Probability of at least one event from a subset happening when choosing 2 out of 4 possible events

We have a complete set of events $\{A,B,C,D\}$.

$p(A)=0.2$
$p(B)=0.3$
$p(C)=0.4$
$p(D)=0.1$

Two events happen in succession, where each event cannot occur twice. (i.e. when one happens the others' probabilities re-scale to sum up to $1$).
So we have a new complete set of events $\{AB,AC,AD,BA,BC,BD,CA,CB,CD,DA,DB,DC\}$

Since they are all mutually exclusive, the probability of, say, $B$ happening is the sum of probabilities of all events that include $B$.
If I calculate correctly, that is
$p(B)=0.608$.
Also, for $C$, it's
$p(C)=0.715$.

So if I want to know the probability of either $B$ or $C$ happening, or both, I'd sum up all the probabilities of events including $B$ or $C$ or both.
And that's $p(B+C)=0.952$.

That feels correct, but there's this formula that's valid for all events, that says $p(B+C)=1-(1-p(B))(1-p(C))$, i.e. the probability of at least one of them happening is one minus the probability of neither of them happening. But that formula yields $p(B+C)=0.888\neq 0.952$.

What am I doing wrong?

And another question - is there a way I can correctly calculate $p(B+C)$ by knowing only $p(B)=0.608$ and $p(C)=0.715$, without iterating over the whole set just to calculate $p(B+C)$ (or any other such composed event, e.g. $A+D$)?

-

The formula that gives you .888 assumes $B$ and $C$ are independent events, doesn't it?
@vedran: $p(B \cap C) \sim 0.37$ which indeed doesn't equal $p(B)p(C) \sim 0.43$. –  Marek Jun 19 '11 at 15:53