{"id":7254,"date":"2011-12-04T18:24:05","date_gmt":"2011-12-04T18:24:05","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=7254"},"modified":"2022-01-26T00:40:38","modified_gmt":"2022-01-26T00:40:38","slug":"dez-maquinas-diferentes","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=7254","title":{"rendered":"Dez m\u00e1quinas diferentes"},"content":{"rendered":"<p><ul id='GTTabs_ul_7254' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_7254' class='GTTabs_curr'><a  id=\"7254_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_7254' ><a  id=\"7254_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_7254'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Numa f\u00e1brica, funcionam dez m\u00e1quinas diferentes.<\/p>\n<p>A probabilidade uma qualquer das m\u00e1quinas avariar durante um m\u00eas \u00e9 $0,1$.<\/p>\n<p>Sabendo que as avarias s\u00e3o independentes, determine a probabilidade dos acontecimentos:<\/p>\n<ol>\n<li>&#8220;Nenhuma m\u00e1quina avariar durante um m\u00eas.&#8221;<\/li>\n<li>&#8220;Pelo menos uma das\u00a0m\u00e1quinas avariar.&#8221;<\/li>\n<li>&#8220;Produzirem-se exatamente duas avarias.&#8221;<\/li>\n<li>&#8220;Produzirem-se pelo menos nove avarias.&#8221;<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_7254' onClick='GTTabs_show(1,7254)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_7254'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<p>A vari\u00e1vel $X$: &#8220;N\u00famero de m\u00e1quinas avariadas, durante um m\u00eas&#8221; tem distribui\u00e7\u00e3o binomial de par\u00e2metros $n=10$ e $p=0,1=\\frac{1}{10}$. Assim, temos:<\/p>\n<ol>\n<li>$$P(X=0)={}^{10}{{C}_{0}}\\times {{\\left( \\frac{1}{10} \\right)}^{0}}\\times {{\\left( \\frac{9}{10} \\right)}^{10}}=\\frac{3486784401}{10000000000}\\approx 0,35$$<\/li>\n<li>$$p=1-P(X=0)=1-{}^{10}{{C}_{0}}\\times {{\\left( \\frac{1}{10} \\right)}^{0}}\\times {{\\left( \\frac{9}{10} \\right)}^{10}}\\approx 0,65$$<\/li>\n<li>$$P(X=2)={}^{10}{{C}_{2}}\\times {{\\left( \\frac{1}{10} \\right)}^{2}}\\times {{\\left( \\frac{9}{10} \\right)}^{8}}=45\\times \\frac{1}{100}\\times \\frac{43046721}{100000000}\\approx 0,19$$<\/li>\n<li>$$p=P(X=9)+P(X=10)={}^{10}{{C}_{9}}\\times {{\\left( \\frac{1}{10} \\right)}^{9}}\\times {{\\left( \\frac{9}{10} \\right)}^{1}}+{}^{10}{{C}_{10}}\\times {{\\left( \\frac{1}{10} \\right)}^{10}}\\times {{\\left( \\frac{9}{10} \\right)}^{0}}\\approx 9,1\\times {{10}^{-9}}$$<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_7254' onClick='GTTabs_show(0,7254)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Numa f\u00e1brica, funcionam dez m\u00e1quinas diferentes. A probabilidade uma qualquer das m\u00e1quinas avariar durante um m\u00eas \u00e9 $0,1$. Sabendo que as avarias s\u00e3o independentes, determine a probabilidade dos acontecimentos: &#8220;Nenhuma m\u00e1quina&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":21065,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[226,97,227],"tags":[427,255,251,215],"series":[],"class_list":["post-7254","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","tag-distribuicao-binomial","tag-probabilidade"],"views":1885,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/12\/12V1Pag180-71_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7254","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=7254"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7254\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/21065"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=7254"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=7254"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=7254"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=7254"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}