"In drawing two cards from a deck (without returning the first card) what is the probability of two aces when you get at least one ace?"
I am aware there is exact same question answered here.
Probability of two aces when you get at least one ace
I know and understand then answers mentioned.
I would like to approach the problem in a different way.
Let us denote A = event that we get ace from first suite in case of success, i.e. when two aces are drawn., or alternately speaking, atleast one ace is from first suite
similarly B,C,D.
What we want to know is $P(A\cup B \cup C\cup D) $
$=\sum_A ^D{P(X)} - {4 \choose 2 }P(A \cap B)$(due to symmetry) + {ignoring the other intersections since we can't have more than two cards}
$= 4*\frac {3}{51} - {4 \choose 2 } *\frac{1}{{52 \choose 2 }}$
$= 3/13$
which is not right answer, obviously I did some mistake, but where?
Your "$P(A)=\frac{3}{51}$" is a conditional probability; the probability that we have a second ace, given that we have the ace of spades. Now you want to add that to $P(B)$, the probability that we have a second ace given that we have the ace of clubs. This sum ... doesn't represent anything meaningful. Those events don't live in the same space, because they don't have the same condition. Then, of course, $P(A\cap B)$ is nonsense, because $A$ and $B$ aren't in the same space. Conditional probabilities create a new probability space by taking a slice of the old one and scaling up the measure - but of course, that means it's a different space for every condition.
Then again, when you wrote
That suggests that $A$ is the event that one of or two cards is the ace of spades given that both are aces. Now we're basically in a deck of four cards - we pick two, so the probability of getting a particular card is $\frac12$. That doesn't seem like a particularly fruitful approach.
I've answered based on my best understanding of the post, but there's enough self-contradiction in there to muddy everything severely. You definitely need to work on writing clearly and saying what you mean.