91Ó°ÊÓ

The term "probability" simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes—how likely they are—when we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

a)

Product Rule for k-Tuples

If we assume that we may pick the first element of an ordered pair in \({{\rm{n}}_{\rm{1}}}\)ways and the second element in \({{\rm{n}}_{\rm{2}}}\)ways for each selected element, then the number of pairings is \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}\).

Similarly, there are \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}{\rm{* \ldots *}}{{\rm{n}}_{\rm{k}}}\)potential\({\rm{k}}\)-tuples given an ordered collection of \({\rm{k}}\)components, where the \({{\rm{k}}^{{\rm{th }}}}\)can be selected in \({{\rm{n}}_{\rm{k}}}\)ways.

We can calculate \({{\rm{n}}_{\rm{1}}}{\rm{ = 3}}\) (three different temperatures), \({{\rm{n}}_{\rm{2}}}{\rm{ = 4}}\) (four different pressures), and \({{\rm{n}}_{\rm{3}}}{\rm{ = 5}}\) (five different catalysts) using this method.

\({{\rm{n}}_{\rm{1}}}{\rm{*}}{{\rm{n}}_{\rm{2}}}{\rm{*}}{{\rm{n}}_{\rm{3}}}{\rm{ = 3*4*5 = 60}}\)

runs.

Therefore, the possible experimental runs are \({\rm{60}}\).

We use the same fundamental counting principle mentioned above, but here we limit \({{\rm{n}}_{\rm{1}}}{\rm{ = 1}}\) (one smallest temperature) and \({{\rm{n}}_{\rm{2}}}{\rm{ = 2}}\) (two smallest pressures), and \({{\rm{n}}_{\rm{3}}}{\rm{ = 5}}\)stays the same, so we have

\({{\rm{n}}_{\rm{1}}}{\rm{*}}{{\rm{n}}_{\rm{2}}}{\rm{*}}{{\rm{n}}_{\rm{3}}}{\rm{ = 1*2*5 = 10}}\)

Therefore, the experimental runs are \({\rm{10}}\).

We want to know how likely it is that a different catalyst will be utilized on each of the five experimental runs on the first day. Assume we have a fixed catalyst, and there are five different catalysts.

\({{\rm{n}}_{\rm{1}}}{\rm{*}}{{\rm{n}}_{\rm{2}}}{\rm{*1 = 3*4 = 12}}\)

runs (pairs of temperature and pressure). Now, we can separate given \({\rm{60}}\) possible runs into \({\rm{5}}\) groups of those \({\rm{12}}\) runs. Now we want to select \({\rm{1}}\) run out of this \({\rm{12}}\) and we can do that in

\(\left( {\begin{aligned}{{\rm{12}}}\\{\rm{1}}\end{aligned}} \right)\)

This is true for each of the five groups in different ways. As a result, the product of the numbers we just calculated is the number of ways to choose one run from each of these five groups.

\(\left( {\begin{aligned}{{\rm{12}}}\\{\rm{1}}\end{aligned}} \right)*\left( {\begin{aligned}{{\rm{12}}}\\{\rm{1}}\end{aligned}} \right)*\left( {\begin{aligned}{{\rm{12}}}\\{\rm{1}}\end{aligned}} \right)*\left( {\begin{aligned}{{\rm{12}}}\\{\rm{1}}\end{aligned}} \right)*\left( {\begin{aligned}{{\rm{12}}}\\{\rm{1}}\end{aligned}} \right)\)

where we used

\(\begin{aligned}{{\rm{n}}_{\rm{1}}}\rm &= {{\rm{n}}_{\rm{2}}}\\\rm &= {{\rm{n}}_{\rm{3}}}{\rm {=}} {{\rm{n}}_{\rm{4}}}\\\rm &= {{\rm{n}}_{\rm{5}}}{\rm{ = }}\left( {\begin{aligned}{{\rm{12}}}\\{\rm{1}}\end{aligned}} \right)\end{aligned}\)

This is our number of favorable outcomes,

\(\begin{aligned}{\left( {\begin{aligned}{{\rm{12}}}\\{\rm{1}}\end{aligned}} \right)^{\rm{5}}}\rm &= 1{{\rm{2}}^{\rm{5}}}\\\rm &= 248,832\end{aligned}\)

Number of all outcomes in the sample space is the number of selecting any \({\rm{5}}\) runs out of \({\rm{60}}\) , and for that we use combinations.

A combination is an unordered subset. For \({\rm{n}}\) individuals in a group, the number of combinations size \({\rm{k}}\) is denoted as \({{\rm{C}}_{{\rm{k,n}}}}\) or

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

which we read as from \({\rm{n}}\)elements we choose \({\rm{k}}\). The proposition:

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

Therefore, the number of ways to select \({\rm{5}}\) runs out of 60 is

\(\begin{aligned}{{\rm{C}}_{{\rm{5,60}}}}\rm &= \left( {\begin{aligned}{{\rm{60}}}\\{\rm{5}}\end{aligned}} \right){\rm{ = }}\frac{{{\rm{60!}}}}{{{\rm{5!(60 - 5)!}}}}\\\rm &= \frac{{{\rm{60!}}}}{{{\rm{5!55!}}}}\\\rm &= 5,461,512\end{aligned}\)

ways.

Finally, the probability \({\rm{P(A) = }}\frac{{{\rm{\# of favorable outcomes in A}}}}{{{\rm{\# of outcomes in the sample space }}}}\)is

\(\begin{aligned}\rm P(A) &= \frac{{{\rm{248,832}}}}{{{\rm{5,461,512}}}}\\\rm &= 0{\rm{.04556}}{\rm{.}}\end{aligned}\)