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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?

- $\left(\dfrac{3}{23}\right)$
- $\left(\dfrac{6}{23}\right)$
- $\left(\dfrac{3}{10}\right)$
- $\left(\dfrac{3}{5}\right)$

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Best answer

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

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

this question is a combination of Total Probability Theorem and bayes theorem . summation of all which is 2.3 is (Total Probability). where as in question randomly picked child belongs to a family with two children asked (which is one path out of three paths) so , for that we use Bayes Theorem,

= Contribution of path / Total

= 0.6 / 2.3

= 6 / 23 is the Answer

= Contribution of path / Total

= 0.6 / 2.3

= 6 / 23 is the Answer

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