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 E₁ be the event that the Indian man is seated  adjacent to his wife and E₂, be the even that each American
man is seated adjacent to his wife. 

Then, 

$P\left(E_1/E_2\right)=\frac{P\left(E_1\cap E_2\right)}{P\left(E_2\right)}$

Now, $E_1\cap E_2$ is the event that all men are seated adjacent to their wives.
Therefore, we can consider the 5 couples as single-single objects which can be arranged in a circle in 4! ways. But for
each couple, husband and wife can interchange their places in 2! ways.
Hence, the umber of ways when all men are seated adjacent to their wives is $4!\times (2!{)}^5$. Also in all, 10 persons can be seated
in a circle in 9 ! ways. 

Therefore,

$P\left(E_1\cap E_2\right)=\frac{4!\times (2!{)}^5}{9!}$

Similarly, if each American man is seated adjacent to his wife, considering each American couple as single object and Indian woman and man as separate objects, there are 6 different objects which can be arranged in a circle in 5 ! ways. 

Also for each American couple, husband and wife can interchange their places in 2! ways.
So, the number of ways in which each American man is seated adjacent to his wife is $5!\times (2!{)}^4.$ 

Therefore,

$P\left(E_2\right)=\frac{5!\times (2!{)}^4}{9!}$

Hence,

$P\left(E_1/E_2\right)=\frac{\left(4!\times (2!{)}^5\right)/9!}{\left(5!\times (2!{)}^4\right)/9!}=\frac{2}{5}$