Exercicios de Probabilidade
DEFINIÇÃO CLÁSSICA DE PROBABILIDADE A probabilidade de A, anotada por P(A), lê-se pe de A, é definida como sendo: P(A) = m / n
Exemplo 1 Calcular a probabilidade de no lançamento de um dado equilibrado obter-se: (a) Um resultado igual a 4. (b) Um resultado ímpar.
Solução:
S = { 1, 2, 3, 4, 5, 6 } n = #(S) = 6 (a) A = { 4 } m = #(A) = 1 então P(A) = m / n = 1 / 6 = 16,67% (b) B = { 1, 3, 5 } m = #(B) = 3 então P(B) = m / n = 3 / 6 = 50%
A DEFINIÇÃO DE PROBABILIDADE COMO FREQÜÊNCIA RELATIVA Então a freqüência relativa do evento A, anotada por fr A, é o quociente: fr A = m / n = (número de vezes que A ocorre) / (número de vezes que E é repetido) Seja E um experimento e A e B dois eventos de um espaço amostra associado S. Sejam fr A e fr B as freqüências relativas de A e B respectivamente. Então. (i) 0 . fr A . 1, isto é, a freqüência relativa do evento A é um número que varia entre 0 e 1. (ii) fr A = 1 se e somente se, A ocorre em todas as “n” repetições de E. (iii) fr A = 0, se e somente se, A nunca ocorre nas “n” repetições de E. AUB = fr A + fr B se A e B forem eventos mutuamente excludentes. (iv) fr AUB
Exemplo 2 (i) Uma moeda foi lançada 200 vezes e forneceu 102 caras. Então a freqüência relativa de “caras” é: fr A = 102 / 200 = 0,51 = 51% (ii) Um dado foi lançado 100 vezes e a face 6 apareceu 18 vezes. Então a freqüência relativa do evento A = { face 6 } é: fr A = 18 / 100 = 0,18 = 18%
PROBABILIDADE CONDICIONADA E INDEPENDÊNCIA Suponha-se que se quer extrair duas peças ao acaso de um lote que contém 100 peças das quais 80 peças são boas e 20 defeituosas, de acordo com os critérios (a) com reposição e (b) sem reposição. Define-se os seguintes eventos: A = { A primeira peça é defeituosa } e B = { A segunda peça é defeituosa }. Então, se a extração for com reposição P(A) = P(B) = 20 / 100 = 1 / 5 = 20%, porque existem 20 peças defeituosas num total de 100. Agora se a extração for sem reposição tem-se ainda que P(A) = 20 / 100 = 20%, mas o mesmo não é verdadeiro para P(B).
P(B/A) = P(A∩B) / P(A)
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TEOREMA DA MULTIPLICAÇÃO Com o conceito de probabilidade condicionada é possível apresentar uma maneira de se calcular a probabilidade da interseção de dois eventos A e B em função destes eventos. Esta expressão é denominada de teorema da multiplicação. P(A∩B) = P(A).P(B/A) = P(A/B).P(B)
INDEPENDÊNCIA DE DOIS EVENTOS Sejam A e B dois eventos de um espaço amostra S. A e B são ditos independentes se a probabilidade de um deles ocorrer não afetar a probabilidade do outro ocorrer, isto é, se:
P(A/B) = P(A) ou P(B/A) = P(B) ou ainda se P(A∩B) = P(A).P(B)
Qualquer uma das 3 relações acima pode ser usada como definição de independência.
Exemplo 3
Três componentes C1, C2, e C3, de um mecanismo são postos em série (em linha reta). Suponha que esses componentes sejam dispostos em ordem aleatória. Seja R o evento { C 2 está à direita de C1 }, e seja S o evento { C 3 está à direita de C1 }. Os eventos R e S são independentes? Por quê?
Solução:
Para que R e S sejam independentes deve-se ter: P(R/S) = P(R).P(S). O espaço amostra para este caso é: S = { C 1C2C3, C1C3C2, C2C1C3, C2C3C1, C3C1C2, C3C2C1 } As seqüências em que C 2 está à direita de C1 são: R = { C1C2C3, C1C3C2, C3C1C2 }. Logo: P(R) = 3/6 = 50% As seqüências em que C 3 está à direita de C1 são: S = { C 1C2C3, C1C3C2, C2C1C3 }. Logo P(S) = 3/6 = 50% As seqüências em que C 2 está à direita de C1 e C3 está também à direita de C1 são: R/S = { C1C2C3, C1C3C2 }. Logo P(R/S ) = 2/6 = 1/3 = 33,33% ≠ P(R).P(S) = 0.5.0,5 = 0,25 = 25% Portanto os eventos R e S não são independentes.
TEOREMAS DA PROBABILIDADE TOTAL E DE BAYES
Exemplo 4 Considere-se o espaço amostra obtido pelos números das faces no lançamento de um dado equilibrado e sejam os eventos: A 1 = { 1, 2, 3 }, A 2 = { 4, 5 } e A 3 = { 6 } Então, pode-se verificar facilmente que, os eventos acima formam um partição do espaço amostral S = { 1, 2, 3, 4, 5, 6 }. obtém-se então o denominado teorema da probabilidade total:
P(B) = P(A1).P(B/A1) + P(A2).P(B/A2) + ... + P(A n).P(B/An) Exemplo 5
Uma determinada peça é manufaturada por 3 fábricas: A, B e C. Sabe-se que A produz o dobro de peças que B e que B e C produzem o mesmo número de peças. Sabe-se ainda que 2% das peças produzidas por A e por B são defeituosas, enquanto que 4% das produzidas por C são defeituosas. Todas as peças produzidas são misturadas e colocadas em um depósito. Se do depósito for retirada uma
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Solução: Considerem-se os seguintes eventos: D = { A peça é defeituosa }, A = { A peça provém da fábrica A }, B = { A peça provém da máquina B } e C = { A peça provém da máquina C }. Tem-se então que: P(A) = 50%, P(B) = P(C) = 25%, uma vez que só existem as 3 fábricas e que A produz o dobro de B e esta por sua vez produz a mesma quantidade que C. Sabe-se também que P(D/A) = P(D/B) = 2% e que P(D/C) = 4%. Pelo teorema da probabilidade total pode-se escrever que: P(D) = P(A).P(D/A) + P(B).P(D/B) + P(C).P(D/C) = 0,5.0,02 + 0,25.0,02 + 0,25.0,04 =2,50%, pois A, B e C formam uma partição do espaço amostra S.
Exemplo 6 Considerando a pergunta acima vem então: P(A / D), isto é a probabilidade de ter sido produzida pela máquina A dado que a peça é defeituosa é: P(A / D) = P(A). P(D / A) / P(D) = 0,02.0,50 / (0,5.0,02 + 0,25.0,02 + 0,25.0,04) = 0,40 = 40%
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EXERCÍCIOS DIVERSOS 1.) Lance dois dados. Descreva o espaço amostral amostral S, e calcule calcule a probabilidade probabilidade de ocorrer ocorrer a soma 7 ? (6/36) 2.) Lance 3 moedas. Descreva o espaço amostral amostral S, utilizando o diagrama da árvore. Qual Qual a probabilid probabilidade ade de a) ocorrerem duas duas caras ? (3/8) b) ocorrer ocorrer pelo menos uma cara ? (7/8) Lance duas duas moedas moedas e um dado. dado. a) Descre Descreva va o espaço espaço amostra amostrall S b) Expresse os 3.) Lance evento eventos: s: A = {aparecem duas caras e um número par } B = {aparece 2} C = c) Expresse claramente o {aparecem exatamente uma cara e um número primo} evento em que : I) A e B ocorrem II) somente B ocorre III) B ou C ocorrem 4.) Das 10 alunas de uma turma 3 delas têm olhos azuis. azuis. Se duas delas forem escolhidas escolhidas ao acaso, qual qual a probabilidade de a) ambas terem terem olhos azuis b) nenhuma nenhuma ter olhos azuis azuis c) pelo menos uma ter olhos azuis. ( 1/15 – 7/15 – 8/15) 5.) Três Três paraf parafus usos os e três três porcas porcas estão estão numa numa cai caixa. xa. Se duas duas peças peças forem retirad retiradas as ao acaso da caixa, qual a probabilidade de uma ser um parafuso e a outra ser uma porca ? (3/5) 6.) Suponha Suponha que numa turma turma há 6 moças e 10 rapazes. rapazes. Se uma comissão comissão de 3 pessoas pessoas é escolhida aleatoriamente, qual a probabilidade de serem selecionados a) 3 rapazes b) exatamente dois rapazes c) pelo menos um rapaz d) exatamente duas moças (3/14 – 27/56 – 27/28 – 15/56) 7.) O serviç serviçoo meteor meteoroló ológi gico co inform informaa que, que, para para o final final de semana semana,, a probab probabili ilidad dadee de chover é de 60%, a de fazer frio é de 70% e a de chover e fazer frio é de 50%. Calcular a probabilidade de que, no final de semana, a) chova ou faça frio; (80%) b) não chova e não faça frio. 20% 8.) A e B jogam 120 partidas partidas de xadrez, xadrez, das quais A ganha 60, B ganha 40 e 20 terminam empatadas. A e B concordam em jogar 3 partidas. Determinar a probabilidade probabilidade de: a) A ganhar todas as três; (1/8) b) duas partidas terminarem empatadas; (5/72) c) A e B ganharem alternadamente. (5/36) 9.) Em uma indústria há 10 pessoas pessoas que ganham ganham mais de 20 salários mínimos(s.m.), mínimos(s.m.), 20 que ganham entre 10 e 20 s.m. e 70 que ganham menos de 10 s.m. Três pessoas desta indústria são selecionadas. Determinar a probabilidade de que pelo menos uma ganhe menos de 10 s.m. ( 0,973) 0,973) 10.) 10.) A probabi probabilid lidade ade de que que um atleta atleta A ultrap ultrapass assee 17,30m 17,30m num num único único salto salto triplo triplo é de 0,7. O atleta dá 4 saltos. Qual a probabilidade de que em pelo menos num dos saltos ultrapasse 17,30m ? (0,9919) 11.) Lançamos um dado duas vezes. Seja a o número de pontos obtidos no primeiro lançamento e b os obtidos no segundo lançamento. Determine a probabilidade de a equação ax – b = 0 ter raiz inteira. ( 38,89%)
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13.) 13.) Com os dígi dígitos tos 1, 2, 3, 4 e 5 são formad formados os númer números os de quatr quatroo algaris algarismos mos disti distinto ntos. s. Se um deles é escolhido ao acaso, qual a probabilidade de ele ser: a) par ? (48/120) b) ímpar ? ( 60%) 14. 14.) Um grup rupo é formad rmadoo por seis eis home homens ns e quatro tro mulh ulheres eres.. Três Três pes pessoas oas são selecionadas ao acaso e sem reposição. Qual a probabilidade de que ao menos duas sejam do do sexo masculino ? (80/120) 15.) Cinco lâmpadas lâmpadas são escolhida escolhidass aleator aleatoriamen iamente te de de um pacote pacote que que contém contém 10 10 lâmpad lâmpadas as das quais três são defeituosas. Seja W o número de lâmpadas defeituosas escolhidas . Determine a probabilidade de W = 2. (105/252)
PROBABILIDADE CONDICIONAL Sejam dois eventos A e B associados a um espaço amostral S. A probabilidade de A ocorrer dado que o evento B ocorreu é definida por: P(A B) P(B)
P(A/B) =
onde P(B)
0.
Portanto, quando calculamos P(A/B), tudo se passa como se o evento B fosse um novo espaço amostral reduzido dentro do qual queremos calcular a probabilidade do evento.
Exemplo: Lançar um par de dados não viciados. Se a soma dos números é 6, qual a probabilidade de ter ocorrido 2 em um deles? S = { (1,1); (1,2); (1,3);...; (6,6) } B = { soma soma é 6 } = { (1,5); (2,4); (2,4); (3,3); (4,2); (4,2); (5,1) } A = { ocorre 2 em um dos dados } = { (1,2); (2,1); (2,2); (2,3); (2,4); (2,5); (3,2); (4,2); (5,2); (6,2) } A
B = {(2,4); (4,2) }
P(A
B) =
2 36
2
P(B) =
5 36
, portanto :
P(A/B) =
36 5 36
Ou n( A
B)
2
2 5
(2,6);
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Da expressão P(A/B) =
P(A B ) P(B )
obtém-se o teorema da multiplicação(ou produto):
P(A B ) = P(A) . P(B/A) que pode ser generalizado para para n eventos: ... N ) P ( A).P (B / A ).P( C / A B )... P(N / A B C...). P(A B C ...
Exemplo: Num lote de 12 peças, 4 são defeituosas. Seja o experimento: retirar 3 peças aleatoriamente uma após a outra. Determinar a probabilidade das 3 peças serem perfeitas. P P1
P2
P3
P( P1 ) . P( P2 / P1 ) . P( P3 / P1
P2 )
8 12
.
7 11
.
6
14
10
55
INDEPENDÊNCIA DE EVENTOS
Se tivermos dois eventos A e B, tais que P(B/A) = P(B), dizemos que A e B são eventos independentes(caso contrário são eventos dependentes). Isto quer dizer que a ocorrência de um não depende( ou não é condicionada, ou não se vincula) da ocorrência do outro, isto é, a informação adicional de que um dos eventos já ocorreu em nada altera a probabilidade de ocorrência do outro. Para Pa ra o caso caso de dois dois even evento toss inde indepe pend nden ente tess o teor teorem emaa da mult multip ipli lica caçã çãoo é simplificado: P(A
Generalizando Generalizando para P(A
B
C
n
B)
P ( A ) . P (B )
eventos independentes entre si, temos: ... N)
P (A). P(B ). P( C ) ... P(N)
Exemplo: Lançar uma moeda 3 vezes. S = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT } A = { 1º lançamento é cara } = {HHH, HHT, HTH, HTT}
P(A) =
4 8
4
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a-) Provar que A e B são independentes. b-) B e C são independentes ? c-) A e C são independentes ? a-)A P (A
B)
b-)B
2
4
8
8
X
4
2
8
8
2
4
8
8
c-) A
P ( A ) X P (B ) INDEPENDEN
C = {HHT, THH }
P( B C ) = P(B) x P(C)
P ( A C ) = P(A) . P(C)
{HHH,HHT}
B
1
4
8
8
.
2
1
8
8
x
P(B
2
1
8
8
C)=
TES
2 8
Assim B e C são dependentes.
C = {HHT} , portanto A e C são independentes.
TEOREMA DA PROBABILIDADE TOTAL
Sejam os eventos A1, A2, A3, ... , An, que constituem uma partição do espaço amostral S, ou seja: A1 U A2 U A3 ... U An = S P(Ai) > 0, para todo i = 1, 2, 3,..., n. para i j Ai A j Então, se B é um evento, temos o seguinte teorema:
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Qual a probabilidade de uma lâmpada defeituosa ser selecionada desse experimento ?
P(D) = P(I D) + P(II∩D) + P(III∩D) P(D) = P(I). P(D/I) + P(II). (D/II) + P(III) . P(D/III) P(D) =
1 3
.
4
1
10
3
.
1
1
6
3
.
3
113
8
360
TEOREMA DE BAYES
Sejam A1, A2, ... , An eventos que formam partição de S e B um evento qualquer. Para qualquer Ai : P (Ai ) . P(B / Ai ) P(Ai/B) = P(A ) . P(B / A ) ... ... P(An ) . P(B / An ) i 1
exempl ploo ante anteri rior or,, se a lâmp lâmpad adaa for for sele seleci cion onad adaa ao acas acasoo e for for Exemplo: Do exem defeituosa, qual a probabilidade probabilidade de ter vindo da caixa I ?
1
4
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2-) Dois dígitos diferentes são selecionados aleatoriamente dos dígitos de 1 a 9. (i) Se a soma é impar, qual a probabilidade probabilidade do 2 ser um do números selecionados selecionados ? (1/4) (ii) Se o 2 é um dos números selecionados, qual a probabilidade da soma ser ímpar ? (5/8) 3-) Numa certa cidade, 40% da população têm cabelos castanhos, 25% olhos castanhos e 15% têm olhos e cabelos castanhos. Uma pessoa da cidade é selecionada aleatoriamente. (i) Se ela têm cabelos castanhos, qual a probabilidade de ter também olhos castanhos ? (3/8) (ii) Se ela têm olhos castanhos, castanhos, qual a probabilidade probabilidade de não ter ter cabelos castanhos ? (2/5) (iii) Qual a probabilidade probabilidade de não ter olhos olhos nem cabelos castanhos castanhos ? (1/2) 4-) São dadas duas urnas. Uma urna A contém 5 bolas vermelhas, 3 brancas e 8 azuis. Uma urna B contém 3 bolas vermelhas e 5 brancas. Lança-se um dado não viciado: se ocorre 3 ou 6 uma bola é escolhida de B, caso contrário uma bola é escolhida de A . Encontre a probab probabili ilida dade de de (i) uma bola vermelh vermelhaa ser escolh escolhid idaa (1/3) (1/3) (ii) (ii) uma bola branca branca ser escolhida (1/3) (1/3) (iii) uma bola bola azul ser escolhida escolhida (1/3) 5-) A caixa A contém 9 cartas numeradas de 1 a 9 e a caixa B contém 5 cartas numeradas de 1 a 5. Uma caixa é escolhida aleatoriamente e uma carta retirada; se a carta indica um número par, outra carta retirada da mesma caixa; se a carta indica um número ímpar, uma carta é retirada da outra caixa. (i) Qual a probabilidade de ambas as cartas indicarem núme número ross pare paress ? (2/1 (2/15) 5) (ii) (ii) Se amba ambass as cart cartas as indi indica cam m núme número ross pare pares, s, qu qual al a probabilid probabilidade ade de terem vindo vindo da caixa A ? (5/8) (5/8) (iii) (iii) Qual a probabilida probabilidade de de ambas as cartas indicarem números ímpares ? (1/3) 6-) Uma caixa contém uma moeda não viciada e uma de duas caras. Uma moeda é selecionada aleatoriamente e lançada. Se ocorre cara, a outra moeda é lançada; se ocorre coroa a mesma moeda é lançada. (i) Encontre a probabilidade de ocorrer cara no segundo lançamento (5/8) (ii) Se ocorreu cara no segundo lançamento, encontre a probabilidade de ter também aparecido no primeiro lançamento. (4/5) 7-) Uma urna A contém 5 bolas vermelhas e 3 brancas e uma urna B contém 2 bolas vermelhas e 6 brancas. (i) Se uma bola é retirada de cada urna, qual a probabilidade de
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Uma segunda bola é então selecionada. Encontre a probabilidade (i) da segunda bola ser vermelha (41/72), (ii) de ambas as bolas serem da mesma cor.(13/36) 10-) A Indústria de Automóveis S.A possui 150 empregados, classificados classificados de acordo com a tabela a seguir: Idade(em anos) Sexo Total Masc Fem < 25 30 5 35 25 |---| 45 40 25 65 > 45 10 40 50 Total 80 70 150 Se um empregado é escolhido ao acaso, determine a probabilidade probabilidade dos seguintes eventos: A : O empregado tem mais de 45 anos (33,33%) B : O empregado tem idade idade igual ou superior superior a 25 anos (76,66%) C : O empregado empregado é do sexo masculino masculino (53,33%) D : O empregado empregado tem menos de 25 anos (23,33%) P(D/C) = ? (37,50%)
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