Question

A party of 23 persons takes their seats at a round table. Find the odds against two persons sitting together. \[\]

Answer 1

Hint: We find the total number of outcomes $ n\left( S \right) $ as number circular permutations for $ n $ distinct objects which is given by $ \left( n-1 \right)! $ .We find number of favourable outcomes $ n\left( A \right) $ as number ways we can arrange 2 persons sitting together and rest 21 persons in circular arrangement. We find the $ P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)},P\left( {{A}^{'}} \right)=1-P\left( A \right) $ . We find odds against $ A $ as $ P\left( {{A}^{'}} \right):P\left( A \right) $ .\[\]

Complete step by step answer:
We know from the definition of probability that if there is $ n\left( A \right) $ number of ways of event $ A $ occurring (or number of favorable outcomes) and $ n\left( S \right) $ is the size of the sample space (number of all possible outcomes) then the probability of the event $ A $ occurring is given by
\[P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}\]

We are given in the question that A party of 23 persons take their seats at a round table. We make them sit in circular arrangement in $ \left( n-1 \right)!=\left( 23-1 \right)!=22! $ ways.
We are asked to find the odds against two persons sitting together. We can fix the sits for those two persons sitting together who can arrange themselves in $ 2! $ ways. We can arrange rest $ 23-2=21 $ persons in circular arrangement in $ \left( 21-1 \right)!=20! $ ways. So the number of ways two persons can sit together is $ n\left( A \right)=21!\times 2! $.
We have the probability of two persons sitting together is ;
\[P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{21!2!}{22!}=\dfrac{2}{22}=\dfrac{1}{11}\]
So the probability of two persons not sitting together is ;
\[P\left( {{A}^{'}} \right)=1-P\left( A \right)=1-\dfrac{1}{11}=\dfrac{10}{11}\]
So the odds against the event $ A $ is;
\[\dfrac{P\left( {{A}^{'}} \right)}{P\left( A \right)}=\dfrac{\dfrac{10}{11}}{\dfrac{1}{11}}=\dfrac{10}{1}=10:1\]

Note:
We can alternatively solve by directly finding the odds against of an event $ A $ happening as the ratio of not favourable ways $ n\left( {{A}^{'}} \right) $ to ratio of favourable ways $ n\left( A \right) $ . If the ratio is $ a:b $ , the odds against are read as $ a $ to $ b $ . We must be careful between odds against and odds in favor which is the reciprocal ratio of odd against.