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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. $\left(\dfrac{3}{23}\right)$
  2. $\left(\dfrac{6}{23}\right)$
  3. $\left(\dfrac{3}{10}\right)$
  4. $\left(\dfrac{3}{5}\right)$
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Answer is (B) $\dfrac{6}{23}$

Let $N$ be the total number of families.

Number of children in a  family of $3$ children $=\left(\dfrac{N}{2}\right)\times {3}$

Number of children in a  family of $2$ children $=\left(\dfrac{3N}{10}\right)\times {2}$

Number of children in a  family of $1$ child $=\left(\dfrac{N}{5}\right)\times {1}$

Probability $=\dfrac{ \text{Favorable cases}} {\text{Total cases}}$

$\quad =\dfrac{\left(\dfrac{3}{10}\right)\times {2}}{\left[\left(\dfrac{1}{2}\right)\times {3} + \left(\dfrac{3}{10}\right)\times {2} +\dfrac{1}{5}\right]} =\dfrac{6}{23}.$

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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 =

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

Answer:

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