Q.4.67 - A First Course in Probability

Question

Repeat the preceding problem when the seating is random but subject to the constraint that the men and women alternate.

Step-by-Step Solution

Answer

1. The probability of $C_i$is $P\left(C_i\right)=\frac{\frac{\left(\right(n-1)!{)}^2}{2n}}{\frac{(n!{)}^2}{2n}}=\frac{1}{n^2}$
2. The probability of $P\left(C_j\mid C_i\right)=\frac{P\left(C_j,C_i\right)}{P\left(C_i\right)}=\frac{\frac{1}{n(n-1{)}^2}}{\frac{1}{n^2}}=\frac{1}{(n-1{)}^2}$
3. For very large $n$ the required probability is$P(X=0)=e^{-\lambda }\approx e^{-1}$
   

Step 1: Given Information (part a)

Given that the $C_i$, denote the event that the members of a couple $i$ are seated next to each other,$i = 1, ..., n.$

We have to find $P\left(C_i\right)$

Step 2: Explanation (part a)

From the combinatorics, we learn that we can permute men on $n !$ paths.

Even, we learn that we can permute women on$n!$paths.

Put them around the table in that demand and alternate. 

Since the table is rounded, we do not differ in every $2n$ seating. Therefore, we have that there exist $\frac{(n!{)}^2}{2n}$ modes to seat them. Suppose that the couple $i$ has been seated somewhere, but together.

Then, we have to remain $n-1$ men and $n-1$ women and we can set them on $\left(\right(n-1)!{)}^2$ routes since they have to alternate, and also, we do not differ every$2n$ seating.

Therefore

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

Step 3: Final Answer (part a)

The probability of $C_i$ is:

width="169" style="max-width: none;" $P\left(C_i\right)=\frac{\frac{\left(\right(n-1)!{)}^2}{2n}}{\frac{(n{)}^2}{2n}}=\frac{1}{n^2}$

Step 4: Given Information (part b)

Given that $j\ne i$.

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

Step 6: Explanation (part b)

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

So, the remaining pairs can place on$\left(\right(n-2)!{)}^2$ modes.

Also, as in (a), we have that

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

Step 6:Final Answer (part b)

Probability of  $C_j|C_i$ is,$P\left(C_j\mid C_i\right)=\frac{P\left(C_j,C_i\right)}{P\left(C_i\right)}=\frac{\frac{1}{n^2(n-1{)}^2}}{\frac{1}{n^3}}=\frac{1}{(n-1{)}^2}$

Step 8: Given Information (part c)

Given that for $n$ large, there are no married couples who are seated next to each other.

Step 8: Explanation (part c)

The probability that some pair sits to each other is $1-P\left(C_i\right)=\frac{n^2-1}{n^2}$. Describe $X$ as the random variable that denotes the number of pairs that do not sit to each other.

Utilizing Poisson approximation, we have that $X~$Pois $\left(\frac{n^2-1}{n^2}\right)$. For very large, the required probability is

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

since 

Step 9:Final Answer(part c)

For $n$ large, there are no married couples who are seated next to each other. The approximate probability is

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