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?
- $3/23$
- $6/23$
- $3/10$
- $3/5$
My attempt :
Using bayes theorem :
Required probability is
$= \frac{1/3*0.3}{1/3*0.5+1/3*0.3+1/3*0.2} = \frac{3}{10}$
But, somewhere answer is given $\frac{6}{23}$
Can you explain in formal way, please ?
Best Answer
Suppose there are ten families. Then five families have three children, which is 15 children; three families have two, which is six more; and the other two families have one child each, for a total of 23 children.