/diy/ - Do It Yourself

I don't understand the point that the Art of Electronics is trying to make.
>most of its collector current comes from Q 3 ’s base (because only ∼0.6 mA of the 4.4 mA collector current comes from R 3 – make sure you understand why).
Okay, if we were to just have circuit (a), the base current is:
Ib = (15-Vbe)/3.3k = (15-0.6)/3.3k = 4.4 mA

In circuit (b), the BE junction of Q1 holds a constant voltage over R3:
IR3 = 0.6/1000 = 6 mA

>That is, R 3 does not have much effect on Q 3 ’s saturation. Another way to say it is that the divider would sit at about +11.6 V (rather than +14.4 V), were it not for Q 3 ’s base–emitter diode, which consequently gets most of Q 2 ’s collector current.
Why would voltage divider be at 11.6V? I don't understand what the authors are trying to say.