De acordo com um estudo feito, a dire\u00e7\u00e3o estima que 30% dos seus funcion\u00e1rios e 20% das suas funcion\u00e1rias, est\u00e3o aptos para desempenhar uma determinada tarefa, que exige rigor e compet\u00eancia.<\/p>\n Escolhendo ao acaso um trabalhador desta empresa, qual \u00e9 a probabilidade de:<\/p>\n Sejam H: “ser homem” e A: “estar apto a desempenhar a tarefa”.<\/p>\n Sabe-se:<\/p>\n Como $P(H)=0,6$, ent\u00e3o $P(\\overline{H})=1-P(H)=1-0,6=0,4$<\/span>.<\/p>\n \u00a0 \u00ad<\/p>\n Ora, $P(H\\cap A)=P(H)\\times P(A|H)=0,6\\times 0,3=0,18$. Ora, $P(\\overline{H}\\cap A)=P(\\overline{H})\\times P(A|\\overline{H})=0,4\\times 0,2=0,08$.<\/p>\n Podemos, entretanto, acabar de preencher a tabela:<\/p>\n $P(A)=P((A\\cap H)\\cup (A\\cap \\overline{H}))=P(A\\cap H)+P(A\\cap \\overline{H})=0,18+0,08=0,26$<\/span><\/p>\n $P(\\overline{A})=1-P(A)=1-0,26=0,74$<\/span><\/p>\n $P(H\\cap \\overline{A})=P(H)-P(H\\cap A)=0,6-0,18=0,42$<\/span><\/p>\n \u00a0$P(\\overline{H}\\cap \\overline{A})=P(\\overline{H})-P(\\overline{H}\\cap A)=0,4-0,08=0,32$ A probabilidade pedida\u00a0\u00e9: \\[\\begin{array}{*{35}{l}} A probabilidade pedida \u00e9: \\[\\begin{array}{*{35}{l}}![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/10\/foto_empresa.jpg\" "\\"Empresa\\"")
<\/a>60% dos funcion\u00e1rios de uma empresa s\u00e3o homens.<\/p>\n\n
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<\/a><\/p>\n\n\n
\n <\/td>\n A<\/td>\n $\\overline{A}$<\/td>\n Total<\/strong><\/td>\n<\/tr>\n \n H<\/td>\n 1) 18%<\/td>\n \u00a042%<\/span><\/td>\n 60%<\/td>\n<\/tr>\n \n $\\overline{H}$<\/td>\n 2) 8%<\/td>\n 32%\u00a0<\/span><\/td>\n 40%<\/span><\/td>\n<\/tr>\n \n Total<\/strong><\/td>\n 26%<\/span><\/td>\n 74%\u00a0<\/span><\/td>\n 100%<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n \n
\n\u00ad<\/p>\n<\/li>\n
\n\u00ad
\n<\/span><\/p>\n<\/li>\n
\nP(H|\\overline{A}) & = & \\frac{P(H\\cap \\overline{A})}{P(\\overline{A})}\u00a0 \\\\
\n{} & = & \\frac{P(H)-P(H\\cap A)}{1-P(A)}\u00a0 \\\\
\n{} & = & \\frac{0,6-0,18}{1-0,26}\u00a0 \\\\
\n{} & = & \\frac{21}{37}\u00a0 \\\\
\n\\end{array}\\]
\n\u00ad<\/p>\n<\/li>\n
\nP(\\overline{H}|A) & = & \\frac{P(\\overline{H}\\cap A)}{P(A)}\u00a0 \\\\
\n{} & = & \\frac{0,08}{0,26}\u00a0 \\\\
\n{} & = & \\frac{4}{13}\u00a0 \\\\
\n\\end{array}\\]<\/p>\n<\/li>\n<\/ol>\n