{"id":7030,"date":"2011-10-12T22:15:50","date_gmt":"2011-10-12T21:15:50","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=7030"},"modified":"2022-01-25T15:57:08","modified_gmt":"2022-01-25T15:57:08","slug":"tres-emissoras-de-radio","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=7030","title":{"rendered":"Tr\u00eas emissoras de r\u00e1dio"},"content":{"rendered":"<p><ul id='GTTabs_ul_7030' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_7030' class='GTTabs_curr'><a  id=\"7030_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_7030' ><a  id=\"7030_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_7030'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/10\/emissoraradio.jpg\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"7031\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=7031\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/10\/emissoraradio.jpg\" data-orig-size=\"191,263\" 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=\"Antena\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/10\/emissoraradio.jpg\" class=\"alignright size-full wp-image-7031\" title=\"Antena\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/10\/emissoraradio.jpg\" alt=\"\" width=\"115\" height=\"158\" \/><\/a>Numa dada regi\u00e3o h\u00e1 tr\u00eas emissoras de r\u00e1dio: a R\u00e1dio Jovem, a R\u00e1dio Moderna e a R\u00e1dio Alegria.<\/p>\n<p>55% dos habitantes ouvem a R\u00e1dio Jovem, 38% ouvem a R\u00e1dio Moderna e 33% ouvem a R\u00e1dio Alegria.<\/p>\n<p>15% ouvem as emissoras Jovem e Moderna, 11% Jovem e Alegria, 9% Moderna e Alegria e 4% ouvem as tr\u00eas emissoras.<\/p>\n<p>Sejam os acontecimentos:<\/p>\n<ul>\n<li>A: &#8220;ouvir a R\u00e1dio Alegria&#8221;;<\/li>\n<li>J: &#8220;ouvir a R\u00e1dio Jovem&#8221;;<\/li>\n<li>M: &#8220;ouvir a R\u00e1dio Moderna&#8221;.<\/li>\n<\/ul>\n<ol>\n<li>Calcule a probabilidade de cada um dos acontecimentos seguintes:\n<p>a) $A\\cup J\\cup M$;<\/p>\n<p>b) $A-(M\\cup J)$.<\/p>\n<\/li>\n<li>Que percentagem de pessoas n\u00e3o ouve r\u00e1dio?<\/li>\n<\/ol>\n<p>\u00a0<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_7030' onClick='GTTabs_show(1,7030)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_7030'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<p>Sejam J, M e A, respetivamente, os conjuntos dos ouvintes que ouvem as R\u00e1dios Jovem, Moderna e Alegria.<\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/10\/diagramaRadios.jpg\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"7032\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=7032\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/10\/diagramaRadios.jpg\" data-orig-size=\"354,223\" 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\/10\/diagramaRadios.jpg\" class=\"alignright size-full wp-image-7032\" title=\"Diagrama\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/10\/diagramaRadios.jpg\" alt=\"\" width=\"354\" height=\"223\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/10\/diagramaRadios.jpg 354w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/10\/diagramaRadios-300x188.jpg 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/10\/diagramaRadios-150x94.jpg 150w\" sizes=\"auto, (max-width: 354px) 100vw, 354px\" \/><\/a>Sabe-se que:<\/p>\n<ul>\n<li>$\\#J\\to 0,55$<\/li>\n<li>$\\#M\\to 0,38$<\/li>\n<li>$\\#A\\to 0,33$<\/li>\n<li>$\\#(J\\cap M)\\to 0,15$<\/li>\n<li>$\\#(J\\cap A)\\to 0,11$<\/li>\n<li>$\\#(M\\cap A)\\to 0,09$<\/li>\n<li>$\\#(J\\cap M\\cap A)\\to 0,04$<\/li>\n<\/ul>\n<p>Com base nestes valores, podemos completar o diagrama ao lado come\u00e7ando por preencher os valores de baixo para cima e calculando os valores interm\u00e9dios, tais como, por exemplo: \\[\\#\\left( (M\\cap A)-(J\\cap M\\cap A) \\right)=\\#M\\cap A)-\\#(J\\cap M\\cap A)\\to 0,09-0,04\\to 0,05\\]<br \/>\n\u00ad<\/p>\n<ol>\n<li>a)<br \/>\nPodemos calcular a probabilidade pedida usando uma propriedade conhecida (tamb\u00e9m confirm\u00e1vel facilmente a partir do diagrama): \\[\\begin{array}{*{35}{l}}<br \/>\nP(A\\cup J\\cup M) &amp; = &amp; P(A)+P(J)+P(M)-P(A\\cap J)-P(A\\cap M)-P(J\\cap M)+P(A\\cap J\\cap M)\u00a0 \\\\<br \/>\n{} &amp; = &amp; 0,33+0,55+0,38-0,11-0,09-0,15+0,04\u00a0 \\\\<br \/>\n{} &amp; = &amp; 0,95\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\nOu, ainda, usando as v\u00e1rias decomposi\u00e7\u00f5es que figuram no diagrama: \\[\\begin{array}{*{35}{l}}<br \/>\nP(A\\cup J\\cup M) &amp; = &amp; 0,33+0,17+0,18+0,07+0,05+0,11+0,04\u00a0 \\\\<br \/>\n{} &amp; = &amp; 0,95\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<p>b)<br \/>\nPodemos calcular a probabilidade pedida usando propriedades conhecidas (tamb\u00e9m confirm\u00e1vel facilmente a partir do diagrama): \\[\\begin{array}{*{35}{l}}<br \/>\nP(A-(M\\cup J)) &amp; = &amp; P(A)-P(A\\cap (M\\cup J))\u00a0 \\\\<br \/>\n{} &amp; = &amp; P(A)-P((A\\cap M)\\cup (A\\cap J))\u00a0 \\\\<br \/>\n{} &amp; = &amp; P(A)-P(A\\cap M)-P(A\\cap J)+P((A\\cap M)\\cap (A\\cap J))\u00a0 \\\\<br \/>\n{} &amp; = &amp; P(A)-P(A\\cap M)-P(A\\cap J)+P(A\\cap M\\cap J)\u00a0 \\\\<br \/>\n{} &amp; = &amp; 0,33-0,09-0,11+0,04\u00a0 \\\\<br \/>\n{} &amp; = &amp; 0,17\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\nOu, ainda, diretamente do diagrama: $P(A-(M\\cup J))=0,17$ (zona rosa).<br \/>\n\u00ad<\/p>\n<\/li>\n<li>\u00c9 de 5% a percentagem de pessoas que n\u00e3o ouvem r\u00e1dio, pois $1-P(A\\cup J\\cup M)=1-0,95=0,05$.<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_7030' onClick='GTTabs_show(0,7030)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Numa dada regi\u00e3o h\u00e1 tr\u00eas emissoras de r\u00e1dio: a R\u00e1dio Jovem, a R\u00e1dio Moderna e a R\u00e1dio Alegria. 55% dos habitantes ouvem a R\u00e1dio Jovem, 38% ouvem a R\u00e1dio Moderna e&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":20981,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[226,97,227],"tags":[427,217,216,215],"series":[],"class_list":["post-7030","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-acontecimento","tag-espaco-de-resultados","tag-probabilidade"],"views":1944,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/10\/12V1Pag167-18_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7030","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=7030"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7030\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20981"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=7030"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=7030"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=7030"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=7030"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}