Question

Assume that in a family, each child is equally likely to be a boy or girl. A family with three children is chosen at random. The probability that the eldest child is a girl given that the family has at least one girl is:
A. $\dfrac{1}{2}$
B. $\dfrac{1}{3}$
C. $\dfrac{2}{3}$
D. $\dfrac{4}{7}$

Answer 1

For solving this question you should know about the concept of probability. In this question we will first calculate all the possibilities which are possible according to our question and that will be our complete result and from that we will select as per our requirement. And at last we will take care of both results. That will be our final answer.

Complete step-by-step solution:
According to the question, it is asked that in a family, each child is equally likely to be a boy or girl and a family with three children is chosen at random. We have to find the probability that the eldest child is a girl given that the family has at least one girl. So, if we find the probability of this, then we have to find the complete result for this. So, here the results for this are:
S=\left\\{ \left( B,B,B \right),\left( G,G,G \right),\left( B,G,G \right),\left( G,B,G \right),\left( G,G,B \right),\left( G,B,B \right),\left( B,G,B \right),\left( B,B,G \right) \right\\}=8
Now according to the question it is asked to find the events that a family has at least one girl. Then,
${{E}_{1}}$ = Event that a family has at least one girl.
\begin{aligned} & {{E}_{1}}=\left\\{ \left( G,G,G \right),\left( B,G,G \right),\left( G,B,G \right),\left( G,G,B \right),\left( G,B,B \right),\left( B,G,B \right),\left( B,B,G \right) \right\\}=7 \\\ & \Rightarrow P\left( {{E}_{1}} \right)=\dfrac{7}{8} \\\ \end{aligned}
${{E}_{2}}$ = Event that the eldest child is a girl.
\begin{aligned} & {{E}_{2}}=\left\\{ \left( G,G,G \right),\left( G,B,G \right),\left( G,G,B \right),\left( G,B,B \right) \right\\}=4 \\\ & \Rightarrow P\left( {{E}_{2}} \right)=\dfrac{4}{8} \\\ \end{aligned}
So, if we find the probability of that a girl in a family is mandatory and that the girl is the eldest child in the family, then,
\begin{aligned} & \therefore {{E}_{1}}\cap {{E}_{2}}=\left\\{ \left( G,G,G \right),\left( G,B,G \right),\left( G,G,B \right),\left( G,B,B \right) \right\\} \\\ & \therefore P\left( \dfrac{{{E}_{1}}}{{{E}_{2}}} \right)=\dfrac{P\left( {{E}_{1}}\cap {{E}_{2}} \right)}{P\left( {{E}_{1}} \right)}=\dfrac{\dfrac{4}{8}}{\dfrac{7}{8}}=\dfrac{4}{7} \\\ \end{aligned}
So, the probability of the eldest child being a girl, if the family has at least a girl child is $\dfrac{4}{7}$.
Therefore, the correct option is option D.

Note: For solving these types of questions always understand the condition clearly. If the conditions are not clear then that will give us a wrong answer. And count every single term according to the given condition and if any term is left, then the probability will be wrong and that will be the wrong answer.