Probability of number of children

Question

If a family has a probability of having children as follows :

$$p(1~\text{child}) = .45$$

$$p(2~\text{children})=.3$$

$$p(3~\text{children}) = .15$$

$$p(4~\text{children}) = .1$$

If a child is selected from the family and the selected child is the eldest what is the conditional probability this family has 1. - 1 child ? 2. - 4 children ?

Is my logic correct ? :

$$p(1~\text{child} \mid \text{eldest}) = .45 x 1 = .45$$

$$p(4~\text{children} \mid \text{eldest}) = .1 x \frac{1}{4} = .025$$

Now using Bayes theorem :

$$P(\text{eldest} \mid 1~\text{child}) = \frac{P(1~\text{child} \mid \text{eldest}) P(\text{eldest})} {P(1~\text{child})}$$

$$P(~\text{eldest} \mid 4~\text{children}) = \frac{P(4~\text{children} \mid \text{eldest}) P(\text{eldest})} {P(4~\text{children})}$$

Answer 1

I think you have reversed the order completely. Remember that the notation: $$ P(A|B) $$ denotes the probability of $A$ given $B$, so it is the relative size of the part of the event set $A$ that is contained within the event set $B$.

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Here is a diagram to support our understanding:

The $x$-axis is divided into portions $.45,\ .3,\ .15,\ .1$ corresponding to $1-4$ children. The horisontal lines divide each case into the relevant number of children so the first red box fills $1=100\%$ on the $y$-axis corrsponding to only 1 child. The last three small blue boxes and the small red box make up the case with 4 children. The question now becomes:

Given that we are in the red region (corresponding to the eldest child being chosen in either case) what is the likelihood that we are in the biggest red box (1 child) or the smallest red box (4 children)?

The areas of the individual red boxes are: $$ \begin{align} P(eldest\mid 1\ child)\times P(1\ child) &= 1\times .45 &&=.45 \\ P(eldest\mid 2\ children)\times P(2\ children)&= \frac12\times .3 &&=.15 \\ P(eldest\mid 3\ children)\times P(3\ children) &= \frac13\times .15 &&=.05 \\ P(eldest\mid 4\ children)\times P(4\ children) &= \frac14\times .1 &&=.025 \\ \end{align} $$ and then $P(eldest)$ is the sum of all those. So what the question asks of you is then to determine: $$ \begin{align} P(1\ child\mid eldest) &= \frac{P(eldest\mid 1\ child)P(1\ child)}{P(eldest)} \\ P(4\ children\mid eldest) &= \frac{P(eldest\mid 4\ children)P(4\ children)}{P(eldest)} \end{align} $$ which should be quite simple by now. You have the red boxes to compare to BTW.

Answer 2

I hope to give a little intuition in confirming your answer. The task is to determine the likelihood of some requested number of children in the family. In probability problems, I have to stop and think about what is absolutely certain and what is unknown. Given that you are told that you have selected the eldest child, then the absolute certainty is that $P(e)=1$. Using Bayes, for any n children,

$$ P(n|e) = \frac{P(e|n)P(n)}{P(e)} $$

The probabilities $P(e)$ and $P(n)$ are given and therefore fixed. $P(e|n)$ means, out of $n$ children, what are the odds you have the oldest? Only one can be the oldest (foregoing twins, and Siamese twins at that), that's $1/n$. Your 4 child example is correct: $P(4|e) = \frac{(1/4)0.1}{1} = 0.025$.

(BTW, to remember Bayes on a test, I think of "ABBA" repeating in the following six positions: $P(1|2)=\frac{P(3|4)P(5)}{P(6)}$ with positions 1..6 corresponding to A,B,B,A,A,B[,B,A])