Consider an Exclusive-OR (XOR) gate which is an electronic circuit (logic gate) with two inputs $X$ and $Y$ and an output $Z$ where $X,Y,Z$ take on values in the discrete set $\{0, 1\}$. Think of these as Boolean variables (or Bernouiii random variables if you like). $Z$ is causally related to $X$ and $Y$ by the Exclusive-OR operation:
$$Z = X\oplus Y = X\bar{Y} \,\vee\, \bar{X}Y$$
if you are a Booleander or
$$Z = X(1-Y)+(1-X)Y= X + Y -2XY$$ if you are a Bernoullist. Be that as it may, suppose that
$X$ and $Y$ are independent (meaning that $P(X=a,Y=b) = P(X=a)P(Y=b)$ for all $a,b$ in $\{0, 1\}$. Then,
\begin{align}P(Z=1) &= P(X\neq Y)\\
&=P(X=1, Y=0) + P(X=0, Y=1)\\ &= P(X=1)P(Y=0) + P(X=0)P(Y=1).\end{align}
Everything OK thus far? Now suppose that $P(X=1) = P(Y=1)= \frac 12$. Then it is easy to verify that $P(Z=1) = \frac 12$ also. Now, $Z$ and $X$ are very definitely causally related: the output of an XOR gate does depend on its input(s). But, the event
$\{Z=1,X=1\}$ occurs if and only if the event $\{X=1, Y=0\}$ occurs and so
$$P(Z=1, X=1) = P(X=1,Y=0) = \frac 14 = P(Z=1)P(X=1) = \frac 12\times \frac 12$$
showing that the causally related events $\{Z=1\}$ and $\{X=1\}$ are in fact probabilistically independent. Similarly, $\{Z=1\}$ and $\{Y=1\}$ independent, in fact, the three events $\{X=1\}$, $\{Y=1\}$, and $\{Z=1\}$ are pairwise independent but not mutually independent since $$P(X=1, Y=1, Z=1) = 0 \neq P(X=1)P(Y=1)P(Z=1) = \frac 18.$$
Thus, causal dependence need not be reflected in probabilistic dependence; it is possible to have causally dependent events be probabilistically independent. I will also say that this probabilistic independence is purely a property of the probability measure: if we take $P(X=1)$ or $P(Y=1)$ to be any number in $(0,1)$ other than the $\frac 12$ that I sneakily chose above, the probabilistic independence disappears and the causally dependent events are also probabilistically dependent.
Lest you think that this is an oddball example that will hardly ever be encountered in real life, consider the gold standard in statistical theory and practice: three standard normal random variables $X,Y,Z$. Now suppose that their joint density
$f_{X,Y,Z}(x,y,z)$ is not $\phi(x)\phi(y)\phi(z)$ where
$\phi(\cdot)$ is the standard normal density (as would be the case if $X,Y,Z$ were mutually independent standard normal random variables), but rather
$$f_{X,Y,Z}(x,y,z) = \begin{cases} 2\phi(x)\phi(y)\phi(z)
& ~~~~\text{if}~ x \geq 0, y\geq 0, z \geq 0,\\
& \text{or if}~ x < 0, y < 0, z \geq 0,\\
& \text{or if}~ x < 0, y\geq 0, z < 0,\\
& \text{or if}~ x \geq 0, y< 0, z < 0,\\
0 & \text{otherwise.}
\end{cases}\tag{1}$$
Note that $X$, $Y$, and $Z$ are not a set of three jointly normal random variables (that is, they don't have a multivariate normal distribution) but it can be shown that any two of these is indeed a pair of independent standard normal random variables. For details of the verification, see the latter half of this answer of mine.