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In a population of N families, 50% of the families have three children, 30% of the families have two children and the remaining families have one child. What is the probability that a randomly picked child belongs to a family with two children?

1. 3/23
2. 6/23
3. 3/10
4. 3/5

Let N be the total number of families.

Number of children in a  family of 3 children = (N/2) * 3

Number of children in a  family of 2 children = (3N /10) * 2

Number of children in a  family of 1 child = (N/5) * 1

Probability = Favorable case / Total cases

=( (3/10)*2 ) / ( (1/2)*3 + (3/10)*2 +1/5 )

= 6/23
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@ Prateeksha ma'am ,
Why Bayes' theorem is not applied here???
Exactly..

Plz someone solve using Bayes' theorem..
What is the requirement for conditional probability in this question?

Suppose,

50 families have three children(150 child's),

30 families have two children(60 childs)

and remaining 20 families have one child(20 childs).

So ,Probability that a randomly picked child belongs to a family with two children =

$\frac{60}{150+60+20}$=$\frac{60}{230}$=$\frac{6}{23}$