Q.4.66 - A First Course in Probability

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.

Step-by-Step Solution

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 1:Given information(part a)

Given in the question that,

A total number of people $2n$ 

Married Couples $n$

We have to find  $P\left(C_i\right)$.

Step 2: Explanation (Part a)

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

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

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

Hence

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

Step 3: Final answer (Part a)

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

Step 4: Given information (part b)

Total Number of people $2n$

Number of married couples $n$

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

Step 5: Explanation (Part b)

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

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

Hence

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

Step 6: Final answer (Part b)

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

Step 7: 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.

Step 8: 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 $\underset{n\to \infty }{\lim }\lambda =1$

Step 9: Final answer (Part c)

The required probability is 

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