Question

If $4$ married couples are arranged in a row, find the probability that no husband sits next to his wife

Short answer

The probability that no husband sits next to his wife is$\frac{12}{35}$.

Step-by-step solution

Given Information

Define four events, for $i=1,2,3,4$ :

$E_i=$ event that $i$-th couple sits together.

Explanation

When $4$couples are seated, what is the probability that no couple sits together?

Find events that are simple so many probabilities can be easily calculated using them.

These events have equal probabilities, and the wanted probability is

$P\left[{\left(E_1\cup E_2\cup E_3\cup E_4\right)}^c\right]$

Remember the Proposition 4.4. (the law of inclusion and exclusion):

$\text{Let}{A}_1,A_2,\dots ,A_n\text{be some events.}$localid="1650015851988" $$\begin{gathered} P\left(A_1\cup A_2,\dots A_n\right)=\sum _{i\in \{1,2,\dots ,n\}}P\left(A_i\right)+(-1{)}^1\sum _{i_1<i_2}P\left(A_{i_1}\cap A_{i_2}\right) \\ +\dots +(-1{)}^{n-1}\sum _{i_1<i_2<\dots <i_n}P\left(A_{i_1}\cap A_{i_2}\dots \cap A_{i_n}\right) \\ {i}_1,i_2,\dots i_n\in \{1,2,\dots ,n\} \end{gathered}$$

Explanation

Using formula for probability on equally likely set of outcomes $\left(P\left(A\right)=\frac{\left|A\right|}{\left|S\right|}\right.$, and calculating $\left|A\right|$from similar problems in chapter $1(1.10,1.30)$

localid="1650015940493" role="math" $$\begin{gathered} P\left(E_i\right)=\frac{2\times 7!}{8!}=\frac{1}{4} \\ P\left(E_{i_1}\cap E_{i_2}\right)=\frac{2\times 2\times 6!}{8!}=\frac{1}{14} \\ P\left(E_{i_1}\cap E_{i_2}\cap E_{i_3}\right)=\frac{2\times 2\times 2\times 5!}{8!}=\frac{1}{42} \\ P\left(E_1\cap E_2\cap E_3\cap E_4\right)=\frac{2\times 2\times 2\times 2\times 4!}{8!}=\frac{1}{105} \\ {i}_1,i_2,i_3\in \{1,2,3,4\} \end{gathered}$$

Explanation

$$\begin{gathered} P\left(E_1\cup E_2\cup E_3\cup E_4\right)=\left(\begin{array}{l}4\\ 1\end{array}\right)P\left(E_1\right)+(-1{)}^1\left(\begin{array}{l}4\\ 2\end{array}\right)P\left(E_1\cap E_2\right)+(-1{)}^2\left(\begin{array}{l}4\\ 3\end{array}\right)P\left(E_1\cap E_2\cap E_3\right)+(-1{)}^3\left(\begin{array}{l}4\\ 4\end{array}\right)P\left(E_1\cap E_2\cap E_3\cap E_4\right) \\ =4P\left(E_1\right)-6P\left(E_1\cap E_2\right)+4P\left(E_1\cap E_2\cap E_3\right)-P\left(E_1\cap E_2\cap E_3\cap E_4\right) \\ =4\frac{1}{4}-6\frac{1}{14}+4\frac{1}{42}-\frac{1}{105} \\ =\frac{23}{35} \end{gathered}$$

And according to Proposition $4.1$:

localid="1650015956946" $$\begin{gathered} P\left[{\left(E_1\cup E_2\cup E_3\cup E_4\right)}^c\right]=1-P\left(E_1\cup E_2\cup E_3\cup E_4\right) \\ =1-\frac{23}{35} \\ =\frac{12}{35} \end{gathered}$$