exercise:B27c69717b: 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> (Suggested by Eisenberg and Ghosh<ref group="Notes" >B. Eisenberg and B. K. Ghosh, “Independent Events in a Discrete Uniform Probability Space,” ''The American Statistician,'' vol. 41, no. 1 (1987), pp. 52--56.</ref>) A deck of playing cards...") |
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<math>(\heartsuit,r)</math> where <math>r</math> is any rank. Similarly, a ''rank event'' is any | <math>(\heartsuit,r)</math> where <math>r</math> is any rank. Similarly, a ''rank event'' is any | ||
event described in terms of rank alone. | event described in terms of rank alone. | ||
<ul><li> Show that if <math>A</math> is any suit event and <math>B</math> any rank event, then <math>A</math> | <ul style="list-style-type:lower-alpha"><li> Show that if <math>A</math> is any suit event and <math>B</math> any rank event, then <math>A</math> | ||
and <math>B</math> are ''independent.'' (We can express this briefly by saying that | and <math>B</math> are ''independent.'' (We can express this briefly by saying that | ||
suit and rank are independent.) | suit and rank are independent.) | ||
| Line 49: | Line 49: | ||
</li> | </li> | ||
</ul> | </ul> | ||
'''Notes''' | '''Notes''' | ||
{{Reflist|group=Notes}} | {{Reflist|group=Notes}} | ||
Latest revision as of 01:04, 13 June 2024
[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]
(Suggested by Eisenberg and Ghosh[Notes 1]) A deck of playing cards can be described as a Cartesian product
[[math]]
\mbox{Deck} = \mbox{Suit} \times \mbox{Rank}\ ,
[[/math]]
where [math]\mbox{Suit} = \{\clubsuit,\diamondsuit,\heartsuit,\spadesuit\}[/math] and [math]\mbox{Rank} = \{2,3,\dots,10,{\mbox J},{\mbox Q},{\mbox K},{\mbox A}\}[/math]. This just means that every card may be thought of as an ordered pair like [math](\diamondsuit,2)[/math]. By a suit event we mean any event [math]A[/math] contained in Deck which is described in terms of Suit alone. For instance, if [math]A[/math] is “the suit is red,” then
[[math]]
A = \{\diamondsuit,\heartsuit\} \times \mbox{Rank}\ ,
[[/math]]
so that [math]A[/math] consists of all cards of the form [math](\diamondsuit,r)[/math] or [math](\heartsuit,r)[/math] where [math]r[/math] is any rank. Similarly, a rank event is any event described in terms of rank alone.
- Show that if [math]A[/math] is any suit event and [math]B[/math] any rank event, then [math]A[/math] and [math]B[/math] are independent. (We can express this briefly by saying that suit and rank are independent.)
- Throw away the ace of spades. Show that now no nontrivial (i.e.,
neither empty nor the whole space) suit event [math]A[/math] is independent of any
nontrivial rank event [math]B[/math]. Hint: Here independence comes down to
[[math]] c/51 = (a/51) \cdot (b/51)\ , [[/math]]where [math]a[/math], [math]b[/math], [math]c[/math] are the respective sizes of [math]A[/math], [math]B[/math] and [math]A \cap B[/math]. It follows that 51 must divide [math]ab[/math], hence that 3 must divide one of [math]a[/math] and [math]b[/math], and 17 the other. But the possible sizes for suit and rank events preclude this.
- Show that the deck in (b) nevertheless does have pairs [math]A[/math], [math]B[/math] of nontrivial independent events. Hint: Find 2 events [math]A[/math] and [math]B[/math] of sizes 3 and 17, respectively, which intersect in a single point.
- Add a joker to a full deck. Show that now there is no pair [math]A[/math], [math]B[/math] of nontrivial independent events. Hint: See the hint in (b); 53 is prime.
Notes