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 \(=\frac{2}{5}\).