known conditions:
$$ P(b|a) = 0.6 \\ P(c|b) = 0.8 \\ P(c|\neg b) = 0.7 $$
What I want to solve:
$$P(b\cup c|a )$$
(in the condition of "a" happen, at least one of b and c happen. )
I tried to break it up into:
$$P(b|a) + P(c|a) - P(bc|a)$$
but the final part $$P(bc|a)$$ stuck me so long ...