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

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.