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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?

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Hint: 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.


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