Question: I know the rule to know if two events are independents. And if you apply the rule, here, you get that those two events are independents, however, it doesn't make sense to me that they're independent.

> 
> A fair six-sided die is thrown twice and the scores are noted. Event X is defined as ‘The total of the
> two scores is 4’. Event Y is defined as ‘The first score is 2 or 5’. Are events X and Y independent?
> Justify your answer.
> 
> 
> 

The answer is that they are independent because p(X/Y) = p(X)

But drawing a probability table we get to know that X can only happen in these cases: (3,1) or (1,3) or (2,2) [p= 3/36]

While Y happens when the first number is equal to 2 or 5

But doesn't it make sense to say that when the first number is 2 or 5 that (1,3)/(3,1)/(2,2) is **less likely** to happen. Or they are independent because of some algebraic coincidence (I've no idea if this even exists)

Answer: 
This is one of those counter-intuitive aspects of the notion of **independence** in probability: that $X$ and $Y$ are *defined* to be independent if $P(X, Y) = P(X)P(Y)$ (or other equivalent statement).

It seems odd that these two events should be independent because the result of the first die roll obviously affects whether the total is even *capable* of being $4$. While that is indeed the case, the definition of independence does not concern itself with general causal relationships, but simply with the relationship between event probabilities and their joint probabilities. Focusing on situations where the first roll is either $2$ or $5$ narrows the scope by a factor of $3$, which (as it so happens) is exactly the factor by which the combinations that total $4$ are narrowed.

So, for example,

* $A =$ "the first roll is $2$ or $5$"
* $B =$ "the two rolls total $5$"

are *not* independent, because the same "coincidence" does not obtain, even though the events are superficially quite similar to $X$ and $Y$ from the original problem, while

* $C =$ "the first roll is $2$, $4$, or $6$
* $D =$ "the two rolls total $5$"

*are* independent, despite the difference in form.
No, I don't think if *does* make more sense to say that $\ \{(1,3),$$(3,1),$$(2,2)\}\ $ is less likely to happen when the first number is $2$ or $5$. You need to realise, though, that, in probability theory, the statement "given that the first number is $2$ or $5$" means that you're still completely in the dark as to *which* of those two numbers it is that has occurred, and since their prior probabilities were both $\ \frac{1}{6}\ $, then it is equally likely (i.e. with probability $\ \frac{1}{2}\ $) that the first number is a $2$, or that it is a $5$

If the first numbet is a $5$, then $\ \{(1,3),$$(3,1),$$(2,2)\}\ $ is impossible, while if it is a $2$ then the probability of $\ \{(1,3),(3,1),(2,2)\}\ $ is just the probability that the second number is a $2$—namely, $\ \frac{1}{6}\ $. So the overall probability of $\ \{(1,3),$$(3,1),$$(2,2)\}\ $ given that the first number is $2$ or $5$ is
\begin{align}
&\frac{1}{2}\text{ (probability first number is $2$)}\times\frac{1}{6}\\
+ &\frac{1}{2}\text{ (probability first number is $5$)}\times0\\
=&\frac{1}{12}\ ,
\end{align}
that is, the same as its prior probability given no information at all.