Conditional Probability of Seating Couple Given Other Couples Adjacent

To find the conditional probability that the Indian man is seated adjacent to his wife given that each of the four American men is seated adjacent to his wife, let us define the following events:

• Let $F$ be the event that each of the four American men is seated adjacent to his wife.
• Let $E$ be the event that the Indian man is seated adjacent to his wife.

We are tasked with finding the conditional probability $P(E \mid F)$, given by:
$P(E \mid F) = \frac{n(E \cap F)}{n(F)}$

Step 1: Finding the number of ways for event $F$ to occur, $n(F)$

In event $F$, each of the four American couples is treated as a single block because each American husband must sit adjacent to his wife.
Let these blocks be denoted by $A_1, A_2, A_3,$ and $A_4$. The Indian man ($I_M$) and his wife ($I_W$) are considered as two separate individuals.

Thus, we have a total of $4 + 2 = 6$ distinct entities to arrange around a circular table:

• $4$ blocks of American couples
• $I_M$ (Indian man)
• $I_W$ (Indian woman)

The number of ways to arrange $6$ entities in a circle is:
$(6 - 1)! = 5!$

Within each of the $4$ American couple blocks, the husband and wife can swap their seats in $2!$ ways. Therefore:
$n(F) = 5! \times (2!)^4 = 5! \times 2^4$

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Step 2: Finding the number of ways for event $E \cap F$ to occur, $n(E \cap F)$

In event $E \cap F$, each American man is seated adjacent to his wife, and the Indian man is also seated adjacent to his wife. Hence, all $5$ couples form distinct blocks:

• $4$ American couple blocks
• $1$ Indian couple block

This gives a total of $5$ blocks to be arranged around the circular table.

The number of ways to arrange $5$ blocks in a circle is:
$(5 - 1)! = 4!$

Within each of the $5$ couple blocks, the spouses can be arranged in $2!$ ways. Therefore:
$n(E \cap F) = 4! \times (2!)^5 = 4! \times 2^5$

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Step 3: Computing the conditional probability $P(E \mid F)$

Using the definition of conditional probability:
$P(E \mid F) = \frac{n(E \cap F)}{n(F)} = \frac{4! \times 2^5}{5! \times 2^4}$

Simplifying the expression:
$P(E \mid F) = \frac{4!}{5!} \times \frac{2^5}{2^4} = \frac{1}{5} \times 2 = \frac{2}{5}$

Thus, the required conditional probability is $\frac{2}{5}$.

Correct Option: C