Question

Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?
(A). 0.33, 0.34
(B). 0.88, 0.98
(C). 0.78, 0.67
(D). 0.5, 0.33

Answer 1

Hint : Probability means the chances of occurring of any event. For finding probability of any experiment for which the outcomes can’t be predicted with certainty (Random experiment), two definitions are there, one is event and the other is sample space.
Event is the favourable outcome of any experiment while Sample space is the set of all possible outcomes of that experiment and we can say that event will be a subset of sample space.
Probability for any random experiment is given by:${\text{Probability of an event}} = \dfrac{{{\text{Number of occurence of event A in S}}}}{{{\text{Total number of cases in S}}}}{\text{ }}$
$= \dfrac{{n\left( A \right)}}{{n\left( S \right)}}$
Conditional Probability is the probability of any event that occurs with some relation with any other event.
It is given by the formula, $P\left( {B|A} \right){\text{ }} = {\text{ }}\dfrac{{P\left( {A \cap B} \right)}}{{{\text{ }}P\left( A \right)}}{\text{ }}$
Where $P(A \cap B)$is defined as the intersection of events A and B.

Complete step by step solution : Given the family the family has two children, now let us denote boy child as ‘B’ and girl child as ‘G’.

Now let us make the sample space for the no. of boys and girls the family can have.
If both the child is boy then sample space would be {(BB)}
One boy and one girl then sample space would be {(BG), (GB)}
If both the child is girl then sample space would be {(GG)}
Then the sample space for the given case would be,
$S = \\{ (BB),(BG),(GB),(GG)\\}$
And thus total No. of events $= 4$

Now let us suppose ‘A’ is the event that both the children are girl
That is $A = \\{ (GG)\\}$
Then $P(A) = \dfrac{{\;No.{\text{ }}of{\text{ }}events{\text{ }}if{\text{ }}both{\text{ }}the{\text{ }}children{\text{ }}are{\text{ }}girl\;}}{{\;Total{\text{ }}No.{\text{ }}of{\text{ }}events}} = \dfrac{1}{4}$

(i). Let B be the condition that the youngest child is girl,
The sample s[ace for this case would be$B = \\{ (BG),(GG)\\}$
$P(B) = \dfrac{{\;No.{\text{ }}of{\text{ }}events{\text{ }}if{\text{ }}the{\text{ youngest is }}girl\;}}{{\;Total{\text{ }}No.{\text{ }}of{\text{ }}events}} = \dfrac{2}{4} = \dfrac{1}{2}$
Now A \cap B = \left\\{ {(GG)} \right\\}
Hence, $P(A \cap B) = \dfrac{1}{4}$ that is if the youngest as well as the elder both are girls.
Then, the conditional probability that both the children are girls, and the youngest child is a girl, is given by,
$P(A\mid B) = \dfrac{{P(A \cap B)}}{{P(B)}} = \dfrac{{\dfrac{1}{4}}}{{\dfrac{1}{2}}} = \dfrac{1}{2} = 0.5$
Which is the required probability.

( ii ). Let C be the event that at least one child is a girl,
Then sample space for this case would be $C = \\{ (BG),(GB),(GG)\\}$
∴$P(C) = \dfrac{{\;No.{\text{ }}of{\text{ }}events\;that{\text{ }}at{\text{ }}least{\text{ }}one{\text{ }}child{\text{ }}is{\text{ }}a{\text{ }}girl\;}}{{\;Total{\text{ }}No.{\text{ }}of{\text{ }}events}} = \dfrac{3}{4}$
Again, A \cap C = \left\\{ {(GG)} \right\\}
Hence, $P(A \cap C) = \dfrac{1}{4}$
Then, the conditional probability that both the children are girls, and at least one of them is girl, is given by,
$P(A\mid C) = \dfrac{{P(A \cap C)}}{{P(C)}} = \dfrac{{\dfrac{1}{4}}}{{\dfrac{3}{2}}} = \dfrac{1}{3} = 0.33$
Which is the required probability.

Thus the answer of the question is option (D) i.e, 0.5, 0.33

Note : Sum of probability of occurrence of an event and non-occurrence is equal to 1 i.e.
$P\left( A \right) + P\left( {\bar A} \right) = 1$
$P(A \cap B)$ can also be represented in a vein diagram.