Question

A family has two children. What is the probability that both the children are boys, given that at least one of them is a boy?

Answer 1

We are asked to find the conditional probability. For that, first, find the possible outcomes for having two children and then we define the events as E- both the children are boys and F- at least one of the children is a boy. We know that $P\left( E|F \right)$ is given as $P\left( E|F \right)=\dfrac{P\left( E\cap F \right)}{P\left( F \right)}.$ We use this to get the required probability.

Complete step-by-step answer :
We are given that a family has two children, it can be a boy or a girl. Let us consider girls to be denoted by ‘g’ and boys to be denoted by ‘b’. So the possible outcome for a family having 2 children are
S=\left\\{ \left( b,b \right),\left( b,g \right),\left( g,b \right),\left( g,g \right) \right\\}
Now, we have to find the probability that the children are both boys given that at least one of them is a boy. To do so, we will consider two events as follows.
E: Both the children are boys
F: At least one of the children is a boy.
We have to find $P\left( E|F \right)$ that says $P\left( E|F \right)$ is the probability of event E given that we have event F.
We know that, $P\left( E|F \right)$ is given as,
$P\left( E|F \right)=\dfrac{P\left( E\cap F \right)}{P\left( F \right)}$
To find the required answer, we will first have to find $P\left( E\cap F \right)$ and P(F).
Now, F = At least one of the children is a boy. So, the outcome for F are \left\\{ \left( a,b \right),\left( b,g \right),\left( b,b \right) \right\\}.
So,
$P\left( F \right)=\dfrac{\text{Favorable Number of Outcomes for F}}{\text{Total Number of Outcomes}}$
$\Rightarrow P\left( F \right)=\dfrac{3}{4}$
Now, let us consider E = Both the children are boys. So, the outcomes for E are \left\\{ \left( b,b \right) \right\\}.
So, E\cap F=\left\\{ \left( b,b \right) \right\\}
Hence,
$P\left( E\cap F \right)=\dfrac{\text{Favorable Number of Outcomes for E}\cap \text{F}}{\text{Total Number of Outcomes}}$
$\Rightarrow P\left( E\cap F \right)=\dfrac{1}{4}$
Now, we have, $P\left( E\cap F \right)=\dfrac{1}{4}$ and $P\left( F \right)=\dfrac{3}{4}.$ Using this in $P\left( E|F \right),$ we get,
$P\left( E|F \right)=\dfrac{P\left( E\cap F \right)}{P\left( F \right)}$
By putting the values in the above equation, we get,
$P\left( E|F \right)=\dfrac{\dfrac{1}{4}}{\dfrac{3}{4}}$
Cancelling out 4 from both numerator and denominator, we get,
$P\left( E|F \right)=\dfrac{1}{3}$
Therefore, the probability of having 2 children both boys given and that at least one of them is a boy is $\dfrac{1}{3}.$

Note : Here, in this question, we are asked a conditional probability. Students should always keep in mind to use the formula $P\left( E|F \right)=\dfrac{P\left( E\cap F \right)}{P\left( F \right)}.$ And not directly apply the formula $P\left( 2\text{ boys} \right)=\dfrac{\text{Outcome of having 1 boy}}{\text{Total Outcomes}}$
This will lead us to a wrong answer as the outcome for having 2 boys are {(b, b)} and the total outcome is 4. So, $P\left( \text{having 2 children both boys} \right)=\dfrac{1}{4}$ which is not the correct solution. Also, keep in mind that at least 1 means that there should be 1 or more than 1. So, all the cases including 1 or more than that will be included.