Did I make a mistake? Checking for independent events I have a feeling that I'm doing something wrong. Say we have a deck of 52 cards, and we have the event Q = pull a Queen and Event H = pull a Hearts, I want to figure out of these two events are independent.
$P(Q) = \frac{4}{52}$ and $P(H) = \frac{13}{52}$
I'm aware that if they are independent, $P(Q\cap H) = P(Q)\cdot P(H) = P(H)\cdot P(Q|H)$. Since a heart could potentially be a queen, that would be $P(H\cap Q) = \frac{1}{52}$, as such: $$P(Q\mid H) \cdot P(H)  = P(H\cap Q) \cdot \frac{1}{P(H)} \cdot P(H)= \frac{1}{52}$$
Now since $P(H)\cdot P(Q) = \frac{1}{52}\cdot \frac{13}{52} = \frac{13}{52^2} \neq \frac{1}{52}$ we can conclude that they are not independent.
Am I doing this right, because I have a feeling I'm overcomplicating it and messed up..
Thanks in advance
 A: This is a confusion of "independent" and "mutually exclusive."
If $Q$ and $H$ are "independent events", that means that knowing whether one happened does not affect the probability of the other one happening; it gives you no information.  Essentially, it is like "the price of tea in China", totally unrelated and irrelevant to the other thing.  An example would be the first roll of a die and the second roll of the die.
Another example would be pulling a card and getting a heart, versus pulling a card and getting a queen.  Note there is only ONE draw.  If you draw a queen, there is a 1/4 chance that it is a heart.  If you don't draw a queen, there is still a 1/4 chance that it is a heart.  Knowing only that you got a queen tells you nothing about whether it also happened to be a heart or not.
If you draw a card, and don't replace it, and make another draw, the probabilities usually shift, as when checking if the first card is a heart, then if the second card is a heart.  When drawing without replacement, draws are not always independent of each other, depending on what information is being looked at on each card.  But this problem has only one draw in the first place.  We are talking about two different details of a single draw, not different draws.
"Mutually exclusive events" cannot both happen.  So if you know one happened, then you know the other did not.  An example of this would be, when rolling a die only once, you cannot get a $2$ and a $3$ at the same time.  They are mutually exclusive outcomes.  If one of them happened, the other did not.  Note it is still possible that neither of them happen.
Choosing a face value and choosing a suit are independent events, and not mutually exclusive, since the queen of hearts exists, as you pointed out.
A: To emphasize and clarify some points,
Mutually Exclusive events are those satisfying $A\cap B=\emptyset$, that is to say it is impossible for an outcome to be in both events simultaneously.  Take for an example "drawing a spade as the first card" versus "drawing a heart as the first card."  Mutually exclusive events will as a result satisfy $\Pr(A\cap B)=0$ and also $\Pr(A\cup B)=\Pr(A)+\Pr(B)$.
Warning: $\Pr(A\cap B)=0$ is not enough to say that two events are mutually exclusive (particularly in infinite contexts such as continuous distributions) and also $\Pr(A\cup B)$ is in general not equal to $\Pr(A)+\Pr(B)$.  Instead, the always true identity is that $\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B)$

Independent Events are those events satisfying $\Pr(A\cap B)=\Pr(A)\cdot \Pr(B)$.  Equivalently, those events satisfying $\Pr(A\mid B)=\Pr(A)$, or equivalently $\Pr(B\mid A)=\Pr(B)$.  That is to say, independent events are those who knowledge (only) of whether one event has happened or not does not influence our expectation of the probability of whether the other event has happened.
Warning: $\Pr(A\cap B)$ is in general not equal to $\Pr(A)\cdot \Pr(B)$.  Instead, the always true identity is that $\Pr(A\cap B)=\Pr(A)\cdot \Pr(B\mid A)$.  Also, be warned that pairwise independence of several events does not imply mutual independence of several events.  That is to say, it is possible that $\Pr(A\cap B)=\Pr(A)\cdot \Pr(B)$ and $\Pr(A\cap C)=\Pr(A)\cdot \Pr(C)$ and $\Pr(B\cap C)=\Pr(B)\cdot \Pr(C)$, that $A,B$ are independent, that $A,C$ are independent, and $B,C$ are independent, while having $\Pr(A\cap B\cap C)\neq \Pr(A)\Pr(B)\Pr(C)$, that $A,B,C$ are not mutually independent.

It is possible for two events to be independent and not mutually exclusive.  It is possible for two events to be dependent and mutually exclusive.  It is possible for two events to be dependent and not be mutually exclusive.
As for independent and mutually exclusive, this occurs only when one or both of the events occur with probability zero.
Altogether, this means that if $\Pr(A)>0$ and $\Pr(B)>0$ you have $A$ and $B$ independent implies that they are not mutually exclusive, and that $A$ and $B$ mutually exclusive implies that they are not independent.  Keep in mind they could still be neither.

On to the specific question about cards.  To recap the comments above, you are confusing two similar problems.  Letting $Q_i$ and $H_j$ be the events that the $i$'th card drawn is a queen and the $j$'th card drawn is a heart respectively for whatever integers $i,j$... the original question being asked was to consider drawing a single card and comparing the events $Q_1,H_1$.
We do have $\Pr(Q_1)=\dfrac{4}{52}$, that $\Pr(H_1)=\dfrac{13}{52}$ and $\Pr(Q_1\cap H_1)=\dfrac{1}{52}$.  We do have $\Pr(Q_1\cap H_1)=\Pr(Q_1)\Pr(H_1)$ thus proving their independence.
The other related but different question you were alluding to is that of comparing the events $Q_1,H_2$... that is we pull two cards in sequence without replacement and ask about the first card being a queen and the second card being a heart.  You voice your concern about how these could possibly be independent since if the first card happened to be the queen of hearts it would reduce the available number of hearts to be pulled for the second card.
That complaint is common, however you failed to account for the fact that the opposite also holds... that if you did not pull a queen of hearts as the first card that the number of available hearts did not decrease along with the total pool of cards and so the probability actually would be increased that the second card would be a heart.
Actually going through the calculations,
$$\Pr(Q_1\cap H_2)=\Pr(Q_1\cap H_1\cap H_2)+\Pr(Q_1\cap H_1^c\cap H_2)$$
$$ = \Pr(Q_1\cap H_1)\Pr(H_2\mid Q_1\cap H_1)+\Pr(Q_1\cap H_1^c)\Pr(H_2\mid Q_1\cap H_1^c)$$
$$=\dfrac{1}{52}\cdot\dfrac{12}{51}+\dfrac{3}{52}\cdot\dfrac{13}{51}=\dfrac{1}{52}=\dfrac{4}{52}\cdot\dfrac{13}{52}=\Pr(Q_1)\cdot\Pr(H_2)$$
This proves their independence.
Imprecise unqualified statements like "When drawing without replacement, draws are not independent of each other" can be incorrect, like shown here.  The individual atomic outcomes may well be dependent, but when looking at them as collections of events like here we find that the only way to be sure is to actually go through with the analysis.
Another faster way to verify their independence is to recall that $\Pr(H_2)=\Pr(H_2\mid Q_1)=\dfrac{1}{4}$ as the second card isn't any more or less likely to be a heart as it would be to be a spade or a club or diamond... since these four suits are just as likely to be the suit of the second card (even if we were to condition on the first card being a queen) it follows that the probability must be $\frac{1}{4}$, thus showing $\Pr(H_2)=\Pr(H_2\mid Q_1)$ thus proving $Q_1$ and $H_2$'s independence.
A: You have correctly evaluated $\mathsf P(Q)=1/13, \mathsf P(H)=1/4,$ and $\mathsf P(Q\cap H)= 1/52$. $~\color{green}\checkmark$
Next you want to establish whether $\mathsf P(Q\cap H)\overset?=\mathsf P(Q)\mathsf P(H)$ when you know $\mathsf P(Q\cap H)=\mathsf P(Q\mid H)\mathsf P(H)$ and $\mathsf P(H)\neq 0$.
Therefore you just need to ascertain whether $\mathsf P(Q\mid H)\overset?=\mathsf P(Q)$.
Well, by definition, $\mathsf P(Q\mid H)~{=\dfrac{\mathsf P(Q\cap H)}{\mathsf P(H)}}$, so... .
