Roll a fair four-sided die once.
Let A be the event that the number obtained is 1 or 2,
B be the event that the number obtained is 1 or 3, and
C be the event that the number obtained is 2 or 3
Thus,
A=\{1,2\}\quad B=\{1,3\}\quad C=\{2,3\}
\Pr(A)=\Pr(B)=\Pr(C)=\frac12
Now,
\Pr(A\cap B)=\frac14=\Pr(A)\Pr(B) \tag{1}
\Pr(A\cap C)=\frac14=\Pr(A)\Pr(C) \tag{2}
\Pr(B\cap C)=\frac14=\Pr(B)\Pr(C) \tag{3}
But,
A\cap B\cap C=\varnothing
Therefore,
\Pr(A\cap B\cap C)=0
\neq
\frac18=\Pr(A)\Pr(B)\Pr(C) \tag{4}
Hence, (1), (2) and (3) are satisfied, but (4) is not.
Therefore, A,B,C are pairwise independent but not mutually independent.