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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. $3/5$

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 ?

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    $\begingroup$ Although the example excludes this, in reality some positive percentage of families has no children at all. What is the probability that a random child comes from such a family? Now modify your example to include 10% of childless families (remove them from the 50%, say) and do your Bayesian computation for the probability of a childless family. Compare. $\endgroup$ – Marc van Leeuwen Nov 18 '15 at 14:01
  • $\begingroup$ Where did the 1/3 come from in your formula? $\endgroup$ – Rick Nov 18 '15 at 17:49
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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.

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  • $\begingroup$ Yes , that's correct ,but what's wrong with my approach ? $\endgroup$ – ً ً Nov 18 '15 at 11:48
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    $\begingroup$ You have made the families equally likely, rather than the children equally likely. If the children are equally likely, then a three-child family is three times more likely to be chosen than a one-child family. $\endgroup$ – Empy2 Nov 18 '15 at 11:51
  • $\begingroup$ Yes. I got it thanks for explanation. $\endgroup$ – ً ً Nov 18 '15 at 12:21
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Let $N$ be the number of families. Then $50\%$ of N have $3$ children, which means that there are $50\%N\cdot 3=\frac{15}{10}N$ children in these families and then $30\%N\cdot 2=\frac{6}{10}N$ children in families with $2$ children and $20\%N\cdot1=\frac{2}{10}N$ children in families of $1$ child. This makes a total of $$N(50\%\cdot3+30\%\cdot2+20\%\cdot1)=\frac{23}{10}N$$ children. So now pick a child. $$P(A_2)=\frac{\frac{6}{10}N}{\frac{23}{10}N}=\frac{6}{23}$$ where $A_2$ denotes the required event that the child belongs to a family with $2$ children.

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  • $\begingroup$ How do you find $\frac 6 {10}$N? $\endgroup$ – Deliss Nov 18 '15 at 12:14
  • $\begingroup$ $30\%\cdot2N= \frac{30}{100}\cdot 2N=\frac{60}{100}N$ $\endgroup$ – Jimmy R. Nov 18 '15 at 12:16
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Multiply the percents of families by N to get number of families:

  • 3 Children: 0.5*N families
  • 2 Children: 0.3*N families
  • 1 Child: 0.2*N families

Then, multiply the number of families by the number of children in each family:

  • 1.5*N children in three-child families
  • 0.6*N children in two-child families
  • 0.2*N children in one-child families

There are (1.5 + 0.6 + 0.2) * N = 2.3*N children total, of which 0.6*N come from two-child families. This reduces to 6/23.

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