Question

In a certain village, it is traditional for the eldest son (or the older son in a two-son family) and his wife to be responsible for taking care of his parents as they age. In recent years, however, the women of this village, not wanting that responsibility, have not looked favorably upon marrying an eldest son.

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

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

Assume that each child is, independently, equally likely to be either a boy or a girl.

Short answer

From the given information,

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

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

Step-by-step solution

Given Information (part a)

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

Explanation (part a)

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$\cap M=E$

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

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

Explanation (part a)

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

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

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

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

The Bayes formula now yields.

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

$=0\cdot \frac{1}{4}+\frac{1}{2}\cdot \frac{1}{2}+\frac{1}{2}\cdot \frac{1}{4}$

$=\frac{3}{8}$

So the final conditional probability is :

$P(E\mid M)=\frac{\frac{3}{8}}{\frac{1}{2}}=\frac{3}{4}$

Step 4: Final Answer (part a)

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

Given information (part b)

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

Explanation (part b)

b) Each family has three children

starts as a:

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

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

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):

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

$P\left(S_1\right)=\dots =\frac{3}{8}$

$P\left(S_2\right)=\dots =\frac{3}{8}$

$P\left(S_3\right)=\dots =\frac{1}{8}$

Explanation (part b)

The Bayes formula now yields.

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

$=0\cdot \frac{1}{8}+\frac{1}{3}\cdot \frac{3}{8}+\frac{1}{3}\cdot \frac{3}{8}+\frac{1}{3}\cdot \frac{1}{8}$

$=\frac{7}{24}$

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

So the final conditional probability is,

$P(E\mid M)=\frac{\frac{7}{24}}{\frac{1}{2}}=\frac{7}{12}$

Final Answer (part b)

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