This was growing too long for a comment. I'm going to construct a similar (although not equivalent) counterexample using three coins so that independence makes more 'sense'.
To determine if a pair of events are independent, the pair need only satisfy that $P(A \cap B) = P(A) P(B)$.
The state space for three coins could be written as:
$$\{(h,h,h), (h,h,t), (h,t,h), (h,t,t), (t,h,h), (t,h,t), (t,t,h), (t,t,t)\}$$
This can also be written as $\{1, 2, \ldots, 8\}$, i.e. by enumerating each outcome.
Consider the following pair of dependet events:
- $A$ the event that the first coin is a head (i.e. state 1, 2, 3 or 4),
- $B$ the event that the first two coins are heads, or the last two are tails
(i.e. state 1, 2, 4 or 8).
The dependence is physically obvious; they both rely on the outcome of the first coin flip. We can also see that they don't obey the definition of independence, as
- $P(A \cap B) = P(\text{state 1, 2, or 4}) = 3/8$ and
- $P(A) P(B) = (1/2) \times (1/2) = 1/4$.
Let's consider a third event:
- $C$ the event that the last two coins are not the same (state 2, 3, 6, 7).
In physical terms, it seems that this event is independent from $A$, because event $C$ doesn't depend on the flip of the first coin, while event $A$ only depends on the first coin. It is easy to show, as well, that they obey the definition of independence.
Meanwhile, event $B$ and $C$ are dependent, in physical terms because they both rely on the outcome of the last two coins; and it is also easy to show that the definition of independence is not satisfied.
Now of course $P(A \cap B \cap C) = P(\text{state 2}) = 1/8$, and $P(A) P(B) P(C) = 1/8$, too.
So the question is, still, are they 'mutually independent'? No. As per the definition of independence for multiple events, we require that each pair of events is independent - which we have already seen is false.