A e B são casados. Têm dois filhos. Um deles é uma rapariga. Qual é a probabilidade de o outro filho ser também uma rapariga?
Este é um problema clássico de probabilidade. O que torna esta pergunta confusa é o facto de a questão ser ambígua, uma vez que pode não existir “a outra criança”.
Se um casal tiver dois filhos, ligue então $\mathcal{X}$ e $\mathcal{Y}$. Quando se diz “um deles é uma menina”, duas interpretações possíveis são:
#1 Alguém me disse que o casal tem pelo menos uma rapariga, mas não sei se ambas as crianças são raparigas, se $\mathcal{X}$ é uma rapariga, ou se $\mathcal{Y}$ é uma rapariga; os casos possíveis são GG, GB, BG. Quero calcular a probabilidade de ambos os filhos do casal serem raparigas (GG).
ou
#2 Alguém me disse que $\mathcal{X}$ é uma rapariga, mas não sei se $\mathcal{Y}$ é um rapaz ou uma rapariga; os casos possíveis são GG, GB, BG. Quero calcular a probabilidade de $\mathcal{Y}$ ser também uma rapariga (GG).
Agora, pode responder a cada uma destas perguntas usando, por exemplo, expressões para probabilidades condicionais (as respostas às interpretações #1 e #2 são 1/3 e 1/2, respectivamente). No entanto, se ficar com uma sensação de “claro, vejo como pode obter esse resultado mas porque é que a minha resposta não está correcta” leia em.
Interpretação #1:
(Esta é a que eu acho interessantemente confusa.)
Quando se pensa pela primeira vez no problema, pode-se usar o seguinte raciocínio (incorrecto):
Eu sei que pelo menos uma das crianças é uma menina, qual é a probabilidade de a outra criança (a criança de sexo desconhecido) ser também uma menina? bem, não deveria ser 1/2? Quer dizer, a outra criança pode ser um menino ou uma menina, independentemente… certo?
Um dos problemas com este raciocínio é que não existe “a criança de sexo desconhecido” (portanto, não existe “a outra criança”). De facto, não se conhece nenhum dos géneros das crianças (nesta interpretação, ambas as crianças têm o sexo desconhecido!). Tudo o que sabe é que há uma rapariga algures, mas não sabe qual das duas crianças é a rapariga. Como pode esperar calcular a probabilidade de a outra criança (a criança de sexo desconhecido) ser uma rapariga se não existe tal coisa como “a criança de sexo desconhecido”? De facto, se por acaso souber que $\mathcal{X}$ (ou $\mathcal{Y}$) é uma rapariga então está a utilizar uma interpretação diferente da questão (interpretação #2).
O outro problema com este raciocínio está relacionado com a independência. Como as pessoas têm salientado, dada a suposição global adicional de que há pelo menos uma rapariga, “$\mathcal{X} Isto é fácil de ver visto que, por exemplo, se $\mathcal{X}$ for um rapaz, então $\mathcal{Y}$ é forçado a ser uma rapariga com probabilidade 1 (e não 1/2). Esta falta de independência é a razão pela qual não se pode dizer que uma das duas crianças é um rapaz ou uma rapariga independentemente / independentemente da outra.
Formal Derivação de Resposta à Interpretação #1:
Usando a fórmula das probabilidades condicionais que temos\begin{align*}P\big(big(big=ambas são raparigas) {No mínimo uma rapariga)_&=frac{P\big(big(big(big=ambas são raparigas) {big)cap)big(big(big=ambas são raparigas) Pelo menos uma raparigaig)big)big)big(texto{No mínimo uma raparigaig)big)&=frac{Pig(texto{ambos are Girls}big){P{P&= {1/4}{3/4}=frac{1}{3}frac{1}{3}.{\i1}end{alinhamento*}
Interpretação #2:
Não creio que haja algo de confuso nesta interpretação. Sabe que o $\mathcal{X}$ é uma rapariga e $\mathcal{Y}$ pode ser uma rapariga ou um rapaz independentemente do $\mathcal{X}$. Neste caso, as variáveis aleatórias “$\mathcal{X}
Derivação formal de resposta à interpretação #2:
Usar a independência do sexo da segunda criança em relação ao sexo da primeira criança:
p>p>begin{alinhamento*}P\big(texto{Segundo é uma Rapariga} {segundo é uma Rapariga} {primeiro é uma Rapariga} =P\big(texto é uma Rapariga)=frac{1}frac{2}.{\i1}end{align*}h2>Segunda pergunta:
p>Agora A e B têm 4 filhos e todos eles são rapazes. B está grávida. Então, qual é a probabilidade de A e B serem dotados com uma menina? Será 1/2 ou haverá alguma probabilidade condicional?
A resposta é que o sexo da próxima criança é independente dos quatro primeiros géneros de crianças (por suposição), isto é porque não há nenhuma suposição global adicional envolvendo a próxima criança. A resposta é então simplesmente $1/2$. (Isto é exactamente como na Interpretação #2.)
s gender” e “$\mathcal{Y} This is easy to see since, for example, if $\mathcal{X}$ is a boy then $\mathcal{Y}$ is forced to be a girl with probability 1 (not 1/2). This lack of independence is the reason why you can’t say that one of the two children is a boy or girl independently/regardless of the other.
Formal Derivation of Answer to Interpretation #1:
Using the formula for conditional probabilities we have\begin{align*}P\big(\text{Both are Girls } \big| \text{ At Least One Girl}\big)&=\frac{P\Big(\big(\text{Both are Girls}\big)\cap\big(\text{At Least One Girl}\big)\Big)}{P\big(\text{At Least One Girl}\big)}\\&=\frac{P\big(\text{Both are Girls}\big)}{P\big(\text{At Least One Girl}\big)}\\&= \frac{1/4}{3/4}=\frac{1}{3}.\end{align*}
Interpretation #2:
I don’t think there is anything confusing about this interpretation. You know the $\mathcal{X}$ is a girl and $\mathcal{Y}$ can be a girl or a boy independently of $\mathcal{X}$. In this case, the random variables “$\mathcal{X}$’s gender” and “$\mathcal{Y}$’s gender” are not linked by any added overarching assumption and are independent.
Formal Derivation of Answer to Interpretation #2:
Using the independence of the second child’s gender relative to the first child’s gender:
\begin{align*}P\big(\text{Second is a Girl } \big| {\text{ First is a Girl}}\big)=P\big(\text{Second is a Girl}\big)=\frac{1}{2}.\end{align*}
Regarding the second question:
Now A and B have 4 children and all of them are boys. B is pregnant. So what is the probability that A and B are gifted with a baby girl? Is it 1/2 or there will be some conditional probability?
The answer is that the next child’s gender is independent of the first four children’s genders (by assumption), this is because there is no added overarching assumption involving the next child. The answer is then simply $1/2$. (This is just like in Interpretation #2.)
s gender” não são variáveis aleatórias independentes. This is easy to see since, for example, if $\mathcal{X}$ is a boy then $\mathcal{Y}$ is forced to be a girl with probability 1 (not 1/2). This lack of independence is the reason why you can’t say that one of the two children is a boy or girl independently/regardless of the other.
Formal Derivation of Answer to Interpretation #1:
Using the formula for conditional probabilities we have\begin{align*}P\big(\text{Both are Girls } \big| \text{ At Least One Girl}\big)&=\frac{P\Big(\big(\text{Both are Girls}\big)\cap\big(\text{At Least One Girl}\big)\Big)}{P\big(\text{At Least One Girl}\big)}\\&=\frac{P\big(\text{Both are Girls}\big)}{P\big(\text{At Least One Girl}\big)}\\&= \frac{1/4}{3/4}=\frac{1}{3}.\end{align*}
Interpretation #2:
I don’t think there is anything confusing about this interpretation. You know the $\mathcal{X}$ is a girl and $\mathcal{Y}$ can be a girl or a boy independently of $\mathcal{X}$. In this case, the random variables “$\mathcal{X}$’s gender” and “$\mathcal{Y}$’s gender” are not linked by any added overarching assumption and are independent.
Formal Derivation of Answer to Interpretation #2:
Using the independence of the second child’s gender relative to the first child’s gender:
\begin{align*}P\big(\text{Second is a Girl } \big| {\text{ First is a Girl}}\big)=P\big(\text{Second is a Girl}\big)=\frac{1}{2}.\end{align*}
Regarding the second question:
Now A and B have 4 children and all of them are boys. B is pregnant. So what is the probability that A and B are gifted with a baby girl? Is it 1/2 or there will be some conditional probability?
The answer is that the next child’s gender is independent of the first four children’s genders (by assumption), this is because there is no added overarching assumption involving the next child. The answer is then simply $1/2$. (This is just like in Interpretation #2.)
s gender” e “$\mathcal{Y}
Formal Derivation of Answer to Interpretation #2:
Using the independence of the second child’s gender relative to the first child’s gender:
\begin{align*}P\big(\text{Second is a Girl } \big| {\text{ First is a Girl}}\big)=P\big(\text{Second is a Girl}\big)=\frac{1}{2}.\end{align*}
Regarding the second question:
Now A and B have 4 children and all of them are boys. B is pregnant. So what is the probability that A and B are gifted with a baby girl? Is it 1/2 or there will be some conditional probability?
The answer is that the next child’s gender is independent of the first four children’s genders (by assumption), this is because there is no added overarching assumption involving the next child. The answer is then simply $1/2$. (This is just like in Interpretation #2.)
s gender” e “$\mathcal{Y} This is easy to see since, for example, if $\mathcal{X}$ is a boy then $\mathcal{Y}$ is forced to be a girl with probability 1 (not 1/2). This lack of independence is the reason why you can’t say that one of the two children is a boy or girl independently/regardless of the other.
Formal Derivation of Answer to Interpretation #1:
Using the formula for conditional probabilities we have\begin{align*}P\big(\text{Both are Girls } \big| \text{ At Least One Girl}\big)&=\frac{P\Big(\big(\text{Both are Girls}\big)\cap\big(\text{At Least One Girl}\big)\Big)}{P\big(\text{At Least One Girl}\big)}\\&=\frac{P\big(\text{Both are Girls}\big)}{P\big(\text{At Least One Girl}\big)}\\&= \frac{1/4}{3/4}=\frac{1}{3}.\end{align*}
Interpretation #2:
I don’t think there is anything confusing about this interpretation. You know the $\mathcal{X}$ is a girl and $\mathcal{Y}$ can be a girl or a boy independently of $\mathcal{X}$. In this case, the random variables “$\mathcal{X}$’s gender” and “$\mathcal{Y}$’s gender” are not linked by any added overarching assumption and are independent.
Formal Derivation of Answer to Interpretation #2:
Using the independence of the second child’s gender relative to the first child’s gender:
\begin{align*}P\big(\text{Second is a Girl } \big| {\text{ First is a Girl}}\big)=P\big(\text{Second is a Girl}\big)=\frac{1}{2}.\end{align*}
Regarding the second question:
Now A and B have 4 children and all of them are boys. B is pregnant. So what is the probability that A and B are gifted with a baby girl? Is it 1/2 or there will be some conditional probability?
The answer is that the next child’s gender is independent of the first four children’s genders (by assumption), this is because there is no added overarching assumption involving the next child. The answer is then simply $1/2$. (This is just like in Interpretation #2.)
s gender” não são variáveis aleatórias independentes. This is easy to see since, for example, if $\mathcal{X}$ is a boy then $\mathcal{Y}$ is forced to be a girl with probability 1 (not 1/2). This lack of independence is the reason why you can’t say that one of the two children is a boy or girl independently/regardless of the other.
Formal Derivation of Answer to Interpretation #1:
Using the formula for conditional probabilities we have\begin{align*}P\big(\text{Both are Girls } \big| \text{ At Least One Girl}\big)&=\frac{P\Big(\big(\text{Both are Girls}\big)\cap\big(\text{At Least One Girl}\big)\Big)}{P\big(\text{At Least One Girl}\big)}\\&=\frac{P\big(\text{Both are Girls}\big)}{P\big(\text{At Least One Girl}\big)}\\&= \frac{1/4}{3/4}=\frac{1}{3}.\end{align*}
Interpretation #2:
I don’t think there is anything confusing about this interpretation. You know the $\mathcal{X}$ is a girl and $\mathcal{Y}$ can be a girl or a boy independently of $\mathcal{X}$. In this case, the random variables “$\mathcal{X}$’s gender” and “$\mathcal{Y}$’s gender” are not linked by any added overarching assumption and are independent.
Formal Derivation of Answer to Interpretation #2:
Using the independence of the second child’s gender relative to the first child’s gender:
\begin{align*}P\big(\text{Second is a Girl } \big| {\text{ First is a Girl}}\big)=P\big(\text{Second is a Girl}\big)=\frac{1}{2}.\end{align*}
Regarding the second question:
Now A and B have 4 children and all of them are boys. B is pregnant. So what is the probability that A and B are gifted with a baby girl? Is it 1/2 or there will be some conditional probability?
The answer is that the next child’s gender is independent of the first four children’s genders (by assumption), this is because there is no added overarching assumption involving the next child. The answer is then simply $1/2$. (This is just like in Interpretation #2.)
s gender” não estão ligadas por nenhuma suposição global adicional e são independentes.
Formal Derivation of Answer to Interpretation #2:
Using the independence of the second child’s gender relative to the first child’s gender:
\begin{align*}P\big(\text{Second is a Girl } \big| {\text{ First is a Girl}}\big)=P\big(\text{Second is a Girl}\big)=\frac{1}{2}.\end{align*}
Regarding the second question:
Now A and B have 4 children and all of them are boys. B is pregnant. So what is the probability that A and B are gifted with a baby girl? Is it 1/2 or there will be some conditional probability?
The answer is that the next child’s gender is independent of the first four children’s genders (by assumption), this is because there is no added overarching assumption involving the next child. The answer is then simply $1/2$. (This is just like in Interpretation #2.)
s gender” e “$\mathcal{Y} This is easy to see since, for example, if $\mathcal{X}$ is a boy then $\mathcal{Y}$ is forced to be a girl with probability 1 (not 1/2). This lack of independence is the reason why you can’t say that one of the two children is a boy or girl independently/regardless of the other.
Formal Derivation of Answer to Interpretation #1:
Using the formula for conditional probabilities we have\begin{align*}P\big(\text{Both are Girls } \big| \text{ At Least One Girl}\big)&=\frac{P\Big(\big(\text{Both are Girls}\big)\cap\big(\text{At Least One Girl}\big)\Big)}{P\big(\text{At Least One Girl}\big)}\\&=\frac{P\big(\text{Both are Girls}\big)}{P\big(\text{At Least One Girl}\big)}\\&= \frac{1/4}{3/4}=\frac{1}{3}.\end{align*}
Interpretation #2:
I don’t think there is anything confusing about this interpretation. You know the $\mathcal{X}$ is a girl and $\mathcal{Y}$ can be a girl or a boy independently of $\mathcal{X}$. In this case, the random variables “$\mathcal{X}$’s gender” and “$\mathcal{Y}$’s gender” are not linked by any added overarching assumption and are independent.
Formal Derivation of Answer to Interpretation #2:
Using the independence of the second child’s gender relative to the first child’s gender:
\begin{align*}P\big(\text{Second is a Girl } \big| {\text{ First is a Girl}}\big)=P\big(\text{Second is a Girl}\big)=\frac{1}{2}.\end{align*}
Regarding the second question:
Now A and B have 4 children and all of them are boys. B is pregnant. So what is the probability that A and B are gifted with a baby girl? Is it 1/2 or there will be some conditional probability?
The answer is that the next child’s gender is independent of the first four children’s genders (by assumption), this is because there is no added overarching assumption involving the next child. The answer is then simply $1/2$. (This is just like in Interpretation #2.)
s gender” não são variáveis aleatórias independentes. This is easy to see since, for example, if $\mathcal{X}$ is a boy then $\mathcal{Y}$ is forced to be a girl with probability 1 (not 1/2). This lack of independence is the reason why you can’t say that one of the two children is a boy or girl independently/regardless of the other.
Formal Derivation of Answer to Interpretation #1:
Using the formula for conditional probabilities we have\begin{align*}P\big(\text{Both are Girls } \big| \text{ At Least One Girl}\big)&=\frac{P\Big(\big(\text{Both are Girls}\big)\cap\big(\text{At Least One Girl}\big)\Big)}{P\big(\text{At Least One Girl}\big)}\\&=\frac{P\big(\text{Both are Girls}\big)}{P\big(\text{At Least One Girl}\big)}\\&= \frac{1/4}{3/4}=\frac{1}{3}.\end{align*}
Interpretation #2:
I don’t think there is anything confusing about this interpretation. You know the $\mathcal{X}$ is a girl and $\mathcal{Y}$ can be a girl or a boy independently of $\mathcal{X}$. In this case, the random variables “$\mathcal{X}$’s gender” and “$\mathcal{Y}$’s gender” are not linked by any added overarching assumption and are independent.
Formal Derivation of Answer to Interpretation #2:
Using the independence of the second child’s gender relative to the first child’s gender:
\begin{align*}P\big(\text{Second is a Girl } \big| {\text{ First is a Girl}}\big)=P\big(\text{Second is a Girl}\big)=\frac{1}{2}.\end{align*}
Regarding the second question:
Now A and B have 4 children and all of them are boys. B is pregnant. So what is the probability that A and B are gifted with a baby girl? Is it 1/2 or there will be some conditional probability?
The answer is that the next child’s gender is independent of the first four children’s genders (by assumption), this is because there is no added overarching assumption involving the next child. The answer is then simply $1/2$. (This is just like in Interpretation #2.)
0 comentários