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

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

by
10 21 37

even i solve it like this.

great method
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
yes

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 =

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

by
123 219 338

Nice approach
Thank you...
That is a better approach.