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Consider independently rolling two fair dice, one red and the other green. Let A be the event that the red die shows \({\rm{3}}\) dots, B be the event that the green die shows \({\rm{4}}\) dots, and C be the event that the total number of dots showing on the two dice is \({\rm{7}}\). Are these events pairwise independent (i.e., are \({\rm{A}}\) and \({\rm{B}}\) independent events, are \({\rm{A}}\) and \({\rm{C}}\) independent, and are \({\rm{B}}\) and \({\rm{C}}\) independent)? Are the three events mutually independent?

Short Answer

Expert verified

The Events are pairwise independent;

The Events are not mutually independent.

Step by step solution

01

Definition of Independent probability

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.

02

Are these events pairwise independent

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 \({{\rm{n}}_{_{\rm{1}}}}{{\rm{n}}_{\rm{2}}}\).

03

Explain the pairwise independent

Therefore,

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

04

Finding the three events mutually independent?

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

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

05

Explanation of the solution

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.

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