91Ó°ÊÓ

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

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

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

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

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

$\text{Let}{A}_1,A_2,…,A_n\text{be some events.}$localid="1650015851988" $$\begin{gathered} P\left(A_1∪A_2,…A_n\right)=\underset{i∈\{1,2,…,n\}}{∑}P\left(A_i\right)+(-1{)}^1\underset{i_1<i_2}{∑}P\left(A_{i_1}âˆ\copyright A_{i_2}\right) \\ +…+(-1{)}^{n-1}\underset{i_1<i_2<…<i_n}{∑}P\left(A_{i_1}âˆ\copyright A_{i_2}…âˆ\copyright A_{i_n}\right) \\ {i}_1,i_2,…i_n∈\{1,2,…,n\} \end{gathered}$$

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" $$\begin{gathered} P\left(E_i\right)=\frac{2×7!}{8!}=\frac{1}{4} \\ P\left(E_{i_1}âˆ\copyright E_{i_2}\right)=\frac{2×2×6!}{8!}=\frac{1}{14} \\ P\left(E_{i_1}âˆ\copyright E_{i_2}âˆ\copyright E_{i_3}\right)=\frac{2×2×2×5!}{8!}=\frac{1}{42} \\ P\left(E_1âˆ\copyright E_2âˆ\copyright E_3âˆ\copyright E_4\right)=\frac{2×2×2×2×4!}{8!}=\frac{1}{105} \\ {i}_1,i_2,i_3∈\{1,2,3,4\} \end{gathered}$$

$$\begin{gathered} P\left(E_1∪E_2∪E_3∪E_4\right)=\left(\begin{array}{l}4\\ 1\end{array}\right)P\left(E_1\right)+(-1{)}^1\left(\begin{array}{l}4\\ 2\end{array}\right)P\left(E_1âˆ\copyright E_2\right)+(-1{)}^2\left(\begin{array}{l}4\\ 3\end{array}\right)P\left(E_1âˆ\copyright E_2âˆ\copyright E_3\right)+(-1{)}^3\left(\begin{array}{l}4\\ 4\end{array}\right)P\left(E_1âˆ\copyright E_2âˆ\copyright E_3âˆ\copyright E_4\right) \\ =4P\left(E_1\right)-6P\left(E_1âˆ\copyright E_2\right)+4P\left(E_1âˆ\copyright E_2âˆ\copyright E_3\right)-P\left(E_1âˆ\copyright E_2âˆ\copyright E_3âˆ\copyright E_4\right) \\ =4\frac{1}{4}-6\frac{1}{14}+4\frac{1}{42}-\frac{1}{105} \\ =\frac{23}{35} \end{gathered}$$

And according to Proposition $4.1$:

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