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.

We have three molecules of type \({\rm{A}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{,}}{{\rm{A}}_{\rm{2}}}{\rm{,}}{{\rm{A}}_{\rm{3}}}} \right)\) three of type \({\rm{B}}\left( {{{\rm{B}}_{\rm{1}}}{\rm{,}}{{\rm{B}}_{\rm{2}}}{\rm{,}}{{\rm{B}}_{\rm{3}}}} \right)\), three of type \({\rm{C}}\left( {{{\rm{C}}_{\rm{1}}}{\rm{,}}{{\rm{C}}_{\rm{2}}}{\rm{,}}{{\rm{C}}_{\rm{3}}}} \right)\), and three of type \({\rm{D}}\left( {{{\rm{D}}_{\rm{1}}}{\rm{,}}{{\rm{D}}_{\rm{2}}}{\rm{,}}{{\rm{D}}_{\rm{3}}}} \right)\).

(a):

First, if If the three \({A^\prime }s,{B^\prime }s,{C^\prime }s\), and \({D^\prime }s\) were distinguishable from one another, we would have \({\rm{3 + 3 + 3 + 3 = 12}}\) places to put the order of a chain molecule, where the order is important. For such, we use permutations.

A permutation is an ordered subset. For \({\rm{n}}\)individuals in a group, the number of permutations size \({\rm{k}}\) is denoted as \({{\rm{P}}_{{\rm{k,n}}}}\). The proposition:

\({{\rm{P}}_{{\rm{k,n}}}}{\rm{ = }}\frac{{{\rm{n!}}}}{{{\rm{(n - k)!}}}}\)

Here we have \({\rm{n = 12}}\) and also a \({\rm{k = 12}}\) because we want a permutation of size \({\rm{12}}\) (a chain molecule is that size), therefore, the number of possible chain molecules is

\(\begin{aligned}{{\rm{P}}_{{\rm{12,12}}}} \rm &= \frac{{{\rm{12!}}}}{{{\rm{(12 - 12)!}}}}\\ \rm &= 12! \\ \rm &= 479,001,600 {\rm{.}}\end{aligned}\)

For the other question, when we remove the subscripts and have only \({{\rm{A}}^{\rm{¢}}}{\rm{s}}\) and not \({\rm{A}}_{\rm{i}}^{\rm{¢}}{\rm{s,}}\)we have that every \({\rm{3*2*1 = 6}}\) samples from \({\rm{479,001,600}}\)become one. This is because we have three\({{\rm{A}}_{\rm{1}}}{\rm{,}}{{\rm{A}}_{\rm{2}}}\), and \({{\rm{A}}_{\rm{3}}}\)for which there are

\(\begin{aligned}{{\rm{P}}_{{\rm{3,3}}}} \rm &= \frac{{{\rm{3!}}}}{{{\rm{(3 - 3)!}}}}\\ \rm &= 3! \\ \rm &= 6 \end{aligned}\)

ways to order them. When we suppress subscripts, all this \({\rm{6}}\) outcomes become just one, because we do not have \({\rm{3}}\) different \({{\rm{A}}^{\rm{¢}}}{\rm{s}}\)but only one. Use this example:

\({{\rm{A}}_{\rm{1}}}{{\rm{A}}_{\rm{2}}}{{\rm{A}}_{\rm{3}}}{{\rm{B}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{B}}_{\rm{1}}}{{\rm{D}}_{\rm{1}}}{{\rm{D}}_{\rm{2}}}{{\rm{B}}_{\rm{3}}}{{\rm{C}}_{\rm{1}}}{{\rm{D}}_{\rm{3}}}{{\rm{C}}_{\rm{3}}}\) where we can order \({\rm{A}}_{\rm{i}}^{\rm{¢}}{\rm{s}}\) is \({\rm{6}}\) different ways to obtain outcomes

\(\begin{array}{l}{{\rm{A}}_{\rm{1}}}{{\rm{A}}_{\rm{2}}}{{\rm{A}}_{\rm{3}}}{{\rm{B}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{B}}_{\rm{1}}}{{\rm{D}}_{\rm{1}}}{{\rm{D}}_{\rm{2}}}{{\rm{B}}_{\rm{3}}}{{\rm{C}}_{\rm{1}}}{{\rm{D}}_{\rm{3}}}{{\rm{C}}_{\rm{3}}}\\{{\rm{A}}_{\rm{2}}}{{\rm{A}}_{\rm{1}}}{{\rm{A}}_{\rm{3}}}{{\rm{B}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{B}}_{\rm{1}}}{{\rm{D}}_{\rm{1}}}{{\rm{D}}_{\rm{2}}}{{\rm{B}}_{\rm{3}}}{{\rm{C}}_{\rm{1}}}{{\rm{D}}_{\rm{3}}}{{\rm{C}}_{\rm{3}}}\\{{\rm{A}}_{\rm{3}}}{{\rm{A}}_{\rm{1}}}{{\rm{A}}_{\rm{2}}}{{\rm{B}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{B}}_{\rm{1}}}{{\rm{D}}_{\rm{1}}}{{\rm{D}}_{\rm{2}}}{{\rm{B}}_{\rm{3}}}{{\rm{C}}_{\rm{1}}}{{\rm{D}}_{\rm{3}}}{{\rm{C}}_{\rm{3}}}\\{{\rm{A}}_{\rm{1}}}{{\rm{A}}_{\rm{3}}}{{\rm{A}}_{\rm{2}}}{{\rm{B}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{B}}_{\rm{1}}}{{\rm{D}}_{\rm{1}}}{{\rm{D}}_{\rm{2}}}{{\rm{B}}_{\rm{3}}}{{\rm{C}}_{\rm{1}}}{{\rm{D}}_{\rm{3}}}{{\rm{C}}_{\rm{3}}}\\{{\rm{A}}_{\rm{2}}}{{\rm{A}}_{\rm{3}}}{{\rm{A}}_{\rm{1}}}{{\rm{B}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{B}}_{\rm{1}}}{{\rm{D}}_{\rm{1}}}{{\rm{D}}_{\rm{2}}}{{\rm{B}}_{\rm{3}}}{{\rm{C}}_{\rm{1}}}{{\rm{D}}_{\rm{3}}}{{\rm{C}}_{\rm{3}}}\\{{\rm{A}}_{\rm{3}}}{{\rm{A}}_{\rm{2}}}{{\rm{A}}_{\rm{1}}}{{\rm{B}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{B}}_{\rm{1}}}{{\rm{D}}_{\rm{1}}}{{\rm{D}}_{\rm{2}}}{{\rm{B}}_{\rm{3}}}{{\rm{C}}_{\rm{1}}}{{\rm{D}}_{\rm{3}}}{{\rm{C}}_{\rm{3}}}\end{array}\)

Now imagine we do not have subscript. All the outcomes become just one

\({\rm{AAA}}{{\rm{B}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{B}}_{\rm{1}}}{{\rm{D}}_{\rm{1}}}{{\rm{D}}_{\rm{2}}}{{\rm{B}}_{\rm{3}}}{{\rm{C}}_{\rm{1}}}{{\rm{D}}_{\rm{3}}}{{\rm{C}}_{\rm{3}}}\)So, each group of \({\rm{6}}\) becomes \({\rm{1}}\) outcome when remove subscripts, therefore just by dividing \({\rm{12}}\) ! (all outcomes) with \({\rm{3 !}}\) \({\rm{ = 6}}\)we will get the number of chain molecules when there is only one \({\rm{A}}\)instead of\({{\rm{A}}_{\rm{1}}}{\rm{,}}{{\rm{A}}_{\rm{2}}}\), and \({{\rm{A}}_{\rm{3}}}\)but we still have tree of type\({\rm{B}}\left( {{{\rm{B}}_{\rm{1}}}{\rm{,}}{{\rm{B}}_{\rm{2}}}{\rm{,}}{{\rm{B}}_{\rm{3}}}} \right)\), three of type\({\rm{C}}\left( {{{\rm{C}}_{\rm{1}}}{\rm{,}}{{\rm{C}}_{\rm{2}}}{\rm{,}}{{\rm{C}}_{\rm{3}}}} \right)\), and three of type\({\rm{D}}\left( {{{\rm{D}}_{\rm{1}}}{\rm{,}}{{\rm{D}}_{\rm{2}}}{\rm{,}}{{\rm{D}}_{\rm{3}}}} \right)\). We have\(\frac{{{\rm{12!}}}}{{{\rm{3!}}}}\).

When we do the same with \({{\rm{B}}^{\rm{¢}}}{\rm{s}}\) or \({C^\prime }s\)or \({D^\prime }s\)we will get the same answer with the same reasoning (next step is when you remove subscript of \({B^\prime }s\) just divide with \({\rm{3!}}\) the number\(\frac{{{\rm{12!}}}}{{{\rm{3!}}}}\), and so on). Therefore, in order to answer the second question, we need to divide number of chain molecules (\({\rm{12!}}\)) with four times \({\rm{3 !}}\) for each of the four letters or

\(\frac{{{\rm{12!}}}}{{{\rm{3!3!3!3!}}}}{\rm{ = 369,600}}{\rm{. }}\) chain molecules.

(b):

We want to use

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

Because all three of the same latter are one next to other, you can look at them as single entity. This means that we want to find the number of ways to put four letters\({\rm{A,B,C}}\), and \({\rm{D}}\)in some order of four elements. Since order is important, we have

\(\begin{aligned}{{\rm{P}}_{{\rm{4,4}}}} \rm &= \frac{{{\rm{4!}}}}{{{\rm{(4 - 4)!}}}}\\ \rm &= 4! \\ \rm &= 24 \end{aligned}\)

ways to order these entities. This is our number of favorable outcomes.

Number of outcomes in the sample space has been calculated in (a) and it is \({\rm{369,600}}\) (we use this because all letters are the same).

The probability that all three molecules of each type end up next to one another is

\(\begin{aligned} \rm P(A) &=& \frac{{{\rm{\# of favorable outcomes in A }}}}{{{\rm{\# of outcomes in the sample space }}}}\\ \rm &=& \frac{{{\rm{24}}}}{{{\rm{396,600}}}}\\ \rm &=& 0{\rm{.00006494}}{\rm{.}}\end{aligned}\)

Therefore, the probabilities are \({\rm{0}}{\rm{.00006494}}\).