91Ó°ÊÓ

A family has $j$children with $p_j$probability , where localid="1646821977097" $p_1=.1,p_2=.25,p_3=.35,p_4=.3$. A child from this family is randomly chosen. Given that this child is the eldest child in the family.

Condition is only$1$child.

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,

$P\left(Eâˆ\pounds F_1\right)=1$

$P\left(Eâˆ\pounds F_2\right)=\frac{1}{2}$

$P\left(Eâˆ\pounds F_3\right)=\frac{1}{3}$

$P\left(Eâˆ\pounds F_4\right)=\frac{1}{4}$

So the Bayes's theorem, tells that,

$P\left(F_jâˆ\pounds E\right)=\frac{P\left(E{F}_j\right)}{P\left(E\right)}$

$=\frac{P\left(Eâˆ\pounds F_j\right)P\left(F_j\right)}{\underset{j}{∑} P\left(Eâˆ\pounds F_j\right)P\left(F_j\right)}$

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

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

$P\left(F_1âˆ\pounds E\right)=\frac{P\left(Eâˆ\pounds F_1\right)P\left(F_1\right)}{\underset{j}{∑} P\left(Eâˆ\pounds F_j\right)P\left(F_j\right)}$

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

$=0.24$

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

A family has $j$children with probability $p_j$, where localid="1646822037449" $p_1=.1,p_2=.25,p_3=.35,p_4=.3$. A child from this family is randomly chosen. Given that this child is the eldest child in the family,

Condition is$4$children.

The conditional probability that the family has four children is,

$P\left(F_4âˆ\pounds E\right)=\frac{P\left(Eâˆ\pounds F_4\right)P\left(F_4\right)}{\underset{j}{∑} P\left(Eâˆ\pounds F_j\right)P\left(F_j\right)}$

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

$=0.18$

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