Probability of raining in the weekend Just trying to see if my reasoning is correct....
If the probability that next saturday will rain is 0.25 and the probability that next sunday will rain is 0.25, what is the probability that during the weekend will rain?
Assuming that A and B are independent events,
P(A∪B)=P(A)+P(B)-P(A∩B)
and
P(A∩B)=P(A)*P(B)
therefore the probability that during the weekend will rain is:
P(A∪B)=0.25+0.25*0.25*0.25=0.265625
It this reasoning correct?
Thanks,
Luca
 A: Almost. For $P(A\cup B)$, you ended up calculating $0.25 + 0.25^3$, not what you actually wanted to calculate. So double-check your last step.
Another way to think about this: What is the probability that it will not rain on Saturday? 0.75. And what is the probability that it will not rain on Sunday? 0.75. So what is the probability that it will rain neither on Saturday nor on Sunday? Assuming independence, it is $0.75^2$. So what is then the probability that it will rain on at least one of those two days? Well, that's just the complement of it not raining on either day, so $1 - 0.75^2$.
A: You are correct, but you made a typo - you wrote:
0.25+0.25*0.25*0.25 = 0.265625

instead of
0.25+0.25-0.25*0.25 = 0.4375

But note that this assumes independence, which is not likely the case of weather.
Relaxing the independence
(This is just an extension of the answer, you don't need to read it :-)
In most general case, you would need this model:
pSat = P(rains on saturday)
pSun = P(rains on sunday)
pSat0_Sun1 = P(rains on sunday | not rained on saturday)
pSat1_Sun1 = P(rains on sunday | rained on saturday)
pSun1_Sat1 = P(rains on saturday | rains on sunday) = pSat1_Sun1 * pSat / pSun 
    //  Bayess rule
...

To be realistic, you probably want to simplify the model by assuming pSat = pSun = p and then pSun1_Sat1 = pSat1_Sun1, pSun1_Sat0 = pSat0_Sun1 etc. so the realistic model would be:
p = P(rains on saturday) = P(rains on sunday)
pRainChange = P(rains one day | not rains the other day) = pSat0_Sun1 = pSun0_Sat1

From these 2 parameters you can also derive:
pRainPersist = P(rains one day | rains the other day) = pSat1_Sun1 = pSun1_Sat1 = 
    = 1 - pRainChange * (1/p - 1)

From this you can derive:
P(no rain over weekend) = (1 - p) * (1 - pRainChange)
P(rain on sunday only) = (1 - p) * pRainChange
P(rain on saturday only) = p * (1 - pRainPersist)
P(rain both days) = p * pRainPersist

So then your desired
P(rain on saturday or sunday) = (1 - p) * pRainChange + p * (1 - pRainPersist) + 
    + p * pRainPersist = (1 - p) * pRainChange + p

When independence, as you assumed, then pRainPersist = pRainChange, which implies both must be equal to p, and then:
P(rain on saturday or sunday) = (1 - p) * p + p = p + p - p * p

But in reality, I'd expect pRainPersist > p > pRainChange.
