Question

A family has three children. Event ‘A’ is that family has at most one boy, Event ′B′ is that family has at least one boy and one girl, Event ′C′ is that the family has at most one girl. Find whether events ′A′ and ′B′ are independent. Also find whether A, B, C are independent or not.

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:
${\text{Probability of an event}} = \dfrac{{{\text{Number of occurence of event A in S}}}}{{{\text{Total number of cases in S}}}}{\text{ }}$
$= \dfrac{{n\left( A \right)}}{{n\left( S \right)}}$
Independent events mean that event has no dependency on the other event.
If P(A) is probability of any event A and P(B) is probability of any event B
Then event A is independent with event B if,

Where, $P(A \cap B)$is intersection of events A and B.

Complete answer : Given,
No. of children in the family $= 3$
Now let us make the sample space for the no. of boys and girls the family can have.
(i). If all the children are boys then sample space would be {(BBB)}
(ii)If there are two boys one girl then sample space would be {(BBG),(BGB),(GBB)}
(iii) If there are One boy and two girls then sample space would be {(GGB), (GBG), (BGG)}
(iv)If all the children are girls then sample space would be {(GGG)}
Now total No. of events will be sum of all events in i, ii, iii, iv$= 8$
Hence,

P(i) = \dfrac{{No.{\text{ }}of{\text{ }}events\;if{\text{ }}all{\text{ }}the{\text{ }}children{\text{ }}are{\text{ }}boys}}{{Total{\text{ }}No.{\text{ }}of{\text{ }}events}} = \dfrac{1}{8} \\\ P(ii) = \dfrac{{No.{\text{ }}of{\text{ }}events\;if\;there{\text{ }}are{\text{ }}two{\text{ }}boys{\text{ }}one{\text{ }}girl}}{{Total{\text{ }}No.{\text{ }}of{\text{ }}events}} = \dfrac{3}{8} \\\ P(iii) = \dfrac{{No.{\text{ }}of{\text{ }}events\;if\;there{\text{ }}are{\text{ }}One{\text{ }}boy\;and{\text{ }}two{\text{ }}girls}}{{Total{\text{ }}No.{\text{ }}of{\text{ }}events}} = \dfrac{3}{8} \\\ P(iv) = \dfrac{{No.{\text{ }}of{\text{ }}events\;if\;all{\text{ }}the{\text{ }}children{\text{ }}are{\text{ }}girls}}{{Total{\text{ }}No.{\text{ }}of{\text{ }}events}} = \dfrac{1}{8} \\\ \end{gathered} $$ Now, *Let us suppose ‘A’ be the event if family has at most one boy that means the family can’t have more than one boy but can have less than one boy. Then Probability of the event $$ = P(A) = P\left( {iii} \right) + P\left( {iv} \right)$$ $$\begin{gathered} = \dfrac{3}{8} + \dfrac{1}{8} \\\ = \dfrac{1}{2} \\\ \end{gathered} $$ * Let us suppose ‘B’ be the event if family has at least one boy and at least one girl that means the family must have atleast one boy and one girl. Then Probability that family has at least one boy and at least one girl $$ = P(B) = P(ii) + P(iii)$$ $$\begin{gathered} = \dfrac{3}{8} + \dfrac{3}{8} \\\ = \dfrac{3}{4} \\\ \end{gathered} $$ * Let us suppose ‘B’ be the event if family has at most one girl that means the family can’t have more than one girl but can have less than one girl. Then the Probability that family has at most one girl $$ = P(C) = P(i) + P(ii)$$ $$\begin{gathered} = \dfrac{1}{8} + \dfrac{3}{8} \\\ = \dfrac{1}{2} \\\ \end{gathered} $$ Now, $$P(A \cap B)$$ means family should have only one boy and two girls i.e.$$ = P(iii) = \dfrac{3}{8}$$ Hence As we know that event A is independent with event B if, ∴ A and B are independent of each other Now there is no intersection of A and C that means occurrence of all the boys and all girls can’t take place, i.e. $$P\left( {A \cap C} \right) = 0 \ne P\left( A \right).P\left( C \right).\;$$ Hence A,B,C are not independent **Note** : If $$P\left( {A \cap C} \right) = 0$$then A and B are called mutually exclusive. Sum of probability of occurrence of an event and non-occurrence is equal to 1 i.e. $$P\left( A \right) + P\left( {\bar A} \right) = 1$$