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Question: Mother, father and son line up at random for a family photo. If A and B are two events given by A...

Mother, father and son line up at random for a family photo. If A and B are two events given by
A=son on one end,
B= father in the middle.
Find P(A∣B)P\left( {A|B} \right)andP(B∣A)P\left( {B|A} \right).

Explanation

Solution

In this question we need to find the conditional probability of two given events. If A and B are two given events then the conditional probability P(A∣B)P\left( {A|B} \right)is defined as the probability of occurrence of A when it is known that event B has already occurred.
The conditional probability of A when B has already occurred is given by
P(A∣B)=P(A∩B)P(B)P\left( {A|B} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}
Similarly, the conditional probability of B when A has already occurred is given by
P(B∣A)=P(B∩A)P(A)P\left( {B|A} \right) = \dfrac{{P\left( {B \cap A} \right)}}{{P\left( A \right)}}.

Complete step by step solution:
Step1: We will find the sample space in case when mother, father and son line up randomly.
Let us denote, Mother by ‘M’, Father by ‘F’ and Son by ‘S’.
Now, all three can arrange themselves in3!3!number of ways. And the sample space is given by:S = \left\\{ {\left( {MFS} \right),\left( {MSF} \right),\left( {FMS} \right),\left( {FSM} \right),\left( {SMF} \right),\left( {SFM} \right)} \right\\}.

Step2: We will findP(A∩B),P(A)and P(B)P\left( {A \cap B} \right),P\left( A \right)and\,P\left( B \right).
Given that, event A= son at one end.
Therefore,A = \left\\{ {\left( {MFS} \right),\left( {FMS} \right),\left( {SMF} \right),\left( {SFM} \right)} \right\\}.
∴P(A)=n(A)n(S)=46\therefore P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}} = \dfrac{4}{6} ……(1)
Similarly, Event B= Father in the middle.
B = \left\\{ {\left( {MFS} \right),\left( {SFM} \right)} \right\\}.
∴P(B)=n(B)n(S)=26\therefore P\left( B \right) = \dfrac{{n\left( B \right)}}{{n\left( S \right)}} = \dfrac{2}{6} ……(2)
And,A \cap B = \left\\{ {\left( {MFS} \right),\left( {SFM} \right)} \right\\}.
So we have, P(A∩B)=n(A∩B)n(S)=26P\left( {A \cap B} \right) = \dfrac{{n\left( {A \cap B} \right)}}{{n\left( S \right)}} = \dfrac{2}{6} ……(3)

Step3: We find P(A∣B)P\left( {A|B} \right)and P(B∣A)P\left( {B|A} \right).
Firstly, we find the conditional probability of A when B has already occurred. It means that among the random arrangements in the sample space, we will just consider the arrangements in which the middle position is fixed and filled by father and then find the probability of son on one end when father stood in the middle. Therefore, the required conditional probability is given by,

P(A∣B)=P(A∩B)P(B) =2626=1 P(A∣B)=1P\left( {A|B} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}\\\ = \dfrac{{\dfrac{2}{6}}}{{\dfrac{2}{6}}} = 1\\\ P\left( {A|B} \right) = 1

Similarly, we can find the conditional probability of father standing in the middle when it is known that the son is standing on one of the sides and it is given by:

P(B∣A)=P(A∩B)P(A) =2646=12 P(B∣A)=12P\left( {B|A} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}\\\ = \dfrac{{\dfrac{2}{6}}}{{\dfrac{4}{6}}} = \dfrac{1}{2}\\\ P\left( {B|A} \right) = \dfrac{1}{2}

Hence we have the required probabilities. P(A∣B)=1 and P(B∣A)=12P\left( {A|B} \right) = 1\,and\,P\left( {B|A} \right) = \dfrac{1}{2}

Note:
Conditional probability leads us to a very important theorem of probability called the multiplication theorem. Like the conditional probability if we want to find the occurrence of an event when one has already occurred, we can find the probability of the occurrence of both the events simultaneously using the conditional probability. And it is given by: P(A∩B)=P(A)P(B∣A)P\left( {A \cap B} \right) = P\left( A \right)P\left( {B|A} \right)