Question

If a party of $n$ persons sit at a round table, then the odds against two specified individuals sitting next to each other are
1. $2:(n - 3)$
2. $(n - 3):2$
3. $(n - 2):2$
4. $2:(n - 2)$

Answer 1

In the given question, we need to find the odds against two specified individuals sitting next to each other. The odds are defined as the probability that the event will occur divided by the probability that the event will not occur. At first, we will find the probability of two persons sitting together or the probability of the event $A$ occurring. For this, we will find the total number of outcomes i.e., number of all possible outcomes, $n\left( S \right)$ and the number of favorable outcomes i.e., number of ways of event $A$ occurring, $n\left( A \right)$. Then the probability of the event $A$ occurring is given by $P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}$.
After this, we will find the probability of two persons not sitting together by using $P\left( {A'} \right) = 1 - P\left( A \right)$ where, $P\left( {A'} \right)$ is the probability of the event not occurring.
Then, we will find the odds against the event $A = \dfrac{{P\left( A \right)}}{{P\left( {A'} \right)}}$.

Complete step-by-step solution:
Here, we are given that $n$ persons are seated at the round table. Therefore, the total number of ways in which these $n$ persons can sit along the round table are, $n\left( S \right) = \left( {n - 1} \right)!$ ways.
Now, we are given that two particular persons sit together so considering these two persons as one. There will be only $n - 1$ persons and these $n - 1$ can be seated along the round table by
$\Rightarrow \left( {\left( {n - 1} \right) - 1} \right)!$ ways
$\Rightarrow \left( {n - 1 - 1} \right)!$ ways
On addition of negative terms we get,
$\Rightarrow \left( {n - 2} \right)!$ ways
As mentioned in the question two specified individuals sitting next to each other. So these two particular persons can arrange themselves by $2!$ ways.
Therefore, the total numbers of favorable outcomes $P\left( A \right) = \left( {n - 2} \right)! \times 2!$ ways.
So, the probability of two persons sitting together is;
$P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}$
On substituting the values, we get
$\Rightarrow P\left( A \right) = \dfrac{{\left( {n - 2} \right)!2!}}{{\left( {n - 1} \right)!}}$
We know that, $\left( {n - 1} \right)!$ can be written as $\left( {n - 1} \right)\left( {n - 2} \right)!$
$\Rightarrow P\left( A \right) = \dfrac{{\left( {n - 2} \right)!2!}}{{\left( {n - 1} \right)\left( {n - 2} \right)!}}$
On cancelling out common factors, we get
$\Rightarrow P\left( A \right) = \dfrac{{2 \times 1}}{{\left( {n - 1} \right)}} = \dfrac{2}{{n - 1}}$
So the probability of two persons not sitting together is; $P\left( {A'} \right) = 1 - P\left( A \right)$
$\Rightarrow P\left( {A'} \right) = 1 - \dfrac{2}{{n - 1}}$
Take LCM on RHS.
$\Rightarrow P\left( {A'} \right) = \dfrac{{1\left( {n - 1} \right) - 2}}{{n - 1}}$
$\Rightarrow P\left( {A'} \right) = \dfrac{{n - 1 - 2}}{{n - 1}}$
On simplification, we get
$\Rightarrow P\left( {A'} \right) = \dfrac{{n - 3}}{{n - 1}}$
So the odds the event $A$ is = $\dfrac{{P\left( {A'} \right)}}{{P\left( A \right)}}$
$=\dfrac{{\dfrac{{n - 3}}{{n - 1}}}}{{\dfrac{2}{{n - 1}}}}$
It can also be written as,
$= \dfrac{{n - 3}}{{n - 1}} \times \dfrac{{n - 1}}{2}$
On cancelling out common factors, we get
$= \dfrac{{n - 3}}{2}$
$= \left( {n - 3} \right):2$
Thus, the odds against two specified individuals sitting next to each other are $\left( {n - 3} \right):2$.
Therefore, the correct option is 2.

Note: Remember to calculate the number of ways specified individuals sitting next to each other because there are different ways of their sitting arrangement and multiply it to the number of ways of sitting arrangement of rest. That will be our total number of favorable outcomes. Students must be careful between odds against and odds in favor. Odds in favor of a particular event are given by the number of favorable outcomes divided by number of unfavorable outcomes. Or we can say that it is reciprocal of odds against.