exercise:90ea19f154: 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> It is desired to find the probability that in a bridge deal each player receives an ace. A student argues as follows. It does not matter where the first ace goes. The second ace must go to one of the other three players and this occurs with pro...")
 
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<div class="d-none"><math>
It is desired to find the probability that in a bridge deal each player receives an ace.  A student argues as follows.  It does not matter where the
\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> It is desired to find the probability that in a bridge deal each player
receives an ace.  A student argues as follows.  It does not matter where the
first ace goes.  The second ace must go to one of the other three players and
first ace goes.  The second ace must go to one of the other three players and
this occurs with probability 3/4.  Then the next must go to one of two, an
this occurs with probability 3/4.  Then the next must go to one of two, an

Latest revision as of 23:46, 12 June 2024

It is desired to find the probability that in a bridge deal each player receives an ace. A student argues as follows. It does not matter where the first ace goes. The second ace must go to one of the other three players and this occurs with probability 3/4. Then the next must go to one of two, an event of probability 1/2, and finally the last ace must go to the player who does not have an ace. This occurs with probability 1/4. The probability that all these events occur is the product [math](3/4)(1/2)(1/4) = 3/32[/math]. Is this argument correct?