{"id":7074,"date":"2011-10-16T19:04:42","date_gmt":"2011-10-16T18:04:42","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=7074"},"modified":"2021-12-28T22:02:49","modified_gmt":"2021-12-28T22:02:49","slug":"determine-o-valor-de-pacup-b","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=7074","title":{"rendered":"Determine o valor de"},"content":{"rendered":"<p><ul id='GTTabs_ul_7074' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_7074' class='GTTabs_curr'><a  id=\"7074_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_7074' ><a  id=\"7074_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_7074'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Seja $\\Omega $ o espa\u00e7o de resultados associado a uma certa experi\u00eancia aleat\u00f3ria.<\/p>\n<p>Sejam A e B dois acontecimentos ($A\\subset \\Omega $ e $B\\subset \\Omega $).<\/p>\n<p>Sabe-se que A e B s\u00e3o acontecimentos independentes, que $P(B)=\\frac{2}{3}$ e $P(A\\cap B)=\\frac{1}{2}$.<\/p>\n<p>Determine o valor de $P(A\\cup B)$. Apresente o resultado na forma de fra\u00e7\u00e3o irredut\u00edvel.<\/p>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_7074' onClick='GTTabs_show(1,7074)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_7074'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<p>Sabe-se:<\/p>\n<ul>\n<li>A e B s\u00e3o acontecimentos independentes<\/li>\n<li>$P(B)=\\frac{2}{3}$<\/li>\n<li>$P(A\\cap B)=\\frac{1}{2}$<\/li>\n<\/ul>\n<p>Como \u00e9 sabido, $P(A\\cup B)=P(A)+P(B)-P(A\\cap B)$\u00a0 (1).<\/p>\n<p>Por outro lado, como A e B s\u00e3o acontecimentos independentes, ent\u00e3o $P(A\\cap B)=P(A)\\times P(B)$.<\/p>\n<p>Daqui e substituindo os valores conhecidos, temos: \\[\\begin{array}{*{35}{l}}<br \/>\n\\frac{1}{2}=P(A)\\times \\frac{2}{3} &amp; \\Leftrightarrow\u00a0 &amp; P(A)=\\frac{3}{4}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<p>Assim, de (1), vem: \\[\\begin{array}{*{35}{l}}<br \/>\nP(A\\cup B)=\\frac{3}{4}+\\frac{2}{3}-\\frac{1}{2} &amp; \\Leftrightarrow\u00a0 &amp; P(A\\cup B)=\\frac{9}{12}+\\frac{8}{12}-\\frac{6}{12}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; P(A\\cup B)=\\frac{11}{12}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p><\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_7074' onClick='GTTabs_show(0,7074)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Seja $\\Omega $ o espa\u00e7o de resultados associado a uma certa experi\u00eancia aleat\u00f3ria. Sejam A e B dois acontecimentos ($A\\subset \\Omega $ e $B\\subset \\Omega $). Sabe-se que A e B&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":19181,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[226,97,227],"tags":[427,236,234,215],"series":[],"class_list":["post-7074","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-acontecimentos-independentes","tag-axiomatica","tag-probabilidade"],"views":1998,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2021\/12\/Mat72.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7074","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=7074"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7074\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/19181"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=7074"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=7074"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=7074"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=7074"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}