event\({{\rm{A}}_{\rm{1}}}{\rm{,}}{{\rm{A}}_{\rm{2}}}\) and \({{\rm{A}}_{\rm{3}}}\)

\({\rm{P(A) = }}\frac{{{\rm{\# \;favorable outcomes in\;A}}}}{{{\rm{\# \;outcomes in the sample space\;}}}}\)

Using \({\rm{N(A)}}\) as a notation for the number of items in event A, we have

\({\rm{N}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ = }}\left( {\begin{aligned}{{\rm{8 + 6}}}\\{\rm{5}}\end{aligned}} \right){\rm{ = }}\left( {\begin{aligned}{{\rm{14}}}\\{\rm{5}}\end{aligned}} \right){\rm{ = }}\frac{{{\rm{14!}}}}{{{\rm{5!(14 - 5)!}}}}{\rm{ = 2002}}\)

We selected combinations since it didn't matter which workers we chose as long as they weren't from the day shift, and the number of workers who weren't from the day shift was small. \({\rm{8 + 6 = 14}}\)

An unordered subset is referred to as a combination. The number of combinations of size \({\rm{k}}\)for \({\rm{n}}\)persons in a group is indicated as \({{\rm{C}}_{{\rm{k,n}}}}\),

\(\left( {\begin{aligned}{\rm{n}}\\{\rm{k}}\end{aligned}} \right)\)

which we read as we pick \({\rm{k}}\)from \({\rm{n}}\)elements. The proposal is this:

\({{\rm{C}}_{{\rm{k,n}}}}{\rm{ = }}\left( {\begin{aligned}{\rm{n}}\\{\rm{k}}\end{aligned}} \right){\rm{ = }}\frac{{{{\rm{P}}_{{\rm{k,n}}}}}}{{{\rm{k!}}}}{\rm{ = }}\frac{{{\rm{n!}}}}{{{\rm{k!(n - k)!}}}}\)

We have \({{\rm{A}}_{\rm{2}}}\) and \({{\rm{A}}_{\rm{3}}}\)

\({\rm{N}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ = }}\left( {\begin{aligned}{{\rm{10 + 6}}}\\{\rm{5}}\end{aligned}} \right){\rm{ = }}\left( {\begin{aligned}{{\rm{16}}}\\{\rm{5}}\end{aligned}} \right){\rm{ = }}\frac{{{\rm{16!}}}}{{{\rm{5!(16 - 5)!}}}}{\rm{ = 4368}}\)

And \({\rm{N}}\left( {{{\rm{A}}_{\rm{3}}}} \right){\rm{ = }}\left( {\begin{aligned}{{\rm{10 + 8}}}\\{\rm{5}}\end{aligned}} \right){\rm{ = }}\left( {\begin{aligned}{{\rm{18}}}\\{\rm{5}}\end{aligned}} \right){\rm{ = }}\frac{{{\rm{18!}}}}{{{\rm{5!(18 - 5)!}}}}{\rm{ = 8568,}}\)

in the same way.

We take 5 workers from the other two shifts, which is why the total number of workers from those shifts is summed.

We may use this to compute the probability of \({{\rm{A}}_{\rm{1}}}{\rm{,}}{{\rm{A}}_{\rm{2}}}\) and \({{\rm{A}}_{\rm{3}}}\)events.

\(\begin{aligned}{{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ = }}\frac{{{\rm{2002}}}}{{{\rm{42504}}}}}\\{{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ = }}\frac{{{\rm{4368}}}}{{{\rm{42504}}}}}\\{{\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}} \right){\rm{ = }}\frac{{{\rm{8568}}}}{{{\rm{42504}}}}}\end{aligned}\)

All possible selections are as:

\(\left( {\begin{aligned}{{\rm{24}}}\\{\rm{5}}\end{aligned}} \right){\rm{ = 42504}}\)

we must first identify the number of favourable outcomes in each junction.

We've already done it in (b); check it out for a more detailed explanation, but we've also done it in (c).

\({\rm{N}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{{C}}}{{\rm{A}}_{\rm{2}}}} \right){\rm{ = }}\left( {\begin{aligned}{\rm{6}}\\{\rm{ - }}\end{aligned}} \right){\rm{ = 6,}}\)

Because the number of positive events is only when all five workers are on the graveyard shift (intersection that there is none from day an non from swing shift indicates that all five must be from graveyard shift).

\({\rm{N}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{{C}}}{{\rm{A}}_{\rm{3}}}} \right){\rm{ = }}\left( {\begin{aligned}{\rm{8}}\\{\rm{5}}\end{aligned}} \right){\rm{ = 56}}\)

Because the number of positive occurrences is only when all five workers are on the swing shift (intersection that there is none from day and no from graveyard shift indicates that all five must be from swing shift).

\({\rm{N}}\left( {{{\rm{A}}_{\rm{2}}}{\rm{{C}}}{{\rm{A}}_{\rm{3}}}} \right){\rm{ = }}\left( {\begin{aligned}{{\rm{10}}}\\{\rm{5}}\end{aligned}} \right){\rm{ = 252}}\)

Because the number of positive occurrences is only when all five workers are on the day shift (intersection that there is none from swing and no from graveyard shift indicates that all five must be from day shift).

\(\begin{aligned}{{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{{C}}}{{\rm{A}}_{\rm{2}}}} \right){\rm{ = }}\frac{{\rm{6}}}{{{\rm{42504}}}}}\\{{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{{C}}}{{\rm{A}}_{\rm{3}}}} \right){\rm{ = }}\frac{{{\rm{56}}}}{{{\rm{42504}}}}}\\{{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}{\rm{{C}}}{{\rm{A}}_{\rm{3}}}} \right){\rm{ = }}\frac{{{\rm{252}}}}{{{\rm{42504}}}}}\end{aligned}\)

It's evident that this isn't possible because at least one of the changes requires a representative. Therefore,

\({\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{{C}}}{{\rm{A}}_{\rm{2}}}{\rm{{C}}}{{\rm{A}}_{\rm{3}}}} \right){\rm{ = 0}}{\rm{.}}\)

Finally, the likelihood is

\(\begin{aligned}P\left( {{A}_{1}}\grave{E}{{A}_{2}}\grave{E}{{A}_{3}} \right)&=P\left( {{A}_{1}} \right)+P\left( {{A}_{2}} \right)+P\left( {{A}_{3}} \right)-P\left( {{A}_{1}}{C}{{A}_{2}} \right)-P\left( {{A}_{1}}{C}{{A}_{3}} \right)-P\left( {{A}_{2}}{C}{{A}_{3}} \right) \\+P\left( {{A}_{1}}{C}{{A}_{2}}{C}{{A}_{3}} \right) \\&=\frac{2002}{42504}+\frac{4368}{42504}+\frac{8568}{42504}-\frac{6}{42504}-\frac{56}{42504}-\frac{252}{42504}+\frac{0}{42504} \\&=0.3441. \\\end{aligned}\)