Question

A parent has two children. If one of them is boy, then the probability that other is also a boy is
A. $\dfrac{1}{2}$
B. $\dfrac{1}{4}$
C. $\dfrac{1}{3}$
D. None of these

Answer 1

Hint:
 Probability means possibility. It is a branch of mathematics that deals with the occurrence of event. It is used to predict how likely events are to happen. To find the probability of a single event to occur, first, we should know the total number of possible outcomes.
Probability can range in from 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event.
Sample space is defined as the set of all possible outcomes of a random experiment. Example: Tossing a head, Sample Space(S) = {H, T}. The probability of all the events in a sample space adds up to 1.
Event is a subset of the sample space i.e. a set of outcomes of the random experiment.
\[{\text{Probability of an event}} = \dfrac{{{\text{Number of occurence of event A in S}}}}{{{\text{Total number of cases in S}}}}{\text{ }}\]
The probability of occurrence of any event A when another event B in relation to A has already occurred is known as conditional probability. This expression is only valid when P(A) is greater than 0.
$P\left( {B|A} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}$
If a parent has two children the number of sample spaces is 4.
Here use of conditional probability
$P(\dfrac{{{E_2}}}{{{E_1}}}) = \dfrac{{P({E_1} \cap {E_2})}}{{P({E_1})}}$

 Complete step by step solution:
Let we take B1, B2 for boys and G1, G2 for girls.
A parent has two children
So, the sample space be
$S = \left\{ {{B_1}{B_2},{B_1}{G_2},{G_1}{B_2},{G_1}{G_2}} \right\}$
Let E1 = Event that one them is boy
${E_1} = \left\{ {{B_1}{B_2},{B_1}{G_2},{G_1}{B_2}} \right\}$ ${E_1} = \left\{ {{B_1}{B_2},{B_1}{G_2},{G_1}{B_2}} \right\}$
E2 = Event that 2nd is boy
${E_2} = \left\{ {{B_1}{B_2}} \right\}$
Using concept of Conditional Probability,
Probability of finding E2 when E1 already occurred i.e.
$P\left( {\dfrac{{{E_2}}}{{{E_1}}}} \right) = \dfrac{{P({E_2} \cap {E_1})}}{{P({E_1})}}$
$ = \dfrac{{\dfrac{1}{4}}}{{\dfrac{3}{4}}}$
$P\left( {\dfrac{{{E_2}}}{{{E_1}}}} \right) = \dfrac{1}{3}$
Hence, in a family of two children, if one of them is a boy then the probability that the other is also a boy is $\dfrac{1}{3}$.
∴ Option (C) is correct.

 Note:
This question can be solved by taking
Sample space = $S = \left\{ {BG,GB,BB} \right\}$
Favorable case (Event) $E = \left\{ {BB} \right\}$
Hence probability$ = \dfrac{{n(E)}}{{n(S)}} = \dfrac{1}{3}$