Revision as of 02:17, 9 June 2024 by Bot (Created page with "<div class="d-none"><math> \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}</math></div> Let <math>A_1</math>, <math>A_2</math>, and <math>A_3</math> be events, and let <math>B_i</math> represent either <math>A_i</math> or its complement <math>\tilde A_i</math>. Then there are eight possible choices for the triple <math>(B_1, B_2, B_...")
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Jun 09'24

Exercise

[math] \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}[/math]

Let [math]A_1[/math], [math]A_2[/math], and [math]A_3[/math] be events, and let [math]B_i[/math] represent

either [math]A_i[/math] or its complement [math]\tilde A_i[/math]. Then there are eight possible choices for the triple [math](B_1, B_2, B_3)[/math]. Prove that the events [math]A_1[/math], [math]A_2[/math], [math]A_3[/math] are independent if and only if

[[math]] P(B_1 \cap B_2 \cap B_3) = P(B_1)P(B_2)P(B_3)\ , [[/math]]

for all eight of the possible choices for the triple [math](B_1, B_2, B_3)[/math].