When to use product or set notations in calculating Probability. Problem: 

The probability that it will rain today is 0.5.The probability that it will rain tomorrow is 0.6.The probability that it will rain either today or tomorrow is 0.7.What is the probability that it will rain today and tomorrow?



*

*Why we just can't multiply the probability of today's and tomorrows' raining, as they are two events and must be followed one after other to get the final event on raining on two days. So the answer is 0.3.

*But the solution say we got to use the equality Pr(A and B) = Pr(A) + Pr(B) - Pr(A or B). This give us the answer 0.4.
It look to me like Independence is to do something in here, but I might be wrong.
Please help me with, why set notation works but not the product rule ?
 A: We don't have the equality $\Pr(A\cap B)=\Pr(A)\Pr(B)$ in general. However, if $A$ and $B$ are independent events, then
$$
\Pr(A\cap B)=\Pr(A)\Pr(B).
$$
So if we don't have independence, we cannot use the equality above.
A: The probability of a union of events is always the sum of probabilities of the events minus the probability of their intersection.   This is how probability measures are required to work (among other things), so: $\Pr(A\cup B)=\Pr(A)+\Pr(B)−\Pr(A\cap B)$.   This is equivalently: 
$$\Pr(A\cap B)=\Pr(A)+\Pr(B)−\Pr(A\cup B)$$ 
This is always the case.   It is only when the events are independent that it is also true that: $$\Pr(A\cap B)=\Pr(A)\cdot\Pr(B)$$
So we can only use the probability rule when we have certainty that the events are independent.   However, we can always use the addition rule when given three of the four probabilities.
In this case, we don't have any way to guarantee that the events of rainfall on subsequent days are independent.   Rather it would seem reasonable that there may be some dependence.   Since we are given three probabilities (of two events and their union), it is best to use the addition rule to find the fourth (their intersection); because that will always work.
And indeed on doing so, we find that these events are in fact not independent because, in this case, $\Pr(A)\cdot\Pr(B)\neq \Pr(A)+\Pr(B)-\Pr(A\cup B) = \Pr(A\cap B)$ .
