Question

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

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{2}{5}$
• D. $\frac{1}{5}$

Answer 1

The correct option is C $\frac{2}{5}$

Let $E_1$ be the event that the Indian man is seated adjacent to his wife and $E_2$ be the even that each American man is seated adjacent to his wife. Then,
$P\left(E_1/E_2\right)=\frac{P(E_1\cap E_2)}{P\left(E_2\right)}$
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 number 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,
$P(E_1\cap E_2)=\frac{4!\times (2!{)}^5}{9!}$
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,
$P\left(E_1/E_2\right)=\frac{(4!\times (2!{)}^5)/9!}{(5!\times (2!{)}^4)/9!}$
$=\frac{2}{5}$