{"id":7263,"date":"2011-12-06T19:27:19","date_gmt":"2011-12-06T19:27:19","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=7263"},"modified":"2022-01-26T01:34:42","modified_gmt":"2022-01-26T01:34:42","slug":"para-um-exame","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=7263","title":{"rendered":"Para um exame"},"content":{"rendered":"<p><ul id='GTTabs_ul_7263' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_7263' class='GTTabs_curr'><a  id=\"7263_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_7263' ><a  id=\"7263_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_7263'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/12\/study.jpg\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"7264\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=7264\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/12\/study.jpg\" data-orig-size=\"250,250\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}\" data-image-title=\"Exame\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/12\/study.jpg\" class=\"alignright size-full wp-image-7264\" title=\"Exame\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/12\/study.jpg\" alt=\"\" width=\"250\" height=\"250\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/12\/study.jpg 250w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/12\/study-150x150.jpg 150w\" sizes=\"auto, (max-width: 250px) 100vw, 250px\" \/><\/a>Para um exame, dez examinadores preparam, cada um, duas quest\u00f5es.<\/p>\n<p>As 20 quest\u00f5es s\u00e3o colocadas em envelopes id\u00eanticos.<\/p>\n<p>Apresentam-se dois candidatos e cada um escolheu, ao acaso, dois envelopes. Os envelopes escolhidos pelo primeiro candidato n\u00e3o ficam dispon\u00edveis para o segundo.<\/p>\n<p>Designe-se por:<\/p>\n<ul>\n<li>${{A}_{1}}$: &#8220;as duas quest\u00f5es obtidas pelo primeiro candidato prov\u00eam do mesmo examinador&#8221;;<\/li>\n<li>${{A}_{2}}$: &#8220;as duas quest\u00f5es obtidas pelo segundo candidato prov\u00eam do mesmo examinador&#8221;.<\/li>\n<\/ul>\n<p>Seja $\\overline{{{A}_{1}}}$ o acontecimento contr\u00e1rio de ${{A}_{1}}$.<\/p>\n<ol>\n<li>a) Mostre que a probabilidade de ${{A}_{1}}$ \u00e9 $\\frac{1}{19}$.\n<p>b) Calcule diretamente a probabilidade: $P({{A}_{2}}|{{A}_{1}})$.<\/p>\n<p>c) Mostre que a probabilidade de dois candidatos obterem cada um duas quest\u00f5es provenientes do mesmo examinador \u00e9 igual a $\\frac{1}{323}$.<\/p>\n<\/li>\n<li>a) Calcule $P({{A}_{2}}|\\overline{{{A}_{1}}})$.\n<p>b) Calcule $P({{A}_{2}})$ e depois mostre que $P({{A}_{1}}\\cup {{A}_{2}})=\\frac{33}{323}$.<\/p>\n<\/li>\n<li>Seja $X$ a vari\u00e1vel aleat\u00f3ria igual ao n\u00famero de candidatos que escolheram duas quest\u00f5es provenientes de um mesmo examinador.\n<p>a) Determine a lei de probabilidade da vari\u00e1vel aleat\u00f3ria $X$.<\/p>\n<p>b) Calcule o valor m\u00e9dio e o desvio padr\u00e3o desta distribui\u00e7\u00e3o.<\/p>\n<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_7263' onClick='GTTabs_show(1,7263)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_7263'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>a)<br \/>\nH\u00e1 vinte envelopes e o primeiro candidato escolhe dois deles.<br \/>\nLogo, o n\u00famero de casos poss\u00edveis \u00e9 $NCP={}^{20}{{C}_{2}}=190$ e o n\u00famero de casos favor\u00e1veis \u00e9 $NCF=10\\times {}^{2}{{C}_{2}}=10$.<br \/>\nPortanto, a probabilidade pedida \u00e9 $$P({{A}_{1}})=\\frac{10\\times {}^{2}{{C}_{2}}}{{}^{10}{{C}_{2}}}=\\frac{10}{190}=\\frac{1}{19}$$b)<br \/>\n$P({{A}_{2}}|{{A}_{1}})$ significa: a probabilidade de &#8220;as duas quest\u00f5es obtidas pelo segundo candidato prov\u00eam do mesmo examinador&#8221;, dado que &#8220;as duas quest\u00f5es obtidas pelo primeiro candidato prov\u00eam do mesmo examinador&#8221;.<br \/>\nSe o primeiro candidato\u00a0retira duas quest\u00f5es provenientes do mesmo examinador, sobram 18 quest\u00f5es, cada duas delas elaboradas por um de nove examinadores diferentes.<br \/>\nAssim, o segundo candidato pode obter as duas quest\u00f5es de $NCP={}^{18}{{C}_{2}}=153$ maneiras diferentes, sendo $NCF=9\\times {}^{2}{{C}_{2}}=9$ delas favor\u00e1veis a serem provenientes do mesmo examinador.<br \/>\nLogo, $$P({{A}_{2}}|{{A}_{1}})=\\frac{9\\times {}^{2}{{C}_{2}}}{{}^{18}{{C}_{2}}}=\\frac{9}{153}=\\frac{1}{17}$$<\/p>\n<p>c)<br \/>\nA probabilidade pedida \u00e9 $$P({{A}_{1}}\\cap {{A}_{2}})=P({{A}_{1}})\\times P({{A}_{2}}|{{A}_{1}})=\\frac{1}{19}\\times \\frac{1}{17}=\\frac{1}{323}$$<\/p>\n<\/li>\n<li>a)<br \/>\n(De forma an\u00e1loga \u00e0 resolu\u00e7\u00e3o de 1-b))<br \/>\nSe o primeiro candidato retira duas quest\u00f5es n\u00e3o provenientes do mesmo examinador, sobram 18 quest\u00f5es, mas apenas 8 pares delas s\u00e3o provenientes de um mesmo examinador.<br \/>\nAssim, $$P({{A}_{2}}|\\overline{{{A}_{1}}})=\\frac{8\\times {}^{2}{{C}_{2}}}{{}^{18}{{C}_{2}}}=\\frac{8}{153}$$b)<br \/>\nOra, $$\\begin{array}{*{35}{l}}<br \/>\nP({{A}_{2}}) &amp; = &amp; P({{A}_{1}}\\cap {{A}_{2}})+P(\\overline{{{A}_{1}}}\\cap {{A}_{2}})\u00a0 \\\\<br \/>\n{} &amp; = &amp; P({{A}_{1}}\\cap {{A}_{2}})+P(\\overline{{{A}_{1}}})\\times P({{A}_{2}}|\\overline{{{A}_{1}}})\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{1}{323}+(1-\\frac{1}{19})\\times \\frac{8}{153}\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{1}{323}+\\frac{18}{19}\\times \\frac{8}{153}\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{1}{19}\u00a0 \\\\<br \/>\n\\end{array}$$<br \/>\nLogo, $$\\begin{array}{*{35}{l}}<br \/>\nP({{A}_{1}}\\cup {{A}_{2}}) &amp; = &amp; P({{A}_{1}})+P({{A}_{2}})-P({{A}_{1}}\\cap {{A}_{2}})\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{1}{19}+\\frac{1}{19}-\\frac{1}{323}\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{33}{323}\u00a0 \\\\<br \/>\n\\end{array}$$<\/li>\n<li>a)<br \/>\nAo exame apresentam-se apenas dois candidatos.<br \/>\nOra, ambos podem retirar quest\u00f5es provenientes de examinadores diferentes; ou, apenas um deles retirar duas quest\u00f5es elaboradas pelo mesmo examinador; ou, por \u00faltimo, ambos retirararem um par de quest\u00f5es elaboradas pelo mesmo examinador.<br \/>\nLogo, a vari\u00e1vel aleat\u00f3ria $X$ pode assumir os valores: $0$, $1$ e $2$.<\/p>\n<p>Comecemos por anotar, numa tabela de dupla entrada, algumas das probabilidades conhecidas:<\/p>\n<table class=\" aligncenter\" style=\"width: 60%;\" border=\"0\" align=\"center\">\n<tbody>\n<tr>\n<td style=\"text-align: center; background-color: #ffe4b5; border: #a9a9a9 1px solid;\"><\/td>\n<td style=\"text-align: center; background-color: #ffe4b5; border: #a9a9a9 1px solid;\">${{A}_{2}}$<\/td>\n<td style=\"text-align: center; background-color: #ffe4b5; border: #a9a9a9 1px solid;\">$\\overline{{{A}_{2}}}$<\/td>\n<td style=\"text-align: center; background-color: #ffe4b5; border: #a9a9a9 1px solid;\"><strong>Total\u00a0<\/strong><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center; background-color: #ffe4b5; border: #a9a9a9 1px solid;\">${{A}_{1}}$<\/td>\n<td style=\"text-align: center; background-color: #ffe4b5; border: #a9a9a9 1px solid;\">$\\frac{1}{323}$<\/td>\n<td style=\"text-align: center; background-color: #ffe4b5; border: #a9a9a9 1px solid;\"><span style=\"color: #0000ff;\">$\\frac{16}{323}$<\/span><\/td>\n<td style=\"text-align: center; background-color: #ffe4b5; border: #a9a9a9 1px solid;\">$\\frac{1}{19}$<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center; background-color: #ffe4b5; border: #a9a9a9 1px solid;\">$\\overline{{{A}_{1}}}$<\/td>\n<td style=\"text-align: center; background-color: #ffe4b5; border: #a9a9a9 1px solid;\"><span style=\"color: #0000ff;\">$\\frac{16}{323}$<\/span><\/td>\n<td style=\"text-align: center; background-color: #ffe4b5; border: #a9a9a9 1px solid;\"><\/td>\n<td style=\"text-align: center; background-color: #ffe4b5; border: #a9a9a9 1px solid;\"><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center; background-color: #ffe4b5; border: #a9a9a9 1px solid;\"><strong>Total<\/strong><\/td>\n<td style=\"text-align: center; background-color: #ffe4b5; border: #a9a9a9 1px solid;\">$\\frac{1}{19}$<\/td>\n<td style=\"text-align: center; background-color: #ffe4b5; border: #a9a9a9 1px solid;\"><\/td>\n<td style=\"text-align: center; background-color: #ffe4b5; border: #a9a9a9 1px solid;\"><strong>$1$<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Entretanto\u00a0(ser\u00e1 \u00fatil mais \u00e0 frente), temos:<span style=\"color: #0000ff;\">$$P(\\overline{{{A}_{1}}}\\cap {{A}_{2}})=P({{A}_{1}}\\cap \\overline{{{A}_{2}}})=\\frac{1}{19}-\\frac{1}{323}=\\frac{16}{323}$$<\/span><\/p>\n<p>C\u00e1lculo das probabilidades:<\/p>\n<p>$$P(X=0)=P(\\overline{{{A}_{1}}}\\cap \\overline{{{A}_{2}}})=P(\\overline{{{A}_{1}}\\cup {{A}_{2}}})=1-P({{A}_{1}}\\cup {{A}_{2}})=1-\\frac{33}{323}=\\frac{290}{323}$$<\/p>\n<p>$$P(X=1)=P({{A}_{1}}\\cap \\overline{{{A}_{2}}})+P(\\overline{{{A}_{1}}}\\cap {{A}_{2}})=\\frac{16}{323}+\\frac{16}{323}=\\frac{32}{323}$$<\/p>\n<p>$$P(X=2)=P({{A}_{1}}\\cap {{A}_{2}})=\\frac{1}{323}$$<\/p>\n<p>Logo, a distribui\u00e7\u00e3o de probabilidades da vari\u00e1vel aleat\u00f3ria \u00e9:<\/p>\n<table class=\" aligncenter\" style=\"width: 60%;\" border=\"0\" align=\"center\">\n<tbody>\n<tr>\n<td style=\"text-align: center; background-color: #e0ffff; border: #00008b 1px solid;\">${{x}_{i}}$<\/td>\n<td style=\"text-align: center; background-color: #e0ffff; border: #00008b 1px solid;\">$0$<\/td>\n<td style=\"text-align: center; background-color: #e0ffff; border: #00008b 1px solid;\">$1$<\/td>\n<td style=\"text-align: center; background-color: #e0ffff; border: #00008b 1px solid;\">$2$<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center; background-color: #e0ffff; border: #00008b 1px solid;\">$P(X={{x}_{i}})$<\/td>\n<td style=\"text-align: center; background-color: #e0ffff; border: #00008b 1px solid;\">$\\frac{290}{323}$<\/td>\n<td style=\"text-align: center; background-color: #e0ffff; border: #00008b 1px solid;\">$\\frac{32}{323}$<\/td>\n<td style=\"text-align: center; background-color: #e0ffff; border: #00008b 1px solid;\">$\\frac{1}{323}$<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>b)<br \/>\nO valor m\u00e9dio e o desvio padr\u00e3o desta distribui\u00e7\u00e3o s\u00e3o, respectivamente:<br \/>\n$$\\mu =\\frac{290}{323}\\times 0+\\frac{32}{323}\\times 1+\\frac{1}{323}\\times 2=\\frac{34}{323}=\\frac{2}{19}$$<br \/>\n$$\\sigma =\\sqrt{\\frac{290}{323}\\times {{\\left( 0-\\frac{2}{19} \\right)}^{2}}+\\frac{32}{323}\\times {{\\left( 1-\\frac{2}{19} \\right)}^{2}}+\\frac{1}{323}\\times {{\\left( 2-\\frac{2}{19} \\right)}^{2}}}=\\sqrt{\\frac{11704}{116603}}=\\sqrt{\\frac{616}{6137}}=\\frac{2\\sqrt{2618}}{323}\\approx 0,32$$<\/p>\n<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_7263' onClick='GTTabs_show(0,7263)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Para um exame, dez examinadores preparam, cada um, duas quest\u00f5es. As 20 quest\u00f5es s\u00e3o colocadas em envelopes id\u00eanticos. Apresentam-se dois candidatos e cada um escolheu, ao acaso, dois envelopes. Os envelopes&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":21070,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[226,97,227],"tags":[427,255,245,215,235],"series":[],"class_list":["post-7263","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-12--ano","category-aplicando","category-probabilidades-e-combinatoria","tag-12-o-ano","tag-analise-combinatoria","tag-distribuicao-de-probabilidades","tag-probabilidade","tag-probabilidade-condicionada"],"views":2797,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/12\/12V1Pag182-75_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7263","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=7263"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7263\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/21070"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=7263"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=7263"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=7263"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=7263"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}