Question

Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then find the conditional probability that all children are girls given that at least two are girls?
(a) $\dfrac{1}{11}$,
(b) $\dfrac{1}{17}$,
(c) $\dfrac{1}{10}$,
(d) $\dfrac{1}{12}$.

Answer 1

We start solving the problem by finding the probability that a new born child is boy and probability that a new born child is boy. We then find the probability that at least two of the four new born are girls and we find the probability that all the four new born are girls. We then divide these two values to get the required result.

Complete step by step answer:
According to the problem, it is given that the new born child is equally likely to be a boy or girl and two families have two children each. We need to find the conditional probability that all children are girls given that at least two are girls.
We have the probability that the new born child is a boy = probability that the new born child is a girl.
We know that the sum of total probability of a new child being born is 1. Let us assume the probability that the new born child is a boy be $P\left( B \right)$ and probability that the new born child is a girl be $P\left( G \right)$.
We have $P\left( B \right)=P\left( G \right)$ and $P\left( B \right)+P\left( G \right)=1$.
$\Rightarrow P\left( B \right)+P\left( B \right)=1$.
$\Rightarrow 2P\left( B \right)=1$.
$\Rightarrow P\left( B \right)=\dfrac{1}{2}$.
$\Rightarrow P\left( G \right)=\dfrac{1}{2}$.
Let us assume event A to be such that all the four new born children are girls and C be such that at least two of the four new born are girls.
We need to find the conditional probability $P\left( A|C \right)$.
We know that conditional probability $P\left( A|C \right)=\dfrac{P\left( A\cap C \right)}{P\left( C \right)}$ ---(1).
Let us find the value of $P\left( C \right)$. We have$P\left( C \right)=P\left( \text{atleast two of the new born children are girls} \right)$.
$P\left( C \right)=1-P\left( \text{all new born children are boys} \right)-P\left( \text{exactly one of the new born children is girl} \right)$.
We know that the event of a new born child is independent. So, let us write the possibilities for each event.
$P\left( C \right)=1-P\left( BBBB \right)-\left( P\left( GBBB \right)+P\left( BGBB \right)+P\left( BBGB \right)+P\left( BBBG \right) \right)$ ---(2).
Let us find the values of $P\left( BBBB \right)$, $P\left( GBBB \right)$, $P\left( BGBB \right)$, $P\left( BBGB \right)$, $P\left( BBBG \right)$ and substitute in equation (2).
We know that the probability of two independent events S and R is $P\left( SR \right)=P\left( S \right).P\left( R \right)$.
So, we have $P\left( BBBB \right)=P\left( B \right)\times P\left( B \right)\times P\left( B \right)\times P\left( B \right)$.
$\Rightarrow P\left( BBBB \right)=\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}$.
$\Rightarrow P\left( BBBB \right)=\dfrac{1}{16}$.
We have $P\left( GBBB \right)=P\left( G \right)\times P\left( B \right)\times P\left( B \right)\times P\left( B \right)$.
$\Rightarrow P\left( GBBB \right)=\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}$.
$\Rightarrow P\left( GBBB \right)=\dfrac{1}{16}$.
We have $P\left( BGBB \right)=P\left( B \right)\times P\left( G \right)\times P\left( B \right)\times P\left( B \right)$.
$\Rightarrow P\left( BGBB \right)=\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}$.
$\Rightarrow P\left( BGBB \right)=\dfrac{1}{16}$.
We have $P\left( BBGB \right)=P\left( B \right)\times P\left( B \right)\times P\left( G \right)\times P\left( B \right)$.
$\Rightarrow P\left( BBGB \right)=\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}$.
$\Rightarrow P\left( BBGB \right)=\dfrac{1}{16}$.
We have $P\left( BBBG \right)=P\left( B \right)\times P\left( B \right)\times P\left( B \right)\times P\left( G \right)$.
$\Rightarrow P\left( BBBG \right)=\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}$.
$\Rightarrow P\left( BBBG \right)=\dfrac{1}{16}$.
Now, $P\left( C \right)=1-\dfrac{1}{16}-\dfrac{1}{16}-\dfrac{1}{16}-\dfrac{1}{16}-\dfrac{1}{16}$.
$P\left( C \right)=1-\dfrac{5}{16}$.
$P\left( C \right)=\dfrac{16-5}{16}$.
$P\left( C \right)=\dfrac{11}{16}$.
Now we find the value of $P\left( A|C \right)=\dfrac{P\left( A\cap C \right)}{P\left( C \right)}$.
$\Rightarrow P\left( A|C \right)=\dfrac{P\left( GGGG \right)}{P\left( C \right)}$.
$\Rightarrow P\left( A|C \right)=\dfrac{P\left( G \right).P\left( G \right).P\left( G \right).P\left( G \right)}{P\left( C \right)}$.
$\Rightarrow P\left( A|C \right)=\dfrac{\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}}{\dfrac{11}{16}}$.
$\Rightarrow P\left( A|C \right)=\dfrac{\dfrac{1}{16}}{\dfrac{11}{16}}$
$\Rightarrow P\left( A|C \right)=\dfrac{1}{11}$.
We found the value of the conditional probability that all children are girls given that at least two are girls as $\dfrac{1}{11}$.
∴ The value of the conditional probability that all children are girls given that at least two are girls is $\dfrac{1}{11}$.
The correct option for the given problem is (a).

Note:
We can also calculate the probability that at least two new born children are girls by taking the sum of probability of two new born girls, three new born girls and four. New borns are girls. We should keep in mind that the probability of any event lies between 0 and 1. We should know about the properties of independent events and distinct events.