Conditional Probability: Birth rank of children in randomly chosen families

Question

(BH 4.7) A certain small town, whose population consists of 100 families, has 30 families with 1 child, 50 families with 2 children, and 20 families with 3 children. The birth rank of one of these children is 1 if the child is the firstborn, 2 if the child is the second-born, and 3 if the child is the third-born.
(a) A random family is chosen (with equal probabilities), and then a random child within that family is chosen (with equal probabilities). Find the PMF, mean, and variance of the child’s birth rank.
(b) A random child is chosen in the town (with equal probabilities). Find the PMF,mean, and variance of the child’s birth rank.

It seems to me (a) and (b) are one and the same. $\newcommand{\P}{\mathbb{P}}$

If we assume that finding the PMF is pretty straightforward. You find $\P(X=x) = \sum_{T=1}^3 \P (X=x|T). \P(T), x=1(1)3$, where T = Family Type indicates number of children in a family and can be Type 1,Type 2, Type 3 with probabilities 30%, 50%, 20% respectively. $\P(X=x|T)$ would of course be $1, 1/2$ or $1/3$ depending on $T=1, 2$ or $3$.

Answer:

PMF (probabilities in percentage terms):

$\P(X=1) = 61.67% $

$\P(X=2) = 31.67%$

$\P(X=3) = 6.67%$

Could anyone please explain if the assumption that (a) and (b) are one and the same is correct and if not why not?

Edit : Following Mr. Graham Kemp's clarifications I retried the second part (b) and then realized when a child is getting chosen randomly, $\P(T)$ is getting weighted by the number of children each family has. So if we follow the above solution method we'd have (in case (b)):

Total number of children $= 1\cdot 30 + 2\cdot 50 + 3\cdot 20 = 190$

$\P(T=1) = \frac{1\cdot30}{190}$ $ = \frac{30}{190}$

$\P(T=2) = \frac{2\cdot50}{190}$ $ = \frac{100}{190}$

$\P(T=1) = \frac{3\cdot20}{190}$ $ = \frac{60}{190}$

Subsequently we can apply the method shown above with the same $\P(X=x|T)$ values $1,1/2$ and $1/3$.

Answer 1

Could anyone please explain if the assumption that (a) and (b) are one and the same is correct and if not why not?

It is not correct.

Let us look at a simpler scenario.

Take two families, one with three children and one with a single child.   Here there are four children: two with birth order 1, and one each with birth orders 2 and 3.

Case 1: First we select a family with probability 1/2, then from that family we select a child.   The pmf for birth order is: $$\mathsf P(X=x) = \begin{cases}\tfrac 12\cdot\tfrac 13+\tfrac 12\cdot1 & : x=1 \\ \tfrac 12\cdot\tfrac 13 &: x\in \{2, 3\}\end{cases}=\begin{cases}\tfrac 23 & : x=1 \\ \tfrac 16 &: x\in \{2, 3\}\end{cases}$$

Case 2: We randomly select a child without first filtering by family.

$$\mathsf P(Y=y) = \begin{cases}\tfrac 12 & : y=1 \\ \tfrac 14 & : y\in \{2,3\}\end{cases}$$

So clearly the different selection methods may yield different probability mass functions.

They are not equivalent.

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PS: Your calculations for "pick a family first" appear to be correct.   Now that you know the methods of selection is important, can you find the pmf for the other method?

Answer 2

Corrected answer for question (b):

Following Mr. Graham Kemp's clarifications I retried the second part (b) and then realized when a child is getting chosen randomly, $P(T)$ is getting weighted by the number of children each family has. So if we follow the above solution method we'd have (in case (b)):

Total number of children $= 1\cdot 30 + 2\cdot 50 + 3\cdot 20 = 190$

$P(T=1) = \frac{1\cdot30}{190}$ $ = \frac{30}{190}$

$P(T=2) = \frac{2\cdot50}{190}$ $ = \frac{100}{190}$

$P(T=1) = \frac{3\cdot20}{190}$ $ = \frac{60}{190}$

Subsequently we can apply the method shown above with the same $P(X=x|T)$ values $1,1/2$ and $1/3$.