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If every family in the village has two children, what proportion of all sons are older sons is?

Events:

M - a person of marriage age is male

F=M^c - a person of marriage age is female

E - a person is the oldest/elder son

S_i - a person is from a family with i = $0,1,2,3$sons.

proportion of x - probability that randomly chosen person is x:

P(M)=$\frac{1}{2}$

a) Each family has two children.

Start with the definition of conditional probability ,note that from the definition of events, E,E$âˆ\copyright M=E$

$P(Eâˆ\pounds M)=\frac{P\left(EM\right)}{P\left(M\right)}$

$=\frac{P\left(E\right)}{P\left(M\right)}$

$ForcalculationofP\left(E\right),usetheBayesformula,withsystemS0,S1andS2thatmakeuptheoutcomespacewithproportions:$

$P\left(S_0\right)=P("F,F")=P\left(F\right)P\left(F\right)=\frac{1}{4}$

$P\left(S_1\right)=P("M,F"or"F,M")=P\left(M\right)P\left(F\right)+P\left(F\right)P\left(M\right)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

$P\left(S_2\right)=P("M,M")=P\left(M\right)P\left(M\right)=\frac{1}{4}$

The Bayes formula now yields.

$P\left(E\right)=P(Eâˆ\pounds S_0)P\left(S_0\right)+P(Eâˆ\pounds S_1)P\left(S_1\right)+P(Eâˆ\pounds S_2)P\left(S_2\right)$

$=0â‹\dots \frac{1}{4}+\frac{1}{2}â‹\dots \frac{1}{2}+\frac{1}{2}â‹\dots \frac{1}{4}$

$=\frac{3}{8}$

So the final conditional probability is :

$P(Eâˆ\pounds M)=\frac{\frac{3}{8}}{\frac{1}{2}}=\frac{3}{4}$

If every family in the village has two children, the proportion of all sons are older sons is$\frac{3}{4}$

If every family in the village has three children, what proportion of all sons are eldest sons?

b) Each family has three children

starts as a:

$P(Eâˆ\pounds M)=\frac{P\left(EM\right)}{P\left(M\right)}$

$=\frac{P\left(E\right)}{P\left(M\right)}$

For calculation of P(E), use the Bayes formula, with system S0, S1, S2 and S3 that make up the outcome space with proportions, calculated as in a):

$P\left(S_0\right)=P("F,F,F")=P\left(F\right)P\left(F\right)P\left(F\right)=\frac{1}{8}$

$P\left(S_1\right)=…=\frac{3}{8}$

$P\left(S_2\right)=…=\frac{3}{8}$

$P\left(S_3\right)=…=\frac{1}{8}$

The Bayes formula now yields.

$P\left(E\right)=P(Eâˆ\pounds S_0)P\left(S_0\right)+P(Eâˆ\pounds S_1)P\left(S_1\right)+P(Eâˆ\pounds S_2)P\left(S_2\right)+P(Eâˆ\pounds S_3)P\left(S_3\right)$

$=0â‹\dots \frac{1}{8}+\frac{1}{3}â‹\dots \frac{3}{8}+\frac{1}{3}â‹\dots \frac{3}{8}+\frac{1}{3}â‹\dots \frac{1}{8}$

$=\frac{7}{24}$

$\text{One of the three children is the oldest son, that is why}P(Eâˆ\pounds S_1)=P(Eâˆ\pounds S_2)=P(Eâˆ\pounds S_3)=1/3$

So the final conditional probability is,

$P(Eâˆ\pounds M)=\frac{\frac{7}{24}}{\frac{1}{2}}=\frac{7}{12}$

If every family in the village has three children, the proportion of all sons are eldest sons is$\frac{7}{12}$