Question

It is given that a family of two children has a girl, what is the probability that the other child is also a girl?
a) $0.50$
b) $0.75$
c) $\dfrac{1}{3}$
d) $0.2$

Answer 1

In this question, we have to find that the probability that the other child is also a girl. For that we are going to solve this problem using a probability formula. Probability formula is given by
${\text{P}}\left( {\text{A}} \right){\text{ = }}\dfrac{{{\text{Number of favourable outcome}}}}{{{\text{Total outcomes}}}}$

Complete step-by-step answer:
Given that family of two children with one being a girl.
To find that the probability that the other child is also a girl.
First possibility of the girl is ${\text{BG}}$, [Here ${\text{B}}$ represents a boy, ${\text{G}}$ represents a girl]
Second possibility of the girl is ${\text{GB}}$, [Here ${\text{B}}$ represents a boy, ${\text{G}}$ represents a girl]
Third possibility of the girl is ${\text{GG}}$, [Here ${\text{B}}$ represents a boy, ${\text{G}}$ represents a girl]
Next, there is no possibility. So we conclude that total number of favorable outcomes are ${\text{BG}}$,${\text{GB}}$,${\text{GG}}$.
Therefore, ${\text{n}}\left( {\text{S}} \right) = 3$. [Here ${\text{n}}\left( {\text{S}} \right)$ represents the total number of favorable outcomes]
${\text{P}}\left( {\text{A}} \right){\text{ = }}\dfrac{{{\text{Number of favourable outcome}}}}{{{\text{Total outcomes}}}}$
${\text{P}}\left( {\text{A}} \right) = \dfrac{1}{3}$
Hence the probability that the other child is also a girl is $= \dfrac{1}{3}$

Option c is the correct answer.

Note: We can solve this problem by another way
It is given that a family of two children has a girl. We have to find the probability that the other child is also a girl.
First possibility of the girl is ${\text{BG}}$, [Here ${\text{B}}$ represents a boy, ${\text{G}}$ represents a girl]
Second possibility of the girl is ${\text{GB}}$, [Here ${\text{B}}$ represents a boy, ${\text{G}}$ represents a girl] $= \dfrac{1}{3}$
Third possibility of the girl is ${\text{GG}}$, [Here ${\text{B}}$ represents a boy, ${\text{G}}$ represents a girl] $= \dfrac{1}{3}$
Next, there is no possibility. So we conclude that total number of favorable outcomes are ${\text{BG}}$,${\text{GB}}$,${\text{GG}}$.
Therefore, ${\text{n}}\left( {\text{S}} \right) = 3$. [Here ${\text{n}}\left( {\text{S}} \right)$ represents the total number of favorable outcomes]
Probability of ${\text{BG}}$ $= \dfrac{1}{3}$
Probability of ${\text{GB}}$ $= \dfrac{1}{3}$
Probability of ${\text{GG}}$ $= \dfrac{1}{3}$
We already know that,
Addition of all possibility events is 1. By using this,
Probability of ${\text{BG}}$ $+$ Probability of ${\text{GB}}$ $+$ Probability of ${\text{GG}}$ $= \dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3} = \dfrac{3}{3} = 1$
Hence we get, addition of all possible events is 1.
Therefore the probability of that other child is also a girl $= \dfrac{1}{3}$. This is the right answer.