{"id":7208,"date":"2011-11-27T17:02:56","date_gmt":"2011-11-27T17:02:56","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=7208"},"modified":"2021-12-28T13:06:39","modified_gmt":"2021-12-28T13:06:39","slug":"uma-urna-com-10-cartoes","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=7208","title":{"rendered":"Uma urna com 10 cart\u00f5es"},"content":{"rendered":"<p><ul id='GTTabs_ul_7208' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_7208' class='GTTabs_curr'><a  id=\"7208_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_7208' ><a  id=\"7208_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_7208'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Numa urna colocaram-se 10 cart\u00f5es indistingu\u00edveis ao tato.<\/p>\n<p>Cada cart\u00e3o tem escrito um n\u00famero, sendo cinco dos n\u00fameros positivos e os restantes negativos.<\/p>\n<p>Tiram-se simultaneamente dois cart\u00f5es, ao acaso. H\u00e1 maior probabilidade do produto dos dois n\u00fameros extra\u00eddos ser positivo ou negativo?<\/p>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_7208' onClick='GTTabs_show(1,7208)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_7208'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<p>A extra\u00e7\u00e3o simult\u00e2nea de dois cart\u00f5es pode ser efetuada de $$NCP={}^{10}{{C}_{2}}=\\frac{10\\times 9\\times 8!}{8!\\times 2!}=45$$ maneiras diferentes.<\/p>\n<p>Para que o produto dos n\u00fameros extra\u00eddos seja positivo, ambos t\u00eam de ser positivos ou ambos t\u00eam de ser negativos. Assim, o n\u00famero de casos favor\u00e1veis a este acontecimento \u00e9 $$NC{{F}_{+}}=\\underbrace{{}^{5}{{C}_{2}}}_{\\text{n}\\text{. }\\!\\!{}^\\text{o}\\!\\!\\text{\u00a0 }\\!\\!\\grave{\\ }\\!\\!\\text{ de maneiras de extrair 2 n }\\!\\!\\acute{\\mathrm{u}}\\!\\!\\text{ meros positivos}}+\\underbrace{{}^{5}{{C}_{2}}}_{\\text{n}\\text{. }\\!\\!{}^\\text{o}\\!\\!\\text{\u00a0 }\\!\\!\\grave{\\ }\\!\\!\\text{ de maneiras de extrair 2 n }\\!\\!\\acute{\\mathrm{u}}\\!\\!\\text{ meros negativos}}=2\\times \\frac{5\\times 4\\times 3!}{3!\\times 2!}=20$$<\/p>\n<p>Para que o produto dos n\u00fameros extra\u00eddos seja negativo, os n\u00fameros extra\u00eddos t\u00eam de ser de sinais contr\u00e1rios. Logo, o n\u00famero de casos favor\u00e1veis a este acontecimento \u00e9 $$NC{{F}_{-}}=\\underbrace{{}^{5}{{C}_{1}}}_{\\text{n}\\text{. }\\!\\!{}^\\text{o}\\!\\!\\text{\u00a0 }\\!\\!\\grave{\\ }\\!\\!\\text{ de maneiras de extrair 1 n }\\!\\!\\acute{\\mathrm{u}}\\!\\!\\text{ mero positivo}}\\times \\underbrace{{}^{5}{{C}_{1}}}_{\\text{n}\\text{. }\\!\\!{}^\\text{o}\\!\\!\\text{\u00a0 }\\!\\!\\grave{\\ }\\!\\!\\text{ de maneiras de extrair 1 n }\\!\\!\\acute{\\mathrm{u}}\\!\\!\\text{ mero negativo}}=5\\times 5=25$$<\/p>\n<p>Como $NC{{F}_{-}}&gt;NC{{F}_{+}}$, ent\u00e3o \u00e9 mais prov\u00e1vel o produto dos n\u00fameros extra\u00eddos ser negativo.<\/p>\n<p>$$p(+)=\\frac{20}{45}=\\frac{4}{9}$$ $$p(-)=\\frac{25}{45}=\\frac{5}{9}$$<\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_7208' onClick='GTTabs_show(0,7208)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Numa urna colocaram-se 10 cart\u00f5es indistingu\u00edveis ao tato. Cada cart\u00e3o tem escrito um n\u00famero, sendo cinco dos n\u00fameros positivos e os restantes negativos. Tiram-se simultaneamente dois cart\u00f5es, ao acaso. H\u00e1 maior&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":19175,"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],"series":[],"class_list":["post-7208","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"],"views":2345,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2021\/12\/Mat66.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7208","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=7208"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7208\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/19175"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=7208"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=7208"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=7208"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=7208"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}