Q.3.32 - A First Course in Probability

Question

A family has $j$ children with probability $p_j$, where $p_1=.1, p_2=.25, p_3=.35, p_4=.3$. A child from this family is randomly chosen. Given that this child is the eldest child in the family, find the conditional probability that the family has

(a) only $1$ child;

(b) $4$ children.

Step-by-Step Solution

Answer

(a) The probability of having a family with only one child is $0.24$.

(b) The probability of having the family with four children is $0.18$.

Step 1: given information (Part a)

A family has $j$ children with $p_j$ probability , where $p_1=.1, p_2=.25, p_3=.35, p_4=.3$. A child from this family is randomly chosen. Given that this child is the eldest child in the family.

Condition is only $1$ child.

Step 2: Solution (Part a)

Let $F_j$ is event that the family has $j$children.

$P\left(F_1\right)=0.1$

$P\left(F_2\right)=0.25$

$P\left(F_3\right)=0.35$

$P\left(F_4\right)=0.3$

A child from this family is haphazardly selected. Given that this child is the elder child in the family. Let $E$ be the possibility the child selected is the eldest.

The probabilities of chose one children from the family of $1,2,3,4$ are,

$P\left(E\mid F_1\right)=1$

$P\left(E\mid F_2\right)=\frac{1}{2}$

$P\left(E\mid F_3\right)=\frac{1}{3}$

$P\left(E\mid F_4\right)=\frac{1}{4}$

Step 3: Final solution (Part a)

So the Bayes's theorem, tells that,

$P\left(F_j\mid E\right)=\frac{P\left(E{F}_j\right)}{P\left(E\right)}$

$=\frac{P\left(E\mid F_j\right)P\left(F_j\right)}{\sum _j P\left(E\mid F_j\right)P\left(F_j\right)}$

$=\frac{P\left(E\mid F_j\right)P\left(F_j\right)}{0.1\left(\frac{1}{1}\right)+0.25\left(\frac{1}{2}\right)+0.35\left(\frac{1}{3}\right)+0.3\left(\frac{1}{4}\right)}$

The conditional probability that the family has only one child is,

$P\left(F_1\mid E\right)=\frac{P\left(E\mid F_1\right)P\left(F_1\right)}{\sum _j P\left(E\mid F_j\right)P\left(F_j\right)}$

$=\frac{\frac{1}{1}\left(0.1\right)}{0.1\left(\frac{1}{1}\right)+0.25\left(\frac{1}{2}\right)+0.35\left(\frac{1}{3}\right)+0.3\left(\frac{1}{4}\right)}$

$=0.24$

Step 4: Final answer (Part a)

The probability of having the family with only one child is $0.24$.

Step 5: Given information (Part b)

A family has $j$ children with probability $p_j$, where $p_1=.1, p_2=.25, p_3=.35, p_4=.3$. A child from this family is randomly chosen. Given that this child is the eldest child in the family,

Condition is $4$ children.

Step 6: Solution (Part b)

The conditional probability that the family has four children is,

$P\left(F_4\mid E\right)=\frac{P\left(E\mid F_4\right)P\left(F_4\right)}{\sum _j P\left(E\mid F_j\right)P\left(F_j\right)}$

$=\frac{\frac{1}{4}\left(0.3\right)}{0.1\left(\frac{1}{1}\right)+0.25\left(\frac{1}{2}\right)+0.35\left(\frac{1}{3}\right)+0.3\left(\frac{1}{4}\right)}$

$=0.18$

Step 7: Final answer (Part b)

The probability of having the family that has four children is $0.18$.