Issues with the Probability formula We have $ \mathbb P ( A \cup B) = \mathbb P(A) + \mathbb P(B) - \mathbb P(A \cap B)$
and $\mathbb P(A \cap B) = \mathbb P(A) \mathbb P(B)$.
These formulas work for one example but not for the other.
Example 1:
There are 10 girls and 20 boys, half of each are blue-eyed.
What is the probability that a chosen child is either a girl or is blue-eyed?
$\mathbb P(\text{girl}) = \frac{10}{30}$
$\mathbb P(\text{blue-eyed}) = \frac{15}{30}$
$\mathbb P(\text{girl and blue-eyed}) = \frac{10}{30}\cdot\frac{15}{30} = \frac 5{30}$
$\mathbb P(\text{girl or blue-eyed}) = \frac{10}{30}+\frac{15}{30}-\frac 5{30}$
All  is fine and correct. Why not for the following example:
Example 2:
Two dice are thrown. what is the probability for the sum to be $4$ or $6$?
$\mathbb P(\text{sum is $4$}) = \frac 3{36}$
$\mathbb P(\text{sum is $6$}) = \frac 5{36}$
$\mathbb P(\text{sum is $4$ and sum is $6$}) = ? $
We know that it cannot be $4$ and $6$ at the same time, so it is zero.
But the above formula gives $\mathbb P(\text{sum is $4$ and sum is $6$}) = \mathbb P(\text{sum is $4$})\mathbb P(\text{sum is $6$}) = \frac 3{36}\cdot\frac 5{36} = \text{not zero}$
What happened? Why does the product formula suddenly stop working in case of no overlap?
Wasn't the product formula supposed to be always true, so the formula should give a zero result?
Please explain.
 A: The 'product formula' only works for independent events. That is, when the probability of $A$ has no bearing on the probability of $B$. To illustrate why, here is a slightly amusing example taken from Why do Buses Come in Threes?: the probability of getting heads when you flip a fair coin is $\frac{1}{2}$, and the probability of rolling a $6$ with a fair die is $\frac{1}{6}$. Notice that these two events are completely unrelated: if the coin lands on heads, then this does not affect the chance of me rolling a $6$ at all. Therefore, the probability of getting heads and rolling a $6$ is $\frac{1}{12}$.
Now consider this example: let's say the probability of a person walking down the street being a man is $\frac{1}{2}$. Also, the probability of a person walking down the street wearing a skirt is $\frac{1}{4}$. But the probability of a person walking down the street being a man wearing a skirt is not $\frac{1}{8}$! You may laugh, but the reason why this is the case is because one's gender has an effect on one's tendency to wear a skirt. Compare that with the coin/die example, and you have your answer.
NB the study of dependent events is closely linked with conditional probability. I'd highly recommend looking at this website for more information.
