{"id":7051,"date":"2011-10-14T21:47:00","date_gmt":"2011-10-14T20:47:00","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=7051"},"modified":"2021-12-28T22:56:42","modified_gmt":"2021-12-28T22:56:42","slug":"prove-que-3","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=7051","title":{"rendered":"Prove que"},"content":{"rendered":"<p><ul id='GTTabs_ul_7051' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_7051' class='GTTabs_curr'><a  id=\"7051_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_7051' ><a  id=\"7051_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_7051'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Seja S o conjunto de resultados associado a uma experi\u00eancia aleat\u00f3ria e A e B ($B\\ne \\left\\{ {} \\right\\}$) dois acontecimentos (A e B s\u00e3o, pois, subconjuntos de S).<\/p>\n<p>Prove que:<\/p>\n<ol>\n<li>$P(A\\cap \\overline{B})=P(A)-P(A\\cap B)$<\/li>\n<li>Se $P(A)&gt;P(B)$, ent\u00e3o $P(A|B)\\ge P(B|A)$<\/li>\n<li>$P(A|B)+P(\\overline{A}|B)=1$<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_7051' onClick='GTTabs_show(1,7051)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_7051'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>$P(A\\cap \\overline{B})=P(A)-P(A\\cap B)$, com $B\\ne \\left\\{ {} \\right\\}$.\n<p>Ora, temos:<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\nP(A\\cap \\overline{B}) &amp; = &amp; P((A\\cap \\overline{B})\\cup (B\\cap \\overline{B}))\u00a0 \\\\<br \/>\n{} &amp; = &amp; P((A\\cup B)\\cap \\overline{B})\u00a0 \\\\<br \/>\n{} &amp; = &amp; 1-P((\\overline{A\\cup B)\\cap \\overline{B}})\u00a0 \\\\<br \/>\n{} &amp; = &amp; 1-P((\\overline{A\\cup B})\\cup B)\u00a0 \\\\<br \/>\n{} &amp; = &amp; 1-P(\\overline{A\\cup B})-P(B)+P((\\overline{A\\cup B})\\cap B)\u00a0 \\\\<br \/>\n{} &amp; = &amp; 1-(1-P(A\\cup B))-P(B)+P(\\overline{A}\\cap \\overline{B}\\cap B)\u00a0 \\\\<br \/>\n{} &amp; = &amp; 1-1+P(A)+P(B)-P(A\\cap B)-P(B)+0\u00a0 \\\\<br \/>\n{} &amp; = &amp; P(A)-P(A\\cap B)\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\nOu, ainda (construa um diagrama):<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\nP(A\\cap \\overline{B}) &amp; = &amp; P(A\\cup B)-P(B)\u00a0 \\\\<br \/>\n{} &amp; = &amp; P(A)+P(B)-P(A\\cap B)-P(B)\u00a0 \\\\<br \/>\n{} &amp; = &amp; P(A)-P(A\\cap B)\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\n\u00ad<\/p>\n<\/li>\n<li>Se $P(A)&gt;P(B)$, ent\u00e3o $P(A|B)\\ge P(B|A)$, com $B\\ne \\left\\{ {} \\right\\}$.\n<p>Ora, temos:<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\nP(A)&gt;P(B) &amp; \\Rightarrow\u00a0 &amp; P(A)\\times \\frac{P(A\\cap B)}{P(B)}&gt;P(B)\\times \\frac{P(A\\cap B)}{P(B)}\u00a0 \\\\<br \/>\n{} &amp; \\Rightarrow\u00a0 &amp; P(A)\\times P(A|B)&gt;P(A\\cap B)\u00a0 \\\\<br \/>\n{} &amp; \\Rightarrow\u00a0 &amp; P(A|B)&gt;\\frac{P(A\\cap B)}{P(A)}\u00a0 \\\\<br \/>\n{} &amp; \\Rightarrow\u00a0 &amp; P(A|B)&gt;P(B|A)\u00a0 \\\\<br \/>\n{} &amp; \\Rightarrow\u00a0 &amp; P(A|B)\\ge P(B|A)\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\n\u00ad<\/p>\n<\/li>\n<li>$P(A|B)+P(\\overline{A}|B)=1$, com\u00a0$B\\ne \\left\\{ {} \\right\\}$.\n<p>Ora, temos:<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\nP(A|B)+P(\\overline{A}|B) &amp; = &amp; \\frac{P(A\\cap B)}{P(B)}+\\frac{P(\\overline{A}\\cap B)}{P(B)}\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{P(A\\cap B)+P(\\overline{A}\\cap B)}{P(B)}\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{P((A\\cap B)\\cup (\\overline{A}\\cap B))}{P(B)}\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{P((A\\cup \\overline{A})\\cap B)}{P(B)}\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{P(S\\cap B)}{P(B)}\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{P(B)}{P(B)}\u00a0 \\\\<br \/>\n{} &amp; = &amp; 1\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_7051' onClick='GTTabs_show(0,7051)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Seja S o conjunto de resultados associado a uma experi\u00eancia aleat\u00f3ria e A e B ($B\\ne \\left\\{ {} \\right\\}$) dois acontecimentos (A e B s\u00e3o, pois, subconjuntos de S). Prove que:&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":19170,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[226,97,227],"tags":[427,215,235],"series":[],"class_list":["post-7051","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-probabilidade","tag-probabilidade-condicionada"],"views":2617,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2021\/12\/Mat61.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7051","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=7051"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7051\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/19170"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=7051"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=7051"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=7051"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=7051"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}