give an example such that $\mathbb P(\lim\inf\: A_n)<\lim \inf\: \mathbb P(A_n)<\lim \sup\mathbb P(A_n) <\mathbb P(\lim\sup A_n)$ 
Let $(\Omega,\mathcal F,\mathbb P)$ be the uniform distribution on $\Omega=\{1,2,3\}$, give an example of a sequence $A_1,A_2,\cdots\in \mathcal F$ such that
$$\mathbb P(\lim\inf\: A_n)<\lim \inf\: \mathbb P(A_n)<\lim \sup\mathbb P(A_n) <\mathbb P(\lim\sup A_n)$$ i.e. such that all three are strict.

I get one example from internet but couldn't get the logic why it fit here?
\begin{cases} 
      \{1,2\} & n\equiv 0\ (\textrm{mod}\ 6) \\
      \{1\} & n\equiv 1\ (\textrm{mod}\ 6) \\
      \{1,3\} & n\equiv 2\ (\textrm{mod}\ 6) \\
       \{2\} & n\equiv 3\ (\textrm{mod}\ 6) \\
      \{2,3\} & n\equiv 4\ (\textrm{mod}\ 6) \\
      \{3\} & n\equiv 5\ (\textrm{mod}\ 6) \\
\end{cases}
 A: With
$$A_n = \begin{cases} 
      \{1,2\} & n\equiv 0\ (\textrm{mod}\ 6) \\
      \{1\} & n\equiv 1\ (\textrm{mod}\ 6) \\
      \{1,3\} & n\equiv 2\ (\textrm{mod}\ 6) \\
       \{2\} & n\equiv 3\ (\textrm{mod}\ 6) \\
      \{2,3\} & n\equiv 4\ (\textrm{mod}\ 6) \\
      \{3\} & n\equiv 5\ (\textrm{mod}\ 6) \\
\end{cases}
$$
you have :
$$\mathbb P(A_n)= \begin{cases}2/3 & \text{if } n\equiv 0\mod 2\\
1/3 &\text{if } n\equiv 1 \mod 2\end{cases}$$
Therefore:
\begin{align}
\limsup\mathbb P(A_n) &= \lim_{n\to +\infty}\sup_{k\geq n} \mathbb P(A_n) = \lim_{n\to +\infty} 2/3 =  2/3 \\
\liminf\mathbb P(A_n) &= \lim_{n\to +\infty}\inf_{k\geq n} \mathbb P(A_n) = \lim_{n\to +\infty} 1/3 =  1/3 
\end{align}
Furthermore, you have :
\begin{align}
\liminf A_n = \bigcup_{n\geq 0}\bigcap_{k\geq n} A_k = \emptyset\\
\limsup A_n = \bigcap_{n\geq 0}\bigcup_{k\geq n} A_k = \Omega
\end{align}
so :
\begin{align}
\mathbb P(\liminf A_n) &=0\\
\mathbb P(\limsup A_n) &= 1
\end{align}
In the end, you see that all three inequalities are strict.
