Question

A total of $2n$ people, consisting of $n$married couples, are randomly seated (all possible orderings being equally likely) at a round table. Let $C_I$denote the event that the members of couple $i$are seated next to each other,$i=1,...,n.$

(a) Find $P\left(C_i\right)$

(b)For $j\ne i$, find $P\left(C_j\mid C_i\right)$

(c) Approximate the probability, for $n$large, that there are no married couples who are seated next to each other.

Short answer

(a) $P\left(C_i\right)=\frac{2}{2n-1}$

(b)$P\left(C_j\mid C_i\right)=\frac{2}{2n-2}$

(c) The required probability$P(X=0)=e^{-\lambda }\approx e^{-1}$

Step-by-step solution

Step 1:Given information(part a)

Given in the question that,

A total number of people $2n$

Married Couples$n$

We have to find $P\left(C_i\right)$.

Explanation (Part a)

From the combinatorics, we have that there are $(2n-1)!$total ways of seating.

Assume that the couple $i$has been seated somewhere.

Then, we have the remaining $(2n-2)$people and we can set them on $(2n-2)!$ways, also, we can alternate them on $2!$ways.

Hence

$P\left(C_i\right)=\frac{2!·(2n-2)!}{(2n-1)!}=\frac{2}{2n-1}$

Final answer (Part a)

$P\left(C_i\right)=$$\frac{2}{2n-1}$

Given information (part b)

Total Number of people $2n$

Number of married couples$n$

We have to determine$P\left(C_j\mid C_i\right)$

Explanation (Part b)

Assume that pair $i$and $j$sit together.

So, they can sit on $2!·2!·(2n-3)!$ways.

Hence

$P\left(C_j\mid C_i\right)=\frac{P\left(C_j,C_i\right)}{P\left(C_i\right)}=\frac{\frac{2!·2!·(2n-3)!}{(2n-1)!}}{\frac{2}{2n-1}}=\frac{2}{2n-2}$

Final answer (Part b)

$P\left(C_j\mid C_i\right)=$$\frac{2}{2n-2}$

Given information (part c)

A total number of people $2n$

Married Couples $n$

$C_I$denote the event that the members of couple $i$are seated next to each other

We need to approximate the probability, for $n$large, that there are no married couples who are seated next to each other.

Explanation (Part c)

The probability that some couple sits to each other is$1-P\left(C_i\right)=\frac{2n-3}{2n-1}$.

Define $X$as the random variable that marks the number of couples that do not sit to each other.

Using Poisson approximation, we have that $X~$$Pois\left(\right(2n-3)/(2n-1\left)\right)$.

For very large $n$, the required probability is$P(X=0)=e^{-\lambda }\approx e^{-1}$

since $\underset{n\to \infty }{\lim }\lambda =1$

Final answer (Part c)

The required probability is

$P(X=0)=e^{-\lambda }\approx e^{-1}$