{"id":6945,"date":"2011-09-29T17:52:45","date_gmt":"2011-09-29T16:52:45","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=6945"},"modified":"2022-01-16T02:52:49","modified_gmt":"2022-01-16T02:52:49","slug":"no-lancamento-de-dois-dados","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=6945","title":{"rendered":"No lan\u00e7amento de dois dados"},"content":{"rendered":"<p><ul id='GTTabs_ul_6945' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_6945' class='GTTabs_curr'><a  id=\"6945_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_6945' ><a  id=\"6945_1\" onMouseOver=\"GTTabsShowLinks('Nota Pr\u00e9via'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Nota Pr\u00e9via<\/a><\/li>\n<li id='GTTabs_li_2_6945' ><a  id=\"6945_2\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<li id='GTTabs_li_3_6945' ><a  id=\"6945_3\" onMouseOver=\"GTTabsShowLinks('Diagrama'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Diagrama<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_6945'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/dadosAV.jpg\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"6947\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=6947\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/dadosAV.jpg\" data-orig-size=\"315,198\" 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=\"Dados\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/dadosAV.jpg\" class=\"alignright size-medium wp-image-6947\" title=\"Dados\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/dadosAV-300x188.jpg\" alt=\"\" width=\"210\" height=\"132\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/dadosAV-300x188.jpg 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/dadosAV-150x94.jpg 150w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/dadosAV.jpg 315w\" sizes=\"auto, (max-width: 210px) 100vw, 210px\" \/><\/a>No lan\u00e7amento de dois dados, um azul e outro vermelho, qual a probabilidade de o produto dos pontos pbtidos ser:<\/p>\n<ol>\n<li>7?<\/li>\n<li>1?<\/li>\n<li>maior que 12?<\/li>\n<li>um n\u00famero par?<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_6945' onClick='GTTabs_show(1,6945)'>Nota Pr\u00e9via &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_6945'>\n<span class='GTTabs_titles'><b>Nota Pr\u00e9via<\/b><\/span><!--more--><\/p>\n<p>Naturalmente, estamos a pensar resolver o problema recorrendo \u00e0 Lei de Laplace. No entanto, devemos ter em aten\u00e7\u00e3o que \u00e9 indispens\u00e1vel que os\u00a0acontecimentos que integram o espa\u00e7o de acontecimentos sejam equiprov\u00e1veis.<\/p>\n<p>Ser\u00e1 tentador considerar resultados poss\u00edveis da experi\u00eancia aleat\u00f3ria descrita os seguintes 18 resultados:<\/p>\n<p>\\[1,\\,2,\\,3,\\,4,\\,5,\\,6,\\,8,\\,9,\\,10,\\,12,\\,15,\\,16,\\,18,\\,20,\\,24,\\,25,\\,30,\\,36.\\]<\/p>\n<p>Isto \u00e9, os 18 produtos poss\u00edveis da multiplica\u00e7\u00e3o de dois quaisquer elementos do conjunto $\\left\\{ 1,2,3,4,5,6 \\right\\}$.<\/p>\n<p>Desta forma, considerando para espa\u00e7o de acontecimentos o conjunto<\/p>\n<p>\\[S=\\left\\{ 1,\\,2,\\,3,\\,4,\\,5,\\,6,\\,8,\\,9,\\,10,\\,12,\\,15,\\,16,\\,18,\\,20,\\,24,\\,25,\\,30,\\,36 \\right\\}\\]<\/p>\n<p>ser\u00e1 tentador responder:<\/p>\n<ul>\n<li>$P(\\text{&#8220;produto dos pontos ser 1&#8221;})=\\frac{1}{18}$<\/li>\n<li>$P(\\text{&#8220;produto dos pontos ser 2&#8221;})=\\frac{1}{18}$<\/li>\n<\/ul>\n<p>No entanto, a primeira resposta est\u00e1 ERRADA; a segunda est\u00e1 certa, mas resulta apenas de mera coincid\u00eancia.<\/p>\n<p>O erro adv\u00e9m de se ter admitido que os acontecimentos considerados no conjunto S s\u00e3o equiprov\u00e1veis, o que n\u00e3o corresponde \u00e0 verdade.<\/p>\n<p>Por exemplo, os acontecimentos &#8220;o produto dos pontos ser 1&#8221; e &#8220;o produto dos pontos ser 2&#8221; n\u00e3o t\u00eam a mesma probabilidade, ali\u00e1s o primeiro tem metade da probabilidade do segundo.<\/p>\n<p>Para entender esta conclus\u00e3o, basta reparar que:<\/p>\n<ul>\n<li>o 1.\u00ba acontecimento apenas pode ser obtido de <span style=\"text-decoration: underline;\">uma maneira<\/span>: 1 ponto no dado azul e 1 ponto no dado vermelho &#8211; <span style=\"color: #ffffff; font-size: large;\"><span style=\"background-color: #0000ff;\">1<\/span><span style=\"background-color: #ff0000;\">1<\/span><\/span>;<\/li>\n<li>o 2.\u00ba acontecimento pode ser obtido de <span style=\"text-decoration: underline;\">duas maneiras<\/span>: 1 ponto no dado azul e\u00a02 pontos no dado vermelho, ou 2 pontos no dado azul e 1 ponto no dado vermelho &#8211; <span style=\"color: #ffffff; font-size: large;\"><span style=\"background-color: #0000ff;\">1<\/span><span style=\"background-color: #ff0000;\">2<\/span><\/span> ou <span style=\"color: #ffffff; font-size: large;\"><span style=\"background-color: #0000ff;\">2<\/span><span style=\"background-color: #ff0000;\">1<\/span><\/span>.<\/li>\n<\/ul>\n<p>\u00a0<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_6945' onClick='GTTabs_show(0,6945)'>&lt;&lt; Enunciado<\/a><\/span><span class='GTTabs_nav_next'><a href='#GTTabs_ul_6945' onClick='GTTabs_show(2,6945)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_2_6945'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><\/p>\n<p style=\"text-align: center;\"><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/dadosAV.jpg\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"6947\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=6947\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/dadosAV.jpg\" data-orig-size=\"315,198\" 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=\"Dados\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/dadosAV.jpg\" class=\"aligncenter size-medium wp-image-6947\" title=\"Dados\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/dadosAV-300x188.jpg\" alt=\"\" width=\"180\" height=\"113\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/dadosAV-300x188.jpg 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/dadosAV-150x94.jpg 150w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/dadosAV.jpg 315w\" sizes=\"auto, (max-width: 180px) 100vw, 180px\" \/><\/a><\/p>\n<p>Para garantir a igualdade de probabilidade dos resultados da experi\u00eancia aleat\u00f3ria, \u00e9 indispens\u00e1vel assegurar a distin\u00e7\u00e3o dos pontos obtidos face \u00e0 cor do dado.<\/p>\n<p>Nesse prop\u00f3sito, podemos construir uma tabela de dupla entrada ou um diagrama de \u00e1rvore (ver diagrama na sec\u00e7\u00e3o seguinte):<\/p>\n<table class=\"aligncenter\" style=\"width: 50%;\" border=\"1\" align=\"center\">\n<tbody>\n<tr>\n<td><\/td>\n<td style=\"text-align: center;\"><span style=\"background-color: #ff0000;\"><strong><span style=\"color: #ffffff;\">\u00a0 1 \u00a0<\/span><\/strong><\/span><\/td>\n<td style=\"text-align: center;\"><span style=\"background-color: #ff0000;\"><strong><span style=\"color: #ffffff;\">\u00a0 2 \u00a0<\/span><\/strong><\/span><\/td>\n<td style=\"text-align: center;\"><span style=\"background-color: #ff0000;\"><strong><span style=\"color: #ffffff;\">\u00a0 3 \u00a0<\/span><\/strong><\/span><\/td>\n<td style=\"text-align: center;\"><span style=\"background-color: #ff0000;\"><strong><span style=\"color: #ffffff;\">\u00a0 4 \u00a0<\/span><\/strong><\/span><\/td>\n<td style=\"text-align: center;\"><span style=\"background-color: #ff0000;\"><strong><span style=\"color: #ffffff;\">\u00a0 5 \u00a0<\/span><\/strong><\/span><\/td>\n<td style=\"text-align: center;\"><span style=\"background-color: #ff0000;\"><strong><span style=\"color: #ffffff;\">\u00a0 6\u00a0\u00a0<\/span><\/strong><\/span><\/td>\n<\/tr>\n<tr>\n<td><span style=\"background-color: #0000ff;\"><strong><span style=\"color: #ffffff;\">\u00a0 1\u00a0\u00a0<\/span><\/strong><\/span><\/td>\n<td>(1,1)<\/td>\n<td>(1,2)<\/td>\n<td>(1,3)<\/td>\n<td>(1,4)<\/td>\n<td>(1,5)<\/td>\n<td>(1,6)<\/td>\n<\/tr>\n<tr>\n<td><span style=\"background-color: #0000ff;\"><strong><span style=\"color: #ffffff;\">\u00a0 2\u00a0 <\/span><\/strong><\/span><\/td>\n<td>(2,1)<\/td>\n<td>(2,2)<\/td>\n<td>(2,3)<\/td>\n<td>(2,4)<\/td>\n<td>(2,5)<\/td>\n<td>(2,6)<\/td>\n<\/tr>\n<tr>\n<td><span style=\"background-color: #0000ff;\"><strong><span style=\"color: #ffffff;\">\u00a0 3\u00a0 <\/span><\/strong><\/span><\/td>\n<td>(3,1)<\/td>\n<td>(3,2)<\/td>\n<td>(3,3)<\/td>\n<td>(3,4)<\/td>\n<td>(3,5)<\/td>\n<td>(3,6)<\/td>\n<\/tr>\n<tr>\n<td><span style=\"background-color: #0000ff;\"><strong><span style=\"color: #ffffff;\">\u00a0 4\u00a0 <\/span><\/strong><\/span><\/td>\n<td>(4,1)<\/td>\n<td>(4,2)<\/td>\n<td>(4,3)<\/td>\n<td>(4,4)<\/td>\n<td>(4,5)<\/td>\n<td>(4,6)<\/td>\n<\/tr>\n<tr>\n<td><span style=\"background-color: #0000ff;\"><strong><span style=\"color: #ffffff;\">\u00a0 5\u00a0 <\/span><\/strong><\/span><\/td>\n<td>(5,1)<\/td>\n<td>(5,2)<\/td>\n<td>(5,3)<\/td>\n<td>(5,4)<\/td>\n<td>(5,5)<\/td>\n<td>(5,6)<\/td>\n<\/tr>\n<tr>\n<td><span style=\"background-color: #0000ff;\"><strong><span style=\"color: #ffffff;\">\u00a0 6\u00a0 <\/span><\/strong><\/span><\/td>\n<td>(6,1)<\/td>\n<td>(6,2)<\/td>\n<td>(6,3)<\/td>\n<td>(6,4)<\/td>\n<td>(6,5)<\/td>\n<td>(6,6)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Assim, o espa\u00e7o de acontecimentos da experi\u00eancia aleat\u00f3ria considerada \u00e9 constitu\u00eddo pelos 36 pares ordenados acima determinados. Logo, o n\u00famero de casos poss\u00edveis \u00e9 36, isto \u00e9, $NCP=36$.<\/p>\n<p>Para facilitar a identifica\u00e7\u00e3o dos casos favor\u00e1veis dos acontecimentos das diferentes al\u00edneas, \u00e9 \u00fatil substituir na tabela acima cada um dos pares ordenados pelo produto obtido entre os dois elementos que os constituem:<\/p>\n<table class=\"aligncenter\" style=\"width: 50%;\" border=\"1\" align=\"center\">\n<tbody>\n<tr>\n<td><\/td>\n<td><span style=\"background-color: #ff0000;\"><strong><span style=\"color: #ffffff;\">\u00a0 1 \u00a0<\/span><\/strong><\/span><\/td>\n<td><span style=\"background-color: #ff0000;\"><strong><span style=\"color: #ffffff;\">\u00a0 2 \u00a0<\/span><\/strong><\/span><\/td>\n<td><span style=\"background-color: #ff0000;\"><strong><span style=\"color: #ffffff;\">\u00a0 3 \u00a0<\/span><\/strong><\/span><\/td>\n<td><span style=\"background-color: #ff0000;\"><strong><span style=\"color: #ffffff;\">\u00a0 4 \u00a0<\/span><\/strong><\/span><\/td>\n<td><span style=\"background-color: #ff0000;\"><strong><span style=\"color: #ffffff;\">\u00a0 5 \u00a0<\/span><\/strong><\/span><\/td>\n<td><span style=\"background-color: #ff0000;\"><strong><span style=\"color: #ffffff;\">\u00a0 6\u00a0\u00a0<\/span><\/strong><\/span><\/td>\n<\/tr>\n<tr>\n<td><span style=\"background-color: #0000ff;\"><strong><span style=\"color: #ffffff;\">\u00a0 1\u00a0\u00a0<\/span><\/strong><\/span><\/td>\n<td style=\"text-align: center;\">1<\/td>\n<td style=\"text-align: center;\">2<\/td>\n<td style=\"text-align: center;\">3<\/td>\n<td style=\"text-align: center;\">4<\/td>\n<td style=\"text-align: center;\">5<\/td>\n<td style=\"text-align: center;\">6<\/td>\n<\/tr>\n<tr>\n<td><span style=\"background-color: #0000ff;\"><strong><span style=\"color: #ffffff;\">\u00a0 2\u00a0 <\/span><\/strong><\/span><\/td>\n<td style=\"text-align: center;\">2<\/td>\n<td style=\"text-align: center;\">4<\/td>\n<td style=\"text-align: center;\">6<\/td>\n<td style=\"text-align: center;\">8<\/td>\n<td style=\"text-align: center;\">10<\/td>\n<td style=\"text-align: center;\">12<\/td>\n<\/tr>\n<tr>\n<td><span style=\"background-color: #0000ff;\"><strong><span style=\"color: #ffffff;\">\u00a0 3\u00a0 <\/span><\/strong><\/span><\/td>\n<td style=\"text-align: center;\">3<\/td>\n<td style=\"text-align: center;\">6<\/td>\n<td style=\"text-align: center;\">9<\/td>\n<td style=\"text-align: center;\">12<\/td>\n<td style=\"text-align: center;\">15<\/td>\n<td style=\"text-align: center;\">18<\/td>\n<\/tr>\n<tr>\n<td><span style=\"background-color: #0000ff;\"><strong><span style=\"color: #ffffff;\">\u00a0 4\u00a0 <\/span><\/strong><\/span><\/td>\n<td style=\"text-align: center;\">4<\/td>\n<td style=\"text-align: center;\">8<\/td>\n<td style=\"text-align: center;\">12<\/td>\n<td style=\"text-align: center;\">16<\/td>\n<td style=\"text-align: center;\">20<\/td>\n<td style=\"text-align: center;\">24<\/td>\n<\/tr>\n<tr>\n<td><span style=\"background-color: #0000ff;\"><strong><span style=\"color: #ffffff;\">\u00a0 5\u00a0 <\/span><\/strong><\/span><\/td>\n<td style=\"text-align: center;\">5<\/td>\n<td style=\"text-align: center;\">10<\/td>\n<td style=\"text-align: center;\">15<\/td>\n<td style=\"text-align: center;\">20<\/td>\n<td style=\"text-align: center;\">25<\/td>\n<td style=\"text-align: center;\">30<\/td>\n<\/tr>\n<tr>\n<td><span style=\"background-color: #0000ff;\"><strong><span style=\"color: #ffffff;\">\u00a0 6\u00a0 <\/span><\/strong><\/span><\/td>\n<td style=\"text-align: center;\">6<\/td>\n<td style=\"text-align: center;\">12<\/td>\n<td style=\"text-align: center;\">18<\/td>\n<td style=\"text-align: center;\">24<\/td>\n<td style=\"text-align: center;\">30<\/td>\n<td style=\"text-align: center;\">36<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ol>\n<li>O acontecimento A: &#8220;o produto dos pontos obtidos ser 7&#8221; \u00e9 um acontecimento imposs\u00edvel, logo a sua probabilidade \u00e9 nula.<br \/>\n\u00ad<\/li>\n<li>O conjunto dos resultados favor\u00e1veis ao acontecimento B: &#8220;o produto dos pontos obtidos ser 1&#8221; \u00e9 $B=\\left\\{ (1,1) \\right\\}$.<br \/>\nLogo, $NCF=\\#B=1$ e, portanto, $P(B)=\\frac{1}{36}$.<br \/>\n\u00ad<\/li>\n<li>Relativamente ao acontecimento C: &#8220;o produto dos pontos obtidos ser maior que 12&#8221;, h\u00e1 13 casos favor\u00e1veis. (Quais s\u00e3o?)<br \/>\nLogo, $P(C)=\\frac{13}{36}$.<br \/>\n\u00ad<\/li>\n<li>S\u00e3o 27 os casos favor\u00e1veis ao acontecimento D: &#8220;o produto dos pontos obtidos \u00e9 um n\u00famero par&#8221;. (Quais s\u00e3o?)<br \/>\nLogo, $P(D)=\\frac{27}{36}=\\frac{3}{4}$.<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_6945' onClick='GTTabs_show(1,6945)'>&lt;&lt; Nota Pr\u00e9via<\/a><\/span><span class='GTTabs_nav_next'><a href='#GTTabs_ul_6945' onClick='GTTabs_show(3,6945)'>Diagrama &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_3_6945'>\n<span class='GTTabs_titles'><b>Diagrama<\/b><\/span><\/p>\n<p style=\"text-align: center;\"><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/dadosAV.jpg\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"6947\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=6947\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/dadosAV.jpg\" data-orig-size=\"315,198\" 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=\"Dados\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/dadosAV.jpg\" class=\"aligncenter size-medium wp-image-6947\" title=\"Dados\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/dadosAV-300x188.jpg\" alt=\"\" width=\"180\" height=\"113\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/dadosAV-300x188.jpg 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/dadosAV-150x94.jpg 150w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/dadosAV.jpg 315w\" sizes=\"auto, (max-width: 180px) 100vw, 180px\" \/><\/a><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/2dadosAV.jpg\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"6948\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=6948\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/2dadosAV.jpg\" data-orig-size=\"580,839\" 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=\"Diagrama\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/2dadosAV.jpg\" class=\"aligncenter size-full wp-image-6948\" title=\"Diagrama\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/2dadosAV.jpg\" alt=\"\" width=\"580\" height=\"839\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/2dadosAV.jpg 580w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/2dadosAV-207x300.jpg 207w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/2dadosAV-103x150.jpg 103w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/09\/2dadosAV-400x578.jpg 400w\" sizes=\"auto, (max-width: 580px) 100vw, 580px\" \/><\/a><\/p><\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_6945' onClick='GTTabs_show(2,6945)'>&lt;&lt; Resolu\u00e7\u00e3o<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Nota Pr\u00e9via Enunciado No lan\u00e7amento de dois dados, um azul e outro vermelho, qual a probabilidade de o produto dos pontos pbtidos ser: 7? 1? maior que 12? um n\u00famero par? 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