{"id":7093,"date":"2011-10-17T16:07:09","date_gmt":"2011-10-17T15:07:09","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=7093"},"modified":"2022-01-25T18:38:56","modified_gmt":"2022-01-25T18:38:56","slug":"iogurtes-e-sumos","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=7093","title":{"rendered":"Iogurtes e sumos"},"content":{"rendered":"<p><ul id='GTTabs_ul_7093' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_7093' class='GTTabs_curr'><a  id=\"7093_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_7093' ><a  id=\"7093_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_7093'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Considere um espa\u00e7o de resultados finito, $\\Omega $, associado a uma certa experi\u00eancia aleat\u00f3ria.<\/p>\n<p>A prop\u00f3sito de dois acontecimentos X e Y ($X\\subset \\Omega $ e $Y\\subset \\Omega $), sabe-se que:<\/p>\n<ul>\n<li>$P(X)=a$<\/li>\n<li>$P(Y)=b$<\/li>\n<li>X e Y s\u00e3o independentes<\/li>\n<\/ul>\n<ol>\n<li>Mostre que a probabilidade de que n\u00e3o ocorra X nem ocorra Y \u00e9 igual a $1-a-b+a\\times b$.<\/li>\n<li><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/10\/Refrigerator2.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"7094\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=7094\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/10\/Refrigerator2.png\" data-orig-size=\"168,248\" 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=\"Frigor\u00edfico\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/10\/Refrigerator2.png\" class=\"alignright size-full wp-image-7094\" title=\"Frigor\u00edfico\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/10\/Refrigerator2.png\" alt=\"\" width=\"168\" height=\"248\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/10\/Refrigerator2.png 168w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/10\/Refrigerator2-101x150.png 101w\" sizes=\"auto, (max-width: 168px) 100vw, 168px\" \/><\/a>Num frigor\u00edfico, h\u00e1 um certo n\u00famero de iogurtes e um certo n\u00famero de sumos.\n<p>Tiram-se do frigor\u00edfico, ao acaso, um iogurte e um sumo.<\/p>\n<p>Sabe-se que a probabilidade de o iogurte ser de p\u00eassego \u00e9 $\\frac{1}{5}$ e a probabilidade de o sumo ser de laranja \u00e9 $\\frac{1}{3}$.<\/p>\n<p>Admita que os acontecimentos \u00abtirar um iogurte de p\u00eassego\u00bb e \u00abtirar um sumo de laranja\u00bb s\u00e3o independentes.<\/p>\n<p>Utilizando a express\u00e3o mencionada na al\u00ednea anterior, determine a probabilidade de, ao tirar, ao acaso, um iogurte e um sumo do frigor\u00edfico, o iogurte\u00a0n\u00e3o ser de p\u00eassego e o sumo n\u00e3o ser de laranja.<br \/>\nApresente o resultado na forma de fra\u00e7\u00e3o irredut\u00edvel.<\/p>\n<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_7093' onClick='GTTabs_show(1,7093)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_7093'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>Aplicando propriedades das opera\u00e7\u00f5es entre conjuntos e das probabilidades, temos:\u00a0\\[\\begin{array}{*{35}{l}}<br \/>\nP(\\overline{X}\\cap \\overline{Y}) &amp; = &amp; P(\\overline{X\\cup Y})\u00a0 \\\\<br \/>\n{} &amp; = &amp; 1-P(X\\cup Y)\u00a0 \\\\<br \/>\n{} &amp; = &amp; 1-P(X)-P(Y)+P(X\\cap Y)\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\nComo os acontecimentos X e Y s\u00e3o independentes, ent\u00e3o $P(X\\cap Y)=P(X)\\times P(Y)$.<\/p>\n<p>Assim, tendo isto em considera\u00e7\u00e3o, que\u00a0\u00a0$P(X)=a$ e que $P(Y)=b$, vem: \\[\\begin{matrix}<br \/>\nP(\\overline{X}\\cap \\overline{Y}) &amp; = &amp; 1-a-b+a\\times b\u00a0 \\\\<br \/>\n\\end{matrix}\\]<br \/>\nPortanto, a probabilidade de que n\u00e3o ocorra X nem ocorra Y \u00e9 igual a $1-a-b+a\\times b$.<br \/>\n\u00ad<\/p>\n<\/li>\n<li>Sejam X: \u00abtirar um iogurte de p\u00eassego\u00bb e Y: \u00abtirar um sumo de laranja\u00bb.\n<p>Portanto, a probabilidade pedida \u00e9 $P(\\overline{X}\\cap \\overline{Y})$, com $P(X)=a=\\frac{1}{5}$ e $P(Y)=b=\\frac{1}{3}$.<\/p>\n<p>Logo, a probabilidade pedida, considerando a express\u00e3o mencionada na al\u00ednea anterior, \u00e9: \\[\\begin{array}{*{35}{l}}<br \/>\nP(\\overline{X}\\cap \\overline{Y}) &amp; = &amp; 1-\\frac{1}{5}-\\frac{1}{3}+\\frac{1}{5}\\times \\frac{1}{3}\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{15-3-5+1}{15}\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{8}{15}\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_7093' onClick='GTTabs_show(0,7093)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Considere um espa\u00e7o de resultados finito, $\\Omega $, associado a uma certa experi\u00eancia aleat\u00f3ria. A prop\u00f3sito de dois acontecimentos X e Y ($X\\subset \\Omega $ e $Y\\subset \\Omega $), sabe-se que:&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":21012,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[226,97,227],"tags":[427,236,215,235],"series":[],"class_list":["post-7093","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-probabilidade","tag-probabilidade-condicionada"],"views":3322,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/10\/12-Iogurtes_e_sumos.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7093","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=7093"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7093\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/21012"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=7093"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=7093"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=7093"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=7093"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}