Question

A couple has \[2\] children. Find the probability that both are boys if it is known that (i) one of them is a boy (ii) the older child is a boy.

Answer 1

Hint: To solve this question first make the sample space of \[2\] children in a family. Make all the possible causes for a sample space and then apply both the conditions according to the question; that at least one child is a boy and the second condition is a boy; the last condition is that the eldest child is a boy. While writing sample space, write the elder child first and then the younger.

Complete answer: \[2\] children are in a family.
To find,
The probability that both are boys applying both of that condition
So to solve the question first we make a sample space
\[S = \{ BB,BG,GB,GG\} \] this is the sample space according to the question
Here, \[B\] stands for boy in a family and
\[G\] stands for a girl in a family
First, we use the first condition that is given in the question that is at least one of them is a boy:
Outcomes of at least one boy.
\[X:\{ BB,BG,GB\} \]
Now the second condition is both the children are boys:
\[Y:\{ BB\} \]
Now the probability of both the children are boy satisfying one of them is a boy
\[P(Y/X) = \dfrac{{P(X \cap Y)}}{{P(X)}}\]
Here, in intersection only one condition is common that is \[\{ BB\} \]
\[P(X \cap Y) = \dfrac{{favourable\,number\,of\,outcomes}}{{Total\,number\,of\,outcomes}}\]
On putting the values
\[P(X \cap Y) = \dfrac{1}{4}\]
Probability of getting one boy
\[P(X) = \dfrac{{favourable\,number\,of\,outcomes}}{{Total\,number\,of\,outcomes}}\]
On putting the values
\[P(X) = \dfrac{3}{4}\]
Putting both the values in conditional formula
\[P(Y/X) = \dfrac{{P(X \cap Y)}}{{P(X)}}\]
\[P(Y/X) = \dfrac{{\dfrac{1}{4}}}{{\dfrac{3}{4}}}\]
On further solving
\[P(Y/X) = \dfrac{1}{3}\]
Probability of getting both boys if one of them is boy \[P(Y/X) = \dfrac{1}{3}\]
Now we use the second condition that is given in the question that is an elder child is a boy:
Outcomes of at least one boy.
\[X:\{ BB,BG\} \]
Now the second condition is both the children are boys:
\[Y:\{ BB\} \]
Now the probability of both the children are boy satisfying one of them is a boy
\[P(Y/X) = \dfrac{{P(X \cap Y)}}{{P(X)}}\]
Here, in intersection only one condition is common that is \[\{ BB\} \]
\[P(X \cap Y) = \dfrac{{favourable\,number\,of\,outcomes}}{{Total\,number\,of\,outcomes}}\]
On putting the values
\[P(X \cap Y) = \dfrac{1}{4}\]
Probability of getting one boy
\[P(X) = \dfrac{{favourable\,number\,of\,outcomes}}{{Total\,number\,of\,outcomes}}\]
On putting the values
\[P(X) = \dfrac{2}{4}\]
Putting both the values in conditional formula
\[P(Y/X) = \dfrac{{P(X \cap Y)}}{{P(X)}}\]
\[P(Y/X) = \dfrac{{\dfrac{1}{4}}}{{\dfrac{2}{4}}}\]
On further solving
\[P(Y/X) = \dfrac{1}{2}\]
Probability of getting both boys if the elder of them is boy \[P(Y/X) = \dfrac{1}{2}\].

Note:
Conditional probability is defined as the likelihood of an event or outcome occurring, based on the occurrence of a previous event or outcome. Conditional probability is calculated by multiplying the probability of the preceding event by the updated probability of the succeeding, or conditional, event.