exercise:77b1769c9e: Difference between revisions

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(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> \begin{sloppypar} Given that <math>P(X = a) = r</math>, <math>P(\max(X,Y) = a) = s</math>, and <math>P(\min(X,Y) = a) = t</math>, show that you can determine <math>u = P(Y = a)</math> in terms of <math>r</math>, <math>s</math>, and <math>t</math...")
 
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<div class="d-none"><math>
Given that <math>P(X = a) = r</math>, <math>P(\max(X,Y) = a) = s</math>,and  <math>P(\min(X,Y) = a) = t</math>, show that you can determine <math>u = P(Y = a)</math> in terms of <math>r</math>, <math>s</math>, and <math>t</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> \begin{sloppypar} Given that <math>P(X = a) = r</math>, <math>P(\max(X,Y) = a) = s</math>,
and   
<math>P(\min(X,Y) = a) = t</math>, show that you can determine <math>u = P(Y = a)</math> in terms of <math>r</math>, <math>s</math>,
and <math>t</math>.\end{sloppypar}

Latest revision as of 00:05, 13 June 2024

Given that [math]P(X = a) = r[/math], [math]P(\max(X,Y) = a) = s[/math],and [math]P(\min(X,Y) = a) = t[/math], show that you can determine [math]u = P(Y = a)[/math] in terms of [math]r[/math], [math]s[/math], and [math]t[/math].