When we talk about a fraction of a thing, we mean a fraction times a thing. $P(D\mid A\cap B\cap C)$ is the portion of the time that $D$ happens, given that we know $A,$ $B,$ and $C$ happen, and we must toss out the complementary portion (when $A,$ $B,$ and $C,$ happen, but $D$ doesn't), since otherwise, $A,B,C,D,E.$ can't all happen. $P(C\mid A\cap B)$ is the portion of the time that $C$ happens, given that we know $A$ and $B$ happen, and we must toss out the complementary portion (when $A$ and $B$ happen but $C$ doesn't), since otherwise, $A,B,C,D,E.$ can't all happen. $P(B\mid A)$ is the portion of the time that $B$ happens, given that we know $A$ happens, and we must toss out the complementary portion (the portion of the time that $A$ happens and that $B$ doesn't), since otherwise, $A,B,C,D,E.$ can't all happen. $P(A)$ is the portion of the time that $A$ happens, and we must toss out the complementary portion, since otherwise, $A,B,C,D,E.$ can't all happen. Think of it this way for $P(A\cap B\cap C\cap D\cap\cdots)$. The question does a service in bringing out the analogy! In each case, we have a form of combined scalings. So $\Delta x$ in a sense playes the same role as the $1200$ people did in the example above.
When we are dealing with probabilities, the number who go to $A$ need not be exactly $400$, but the multiplication of probabilities takes place for essentially the same reason as in our people example. So to determine what fraction of the $1200$ end up at $C$, we multiply $1/3$ by $5/8$. So $5/8$ of the $400$ people end up at $D$. Now at $A$ they split, some going to $D$ and some to $E$, according to the probabilities written on the branches. In particular, $400$ of them end up at $A$. They split along the $3$ branches coming out of $R$, with $1/3$ of them going to $A$, $1/6$ of them to $B$, and $1/2$ to $C$. Now imagine say $1200$ people starting at $R$. Let's stop here, though if you want to make the thing more tree-like, go ahead. The $5/8$ and $3/8$ mean that if we have reached node $A$, we have probability $5/8$ of going to $D$, and probability $3/8$ of going to $E$. As you know, the $1/3$ means that if we are at $R$, we have probability $1/3$ of going to $A$. The branch that goes from $R$ to $A$ has $1/3$ beside it.
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