Question

Mother, father and son line up at random for a family photo. Find$P\left( {A|B} \right)$, if A and B are defined as follows:
A= son on one end.
B = Father in middle.

Answer 1

If we are given ‘n’ distinct elements then those ‘n’ elements can be arranged in $n!$ number of ways, so we can use this concept of permutations to find the number of elements in the sample space for the given question. Also we are asked to find the conditional probability of two given events. If A and B are two given events then the conditional probability $P\left( {A|B} \right)$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: $P\left( {A|B} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$

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 in$3!$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 find $P\left( {A \cap B} \right)and\,P\left( B \right)$.
Given that, event A= son on one end.
Therefore,A = \left\\{ {\left( {MFS} \right),\left( {FMS} \right),\left( {SMF} \right),\left( {SFM} \right)} \right\\}.
$\therefore P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}} = \dfrac{4}{6}$
Similarly, Event B= Father in the middle.
B = \left\\{ {\left( {MFS} \right),\left( {SFM} \right)} \right\\}.
And,$P\left( B \right) = \dfrac{{n\left( B \right)}}{{n\left( S \right)}} = \dfrac{2}{6}$
Also,A \cap B = \left\\{ {\left( {MFS} \right),\left( {SFM} \right)} \right\\}.
So we have, $P\left( {A \cap B} \right) = \dfrac{{n\left( {A \cap B} \right)}}{{n\left( S \right)}} = \dfrac{2}{6}$

Step3 : We find $P\left( {A|B} \right)$.
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,

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

Hence we have the required probability. $P\left( {A|B} \right) = 1$

Note:
We can use the conditional probability to find the probability of occurrence of two events simultaneously and it is given by: $P\left( {A \cap B} \right) = P\left( A \right)P\left( {B|A} \right)$
This application of conditional probability is called the multiplication theorem of probability. This can also be extended to find the simultaneous occurrence of more than two events.