One Indian and four American men and their wives are to be seated randomly around a circular table. Then, the conditional probability that the Indian man is seated adjacent to his wife given that each American man is seated adjacent to his wife, is

Let $E=$ event when each American man is seated adjacent to his wife and $A=$ event when Indian man is seated adjacent to his wife. Now, $n(A \cap E)=(4 !) \times(2 !)^5$ Even when each American man is seated adjacent to his wife. Again $$ \begin{aligned} n(E) & =(5 !) \times(2 !)^4 \\ P\left(\frac{A}{E}\right) & =\frac{n(A \cap E)}{n(E)} \\ & =\frac{(4 !) \times(2 !)^5}{(5 !) \times(2 !)^4}=\frac{2}{5} \end{aligned} $$ $$ \Rightarrow \quad P\left(\frac{A}{E}\right)=\frac{n(A \cap E)}{n(E)} $$ ALITER Fixing four American couples and one Indian man in between any two couples, we have 5 different ways in which his wife can be seated, of which 2 cases are favourable. $\therefore$ Required probability $=\frac{2}{5}$.