{"id":703,"date":"2023-12-18T18:29:04","date_gmt":"2023-12-18T18:29:04","guid":{"rendered":"https:\/\/pressbooks.library.torontomu.ca\/bearguide\/?post_type=chapter&p=703"},"modified":"2023-12-18T20:12:36","modified_gmt":"2023-12-18T20:12:36","slug":"formula-test","status":"publish","type":"chapter","link":"https:\/\/pressbooks.library.torontomu.ca\/bearguide\/chapter\/formula-test\/","title":{"raw":"Formula Test","rendered":"Formula Test"},"content":{"raw":"[latexpage]\r\n\\begin{tikzpicture}\r\n[+preamble]\r\n\\usepackage{tikz}\r\n\\usepackage{pgfplots}\r\n\\pgfplotsset{compat=newest}\r\n[\/preamble]\r\n\\begin{axis}\r\n\\addplot3[surf,domain=0:360,samples=40] {cos(x)*cos(y)};\r\n\\end{axis}\r\n\\end{tikzpicture}\r\n
<\/div>\r\n
<\/div>\r\nExample 7.3.5\r\n\r\nA person has four keys and only one key fits to the lock of a door. What is the probability that the locked door can be unlocked in at most three tries?\r\nSolution\r\n\r\nLet U be the event that the door has been unlocked and L be the event that the door has not been unlocked. We illustrate with a tree diagram.\r\n\r\n\\documentclass{article}\r\n\\usepackage{tikz}\r\n\r\n\\begin{document}\r\n\\begin{tikzpicture}[grow=right, sloped]\r\n\\tikzstyle{level 1}=[level distance=2cm, sibling distance=4cm]\r\n\\tikzstyle{level 2}=[level distance=2cm, sibling distance=2cm]\r\n\\node {}\r\nchild {\r\nnode[label=right:\r\n{$L_1$}] {}\r\nedge from parent\r\nnode[below] {$3\/4$}\r\nchild {\r\nnode[label=right:\r\n{$L_2$}] {}\r\nedge from parent\r\nnode[below] {$2\/3$}\r\nchild {\r\nnode[label=right:\r\n{$L_3$}] {}\r\nedge from parent\r\nnode[below] {$1\/2$}\r\n}\r\nchild {\r\nnode[label=right:\r\n{$U_3$}] {}\r\nedge from parent\r\nnode[above] {$1\/2$}\r\n}\r\n}\r\nchild {\r\nnode[label=right:\r\n{$U_2$}] {}\r\nedge from parent\r\nnode[above] {$1\/3$}\r\n}\r\n}\r\nchild {\r\nnode[label=right:\r\n{$U_1$}] {}\r\nedge from parent\r\nnode[above] {$1\/4$}\r\n};\r\n\\end{tikzpicture}\r\n\\end{document}\r\n\r\nFirst Try:\r\n\r\n\\begin{tikzpicture}[grow=right, sloped]\r\n\\tikzstyle{level 1}=[level distance=2cm, sibling distance=3cm]\r\n\\node {}\r\nchild {\r\nnode[label=right:\r\n{$L$}] {}\r\nedge from parent\r\nnode[below] {$3\/4$}\r\n}\r\nchild {\r\nnode[label=right:\r\n{$U=1\/4$}] {}\r\nedge from parent\r\nnode[above] {$1\/4$}\r\n};\r\n\\end{tikzpicture}\r\n\r\nSecond Try:\r\n\r\n$\\begin{tikzpicture}[grow=right, sloped]\r\n\\tikzstyle{level 1}=[level distance=2cm, sibling distance=3cm]\r\n\\node {}\r\nchild {\r\nnode[label=right:\r\n{$L$}] {}\r\nedge from parent\r\nnode[below] {$2\/3$}\r\n}\r\nchild {\r\nnode[label=right:\r\n{$U=(3\/4)(1\/3)$}] {}\r\nedge from parent\r\nnode[above] {$1\/3$}\r\n};\r\n\\end{tikzpicture}$\r\n\r\nThird Try:\r\n\r\n\\begin{tikzpicture}[grow=right, sloped]\r\n\\tikzstyle{level 1}=[level distance=2cm, sibling distance=3cm]\r\n\\node {}\r\nchild {\r\nnode[label=right:\r\n{$L$}] {}\r\nedge from parent\r\nnode[below] {$1\/2$}\r\n}\r\nchild {\r\nnode[label=right:\r\n{$U=(3\/4)(2\/3)(1\/2)$}] {}\r\nedge from parent\r\nnode[above] {$1\/2$}\r\n};\r\n\\end{tikzpicture}\r\n\r\nThe probability of unlocking the door in the first try = 1\/4.\r\nThe probability of unlocking the door in the second try = (3\/4)(1\/3) = 1\/4.\r\nThe probability of unlocking the door in the third try = (3\/4)(2\/3)(1\/2) = 1\/4.\r\nTherefore, the probability of unlocking the door in at most three tries = 1\/4 + 1\/4 + 1\/4 = 3\/4.","rendered":"

[latexpage]
\n\\begin{tikzpicture}
\n[+preamble]
\n\\usepackage{tikz}
\n\\usepackage{pgfplots}
\n\\pgfplotsset{compat=newest}
\n[\/preamble]
\n\\begin{axis}
\n\\addplot3[surf,domain=0:360,samples=40] {cos(x)*cos(y)};
\n\\end{axis}
\n\\end{tikzpicture}<\/p>\n

<\/div>\n
<\/div>\n

Example 7.3.5<\/p>\n

A person has four keys and only one key fits to the lock of a door. What is the probability that the locked door can be unlocked in at most three tries?
\nSolution<\/p>\n

Let U be the event that the door has been unlocked and L be the event that the door has not been unlocked. We illustrate with a tree diagram.<\/p>\n

\\documentclass{article}
\n\\usepackage{tikz}<\/p>\n

\\begin{document}
\n\\begin{tikzpicture}[grow=right, sloped]
\n\\tikzstyle{level 1}=[level distance=2cm, sibling distance=4cm]
\n\\tikzstyle{level 2}=[level distance=2cm, sibling distance=2cm]
\n\\node {}
\nchild {
\nnode[label=right:
\n{$L_1$}] {}
\nedge from parent
\nnode[below] {$3\/4$}
\nchild {
\nnode[label=right:
\n{$L_2$}] {}
\nedge from parent
\nnode[below] {$2\/3$}
\nchild {
\nnode[label=right:
\n{$L_3$}] {}
\nedge from parent
\nnode[below] {$1\/2$}
\n}
\nchild {
\nnode[label=right:
\n{$U_3$}] {}
\nedge from parent
\nnode[above] {$1\/2$}
\n}
\n}
\nchild {
\nnode[label=right:
\n{$U_2$}] {}
\nedge from parent
\nnode[above] {$1\/3$}
\n}
\n}
\nchild {
\nnode[label=right:
\n{$U_1$}] {}
\nedge from parent
\nnode[above] {$1\/4$}
\n};
\n\\end{tikzpicture}
\n\\end{document}<\/p>\n

First Try:<\/p>\n

\\begin{tikzpicture}[grow=right, sloped]
\n\\tikzstyle{level 1}=[level distance=2cm, sibling distance=3cm]
\n\\node {}
\nchild {
\nnode[label=right:
\n{$L$}] {}
\nedge from parent
\nnode[below] {$3\/4$}
\n}
\nchild {
\nnode[label=right:
\n{$U=1\/4$}] {}
\nedge from parent
\nnode[above] {$1\/4$}
\n};
\n\\end{tikzpicture}<\/p>\n

Second Try:<\/p>\n

$\\begin{tikzpicture}[grow=right, sloped]
\n\\tikzstyle{level 1}=[level distance=2cm, sibling distance=3cm]
\n\\node {}
\nchild {
\nnode[label=right:
\n{$L$}] {}
\nedge from parent
\nnode[below] {$2\/3$}
\n}
\nchild {
\nnode[label=right:
\n{$U=(3\/4)(1\/3)$}] {}
\nedge from parent
\nnode[above] {$1\/3$}
\n};
\n\\end{tikzpicture}$<\/p>\n

Third Try:<\/p>\n

\\begin{tikzpicture}[grow=right, sloped]
\n\\tikzstyle{level 1}=[level distance=2cm, sibling distance=3cm]
\n\\node {}
\nchild {
\nnode[label=right:
\n{$L$}] {}
\nedge from parent
\nnode[below] {$1\/2$}
\n}
\nchild {
\nnode[label=right:
\n{$U=(3\/4)(2\/3)(1\/2)$}] {}
\nedge from parent
\nnode[above] {$1\/2$}
\n};
\n\\end{tikzpicture}<\/p>\n

The probability of unlocking the door in the first try = 1\/4.
\nThe probability of unlocking the door in the second try = (3\/4)(1\/3) = 1\/4.
\nThe probability of unlocking the door in the third try = (3\/4)(2\/3)(1\/2) = 1\/4.
\nTherefore, the probability of unlocking the door in at most three tries = 1\/4 + 1\/4 + 1\/4 = 3\/4.<\/p>\n","protected":false},"author":2,"menu_order":6,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-703","chapter","type-chapter","status-publish","hentry"],"part":115,"_links":{"self":[{"href":"https:\/\/pressbooks.library.torontomu.ca\/bearguide\/wp-json\/pressbooks\/v2\/chapters\/703","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.library.torontomu.ca\/bearguide\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.library.torontomu.ca\/bearguide\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.library.torontomu.ca\/bearguide\/wp-json\/wp\/v2\/users\/2"}],"version-history":[{"count":3,"href":"https:\/\/pressbooks.library.torontomu.ca\/bearguide\/wp-json\/pressbooks\/v2\/chapters\/703\/revisions"}],"predecessor-version":[{"id":705,"href":"https:\/\/pressbooks.library.torontomu.ca\/bearguide\/wp-json\/pressbooks\/v2\/chapters\/703\/revisions\/705"}],"part":[{"href":"https:\/\/pressbooks.library.torontomu.ca\/bearguide\/wp-json\/pressbooks\/v2\/parts\/115"}],"metadata":[{"href":"https:\/\/pressbooks.library.torontomu.ca\/bearguide\/wp-json\/pressbooks\/v2\/chapters\/703\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.library.torontomu.ca\/bearguide\/wp-json\/wp\/v2\/media?parent=703"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.library.torontomu.ca\/bearguide\/wp-json\/pressbooks\/v2\/chapter-type?post=703"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.library.torontomu.ca\/bearguide\/wp-json\/wp\/v2\/contributor?post=703"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.library.torontomu.ca\/bearguide\/wp-json\/wp\/v2\/license?post=703"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}