Weather forecast and probabilities There are two weather stations, station A and station B which are independent of each other. On average, the weather forecast accuracy of station A is $80\%$ and that of station B is $90\%$. Station A predicts that tomorrow will be sunny, whereas station B predicts rain. What is the probability that it rains tomorrow? We are not asking for the exact probability; we are just asking whether it is more likely to rain or not.
OK I suppose we must examine the following $4$ cases:
a) A and B make the same forecast and both are right 
b) A and B make the same forecast and both are wrong 
c) A and B make different forecasts and A is right 
d) A and B make different forecasts and B is right
and of course we are in one of the cases c) or d), since we know they make different forecasts .
 A: It is not clear what you mean by independent concerning A and B.
It is clear that the forecast of A and B should not be independent as they are both linked to the actual weather of tomorrow.
What I understand is that the event (A fails) is not linked to the event (B fails). Furthermore I interpret your percentage as the probability of forecasting the weather for instance P(Forecast=sunny | weather=sunny)
which is different from P(weather=sunny|Forecast=sunny)
You need to know the marginal probability P(W) of the weather (sunny/rain) tomorrow.
let $f_a$ and $f_b$ stand for the forecast of A and B
You want to know So $P(W=rain|f_a,f_b)$.
You have that :
$P(W|f_a,f_b) \sim P(f_a,f_b|W)P(W)$ (Bayes theorem)
As the error probability of A and B are independent : 
$P(f_a,f_b | W=rain)=P(f_a|W=rain)P(f_b|W=rain)=0.20*0.90 = 0.18$
$P(f_a,f_b | W=sunny)=P(f_a|W=sunny)P(f_b|W=sunny)=0.80*0.10 = 0.08$
So $P(W=rain|f_a,f_b)=\frac{0.18 P(W=rain)}{0.18 P(W=rain)+0.08 P(W=sunny)}$
A: Denote by $r$ the a-priori probability of rain. Then the probability $P[A:S, B:R]$ of the forecasts A: sunny and B: rain amounts to $r\cdot0.2\cdot 0.9+(1-r)\cdot0.8\cdot 0.1$, and the probability that it actually rains under these premises is the first part of this sum. Therefore we obtain
$$P[R\>| A:S, B:R]={0.18 r\over 0.1 r+0.08}\ .$$
