91Ó°ÊÓ

If the probability of one event is unaffected by the occurrence or non-occurrence of the other, two events are said to be independent in probability. Consider the following example for a better understanding of this definition.

We'll utilise the ordered pair \({\rm{(r,g)}}\), with \({\rm{r}}\) representing the number of dots on the red dice and \({\rm{g}}\) representing the number of dots on the green dice. \({\rm{(1,2)}}\), for example, means that the red dice showed \({\rm{1}}\) dot and the green dice showed \({\rm{2}}\)dots.

Using this, events \({\rm{A,B}}\), and \({\rm{C}}\) would be

\(\begin{array}{c}{\rm{A = \{ (3,1)(3,2)(3,3)(3,4)(3,5)(3,6)\} }}\\{\rm{B = \{ (1,4)(2,4)(3,4)(4,4)(5,4)(6,4)\} }}\\{\rm{C = \{ (1,6)(2,5)(3,4)(4,3)(5,2)(6,1)\} }}\end{array}\)

There are total of

\({\rm{6*6 = 36}}\)

number of ordered pairs.

If we assume that we may select 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 \({{\rm{n}}_{_{\rm{1}}}}{{\rm{n}}_{\rm{2}}}\).

Therefore,

\(\begin{array}{c}{\rm{P(A) = }}\frac{{{\rm{\# of favorable outcomes in A }}}}{{{\rm{\# of outcomes in the sample space }}}}\\{\rm{ = }}\frac{{\rm{6}}}{{{\rm{36}}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{6}}}\\{\rm{P(B) = }}\frac{{{\rm{\# of favorable outcomes in B }}}}{{{\rm{\# of outcomes in the sample space }}}}\\{\rm{ = }}\frac{{\rm{6}}}{{{\rm{36}}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{6}}}\\{\rm{P(C) = }}\frac{{{\rm{\# of favorable outcomes in C}}}}{{{\rm{\# of outcomes in the sample space }}}}\\{\rm{ = }}\frac{{\rm{6}}}{{{\rm{36}}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{6}}}\end{array}\)

We must determine whether these occurrences are pairwise independent, as well as whether all three events are mutually independent. The following definitions and propositions are required.

If for two events \({\rm{A}}\)and \({\rm{B}}\)stands

\({\rm{P(A}}\mid {\rm{B) = P(A)}}\)

We call them self-sufficient. Otherwise, they are reliant.

Property of Multiplication: Two events If and only if, \({\rm{A}}\) and \({\rm{B}}\) are independent.

\({\rm{P(A}} \cap {\rm{B) = P(A)*P(B)}}\)

Multiplication Property:

For events \({{\rm{A}}_{\rm{1}}}{\rm{,}}{{\rm{A}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{A}}_{\rm{n}}}{\rm{,n}} \in {\rm{N}}\) we say that they are mutually independent if

\({\rm{P}}\left( {{A_{{i_1}}} \cap {A_{{i_2}}} \ldots {A_{{i_k}}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{{{\rm{i}}_{\rm{1}}}}}} \right){\rm{*P}}\left( {{{\rm{A}}_{{{\rm{i}}_{\rm{2}}}}}} \right){\rm{* \ldots *P}}\left( {{{\rm{A}}_{{{\rm{i}}_{\rm{k}}}}}} \right)\)

for every \({\rm{k}} \in {\rm{\{ 2,3, \ldots ,n\} }}\), and every subset of indices \({{\rm{i}}_{\rm{1}}}{\rm{,}}{{\rm{i}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{i}}_{\rm{k}}}\). First, notice the following intersections.

\({\rm{A}} \cap {\rm{B = \{ (3,4)\} }}\)

\(\begin{array}{c}A \cap {\rm{B = \{ (3,4)\} }};\\A \cap {\rm{C = \{ (3,4)\} }};\\B \cap {\rm{C = \{ (3,4)\} }};\\A \cap B \cap {\rm{C = \{ (3,4)\} }}.\end{array}\)

We have that

\(\begin{array}{c}{\rm{P(A}} \cap {\rm{B) = P(A}} \cap C)\\ = P(B \cap {\rm{C) = P(A}} \cap B \cap {\rm{C) = }}\frac{{\rm{1}}}{{{\rm{36}}}}\end{array}\)

The following is true

\(\begin{array}{l}{\rm{P(A)*P(B) = }}\frac{{\rm{1}}}{{\rm{6}}}*\frac{{\rm{1}}}{{\rm{6}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{36}}}}{\rm{ = }}P(A \cap B)\\{\rm{P(A)*P(C) = }}\frac{{\rm{1}}}{{\rm{6}}}*\frac{{\rm{1}}}{{\rm{6}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{36}}}}{\rm{ = }}P(A \cap C)\\{\rm{P(B)*P(C) = }}\frac{{\rm{1}}}{{\rm{6}}}*\frac{{\rm{1}}}{{\rm{6}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{36}}}}{\rm{ = }}P(B \cap C)\end{array}\)

The events are pairwise independent, according to the multiplication rule.

It is not enough, however, for all three to be mutually independent. We still need to establish one more item from the multiplication rule (for \({\rm{n}}\)events) in order for all three events to be mutually independent.

Looking at

\(\begin{array}{c}{\rm{P(A)*P(B)*P(C) = }}\frac{{\rm{1}}}{{\rm{6}}}*\frac{{\rm{1}}}{{\rm{6}}}*\frac{{\rm{1}}}{{\rm{6}}}\\{\rm{ = }}\frac{{\rm{1}}}{{{\rm{216}}}}*\frac{{\rm{1}}}{{{\rm{36}}}}{\rm{ = P(A}} \cap {\rm{B}} \cap {\rm{C)}}\end{array}\)

Because they are uneven, events A,B, and C are not mutually independent, according to the multiplication rule for n events.

NOTE: if \({\rm{P(A)*P(B)*P(C) = P(A}} \cap {\rm{B}} \cap {\rm{C)}}\)it does not necessary mean that \({\rm{A,B}}\), and \({\rm{C}}\)are mutually independent, perhaps \({\rm{A}}\) and \({\rm{B}}\)are dependent.

Therefore, The Events are pairwise independent;

The Events are not mutually independent.